Do you know how the human body maintains a constant pH level despite changes that occur every day and why chemical reactions didn’t always go to completion but instead decided to settle halfway? The answer to all these questions lies in Equilibrium, a state of reversible reactions, where the forward and backwards processes occur simultaneously. NCERT Class 11 Chemistry Chapter 6 Equilibrium is one of the most crucial chapters in chemistry, as it explains how chemical and physical processes reach a state of balance.
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Equilibrium Class 11 Chemistry covers important topics like Le Chatelier's Principle, equilibrium constants, ionic equilibrium, acid base-reactions. NCERT Class 11 Chemistry Solutions are structured according to the latest NCERT syllabus. These NCERT Solutions include step by step answers, selected Higher order thinking skills questions and approaches to solve questions. NCERT Solutions for Class 11 will help you in the preparation of final examination as well as in various competitive exams. These solutions will help you increase your accuracy and speed.
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These Solutions for Chemistry Chapter 6 Equilibrium are provided in a detailed manner, covering all key concepts. Each solution is explained step by step to help students understand the logic behind the concepts and prepare effectively for CBSE board exams and competitive exams like JEE and NEET
Question 6.1(a) A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
What is the initial effect of the change on vapour pressure?
Answer :
By increasing the volume of the container suddenly, initially, the vapour pressure would decrease. It is due to, the amount of vapour has remained the same but it is distributed in a larger volume.
Question 6.1(b) A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
How do rates of evaporation and condensation change initially?
Answer :
Here the temperature is constant so that the rate of evaporation is also the same as before. On increasing the volume of the container, the density of vapour decreases due to which the rate of collision between vapour particles decreases. Hence the condensation rate also decreases initially.
Question 6.1(c) A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased .
What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer :
When equilibrium is restored, the rate of evaporation is equal to the rate of condensation. The temperature is constant and the volume is changed. The vapour pressure is temperature-dependent, not volumes. So, that final vapour pressure is equal to the initial vapour pressure.
$2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)$
Answer :
Followings are the given information values to solve the above problems-$\left[\mathrm{SO}_2\right]=0.60 \mathrm{M}\left[\mathrm{O}_2\right]$
$=0.82 \mathrm{M}\left[\mathrm{SO}_3\right]=1.90 \mathrm{M}$
The given chemical equation is -
$2 \mathrm{SO}_2(g)+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(g)$
The equilibrium constant for this reaction is expressed as;
$K_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}$
$=\frac{[1.90]^2}{[0.60]^2[0.82]}=12.22 M^{-1}($ approx $)$
Question 6.3 At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms
Calculate Kp for the equilibrium.
Answer :
It is given that total pressure (P T )is 105 Pa and partial pressure of I atom = 40 % of P T
So, the partial pressure of I atom $=\frac{40}{100} \times 10^5=4 \times 10^4$
The partial pressure of I2 = 60% of P T
So, the partial pressure of I2$=\frac{60}{100} \times 10^5=6 \times 10^4$
Now, for the reaction $I_{(g)} \rightleftharpoons 2 I_{(g)}$
$
\begin{aligned}
& K_p=\frac{\left(p_I\right)^2}{p_{l_2}} \\
& =\frac{\left(4 \times 10^4\right)^2}{6 \times 10^4} \mathrm{~Pa} \\
& =2.67 \times 10^4 \mathrm{~Pa}
\end{aligned}
$
Question 6.4 Write the expression for the equilibrium constant, Kc, for each of the following reactions:
(i) $2 \mathrm{NOCl}_{(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}+\mathrm{Cl}_{(g)}$
(ii) $2 \mathrm{Cu}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \rightleftharpoons 2 \mathrm{CuOS}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
(iii) $\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(a q)}+\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}_{(a q)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(a q)}$
(iv) $\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(a q)}+\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}_{(a q)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(a q)}$
(v) $I_{2(s)}+5 F_2 \rightleftharpoons 2 I F_5$
Answer :
The equilibrium constant for any reaction can be written as (concentration of products) / (concentration of reactants). And we considered constant values for the solids and liquids because their density per unit volume or mass per unit volume does not change.
Thus,
(i)$K_c=\frac{[\mathrm{NO}]^2\left[\mathrm{Cl}_2\right]}{[\mathrm{NOCl}]^2}$
(ii)
$K_c=\frac{[\mathrm{CuO}]^2\left[\mathrm{NO}_2\right]^4\left[\mathrm{O}_2\right]}{\left[\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\right]^2}$
(iii)
$K_c=\frac{\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H} 5\right]}$
(iv)$K_C=\frac{1}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]}$
(v)$K_c=\frac{\left[I F_5\right]^2}{\left[F_2\right]^5}$
Question 6.5 Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) $2 \mathrm{NOCl}_{(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}+\mathrm{Cl}_{2(g)} ; K_p=1.8 \times 10^{-2}$ at500K
(ii) $\mathrm{CaCO}_{3(s)} \rightleftharpoons \mathrm{CaO}_{(s)}+\mathrm{CO}_{2(g)} ; \mathrm{K}_p=167$ at 1073 K
Answer :
We know that the relation between Kp and Kc is expressed as;
$
K_p=K_c(R T)^{\Delta n}
$
here Δn = (no. of moles of product) - (no. of moles of reactants)
R = 0.0831 bar L /mol/K, and
For (i)
$K_p=1.8 \times 10^{-2}$and Temp (T) = 500K
Δn = 3 - 2 = 1
By putting the all values in eq (i) we get
$K_c=\frac{167}{0.0831 \times 1073}=1.87$
For (ii)
Kp = 167 and temp(T) = 1073 K
Δn = 2 - 1 = 1
Now, by putting all values in eq (i) we get,
$K_c=\frac{167}{0.0831 \times 1073}=1.87$
Question 6.6 For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
$\mathrm{NO}_{(g)}+\mathrm{O}_{3(g)} \rightleftharpoons \mathrm{NO}_{2(g)}+\mathrm{O}_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?
Answer :
It is given that,
$K_c=6.3 \times 10^{14}$
we know that Kc′ for the reverse reaction is the inverse of the forward equilibrium constant. Thus it can be calculated as:
$\begin{aligned} & K_c^{\prime}=\frac{1}{K_c} \\ & =\frac{1}{6.3 \times 10^{14}} \\ & =1.58 \times 10^{-15}\end{aligned}$
Question 6.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Answer :
For the pure liquids and solids, the molecular mass and the density at a particular temperature is always fixed and it is considered as a constant. Thus they can be ignored while writing the equilibrium constant expression
Question 6.8 Reaction between N2 and O2- takes place as follows:
$2 \mathrm{~N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{(g)}$
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which $K_c=2.0 \times 10^{-37}$ , determine the composition of equilibrium mixture.
Answer :
It is given that,
$K_c=2.0 \times 10^{-37}$
Let the concentration of N2O at equilibrium be x . $$
\begin{array}{lcccc}
& 2 \mathrm{~N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{(g)} & \\
\text { initial conc } & 0.482 & 0.933 & 0 & \text { (in moles) } \\
\text { at equilibrium } & 0.482-x & 0.933-x & x & \text { (in moles) }
\end{array}
$$
The equilibrium constant is very small. So, we can assume $0.482-x=0.482$ and $0.933-x=0.933$
We know that,
$K_c=\frac{\left[\mathrm{N}_2 \mathrm{O}\right]^2}{\left[\mathrm{~N}_2\right]^2\left[\mathrm{O}_2\right]}$
$2 \times 10^{-37}=\frac{(x / 10)^2}{(0.0482)^2(0.0933)}$ {dividing the moles by 10 to get concentration of ions)
$\frac{x^2}{100}=2 \times 10^{-37} \times(0.0482)^2 \times(0.0933) x^2$
$
=43.35 \times 10^{-40} x
$
$
=\sqrt{43.35 \times 10^{-40}} x=6.6 \times 10^{-20}
$
So, the concentration of $\left[\mathrm{N}_2 \mathrm{O}\right]=\frac{x}{10}=6.6 \times 10^{-20}$
Question 6.9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
$2 \mathrm{NO}_{(g)}+\mathrm{Br}_{2(g)} \rightleftharpoons 2 \mathrm{NOBr}_{(g)}$
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br2.
Answer :
The initial concentration of NO and Br2 is 0.087 mol and 0.0437 mol respectively.
The given chemical reaction is-$2 \mathrm{NO}_{(g)}+\mathrm{Br}_{2(g)} \rightleftharpoons 2 \mathrm{NOBr}_{(g)}$
Here, 2 mol of NOBr produces from 2 mol of NO . So, 0.0518 mol of NOBr is obtained from 0.518 mol of NO .
Again, From 1 mol of Br2 two mol of NOBr produced. So, to produce 0.518 mol of NOBr we need $\frac{0.518}{2}=0.0259 \mathrm{~mol}$ of $B r_2$.
Thus, the amount of NO present at equilibrium = 0.087 - 0.0518 = 0.0352 mol
and the amount of Br2 present at the equilibrium = 0.0437-0.0259 = 0.0178 mol
Question 6.10 At 450K, Kp = 2.0 × 1010 /bar for the given reaction at equilibrium.
$2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{SO}_{3(g)}$
What is Kc at this temperature?
Answer :
We have,
$K_p=2 \times 10^{10} /$ bar
We know that the relation between Kp and Kc ;
$K_p=K_c(R T)^{\Delta n}$
Here Δn = ( moles of product) - (moles of reactants)
$2 \mathrm{SO}_{2(g)}+O_{2(g)} \rightleftharpoons 2 \mathrm{SO}_{3(g)}$
So. here Δn = 2-3 = -1
By applying the formula we get;
$2 \times 10^{10}=K_c(0.0831 \mathrm{Lbar} / \mathrm{K} / \mathrm{mol}) \times 450 \mathrm{~K}$$K_c=\frac{2 \times 10^{10}}{(0.0831 \mathrm{Lbar} / \mathrm{K} / \mathrm{mol}) \times 450 \mathrm{~K}}$
$\begin{aligned} & =74.79 \times 10^{10} \mathrm{Lmol}^{-1} \\ & =7.48 \times 10^{10} \mathrm{Lmol}^{-1}\end{aligned}$
$2 H I_{(g)} \rightleftharpoons H_{2(g)}+I_{2(g)}$
Answer :
The initial pressure of HI is 0.2 atm . At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 - 0.04 = 0.16 .
The given reaction is:
$\begin{gathered}2 H I_{(g)} \rightleftharpoons H_{2(g)}+I_{2(g)} \\ H I_{(g)} \rightleftharpoons 1 / 2 H_{2(g)}+1 / 2 I_{2(g)}\end{gathered}$
At equilibrium,
$\begin{aligned} & \text { pHI }=0.04 p H_2=\frac{0.16}{2} \\ & =0.08 p I_2=\frac{0.16}{2}=0.08\end{aligned}$
Therefore,
$\begin{aligned} & K_p=\frac{p H_2 \times p I_2}{p^2 H I} \\ & =\frac{0.08 \times 0.08}{(0.04)^2}=4\end{aligned}$
Answer :
We have,$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}$
Kc=1.7×102
The concentration of species are-$\left[\mathrm{N}_2\right]=1.57 / 20 \mathrm{molL}^{-1}\left[\mathrm{H}_2\right]=1.92 / 20 \mathrm{molL}^{-1}\left[\mathrm{NH}_3\right]=8.13 / 20 \mathrm{molL}^{-1}$
We know the formula of
$\begin{aligned} & Q_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \\ & =\frac{(8.13 / 20)^2}{(1.57 / 20)(1.92 / 20)^3}=2.4 \times 10^{-3}\end{aligned}$
The reaction is not in equilibrium. Since $Q c>K_c$, the equilibrium proceeds in reverse direction.
Write the balanced chemical equation corresponding to this expression.
Answer :
The balanced chemical equation corresponding to the given expression can be written as:
$4 \mathrm{NO}+6 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 4 \mathrm{NH}_3+5 \mathrm{O}_2$
$\mathrm{H}_2 \mathrm{O}_{(g)}+\mathrm{CO}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)}+\mathrm{CO}_{2(g)}$
Calculate the equilibrium constant for the reaction.
Answer :
The given reaction is-
$\begin{array}{ccccc}\mathrm{H}_2 \mathrm{O}_{(g)}+\mathrm{CO}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)}+\mathrm{CO}_{2(g)} & & & & \\ \text { initial conc } & 1 / 10 & 1 / 10 & 0 & 0 \\ \text { At equilibrium } & 0.6 / 10 & 0.6 / 10 & 0.04 & 0.04\end{array}$
Now, the equilibrium constant for the reaction can be calculated as;
$\begin{aligned} & K_C=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{CO}_2\right]}{\left[\mathrm{H}_2 \mathrm{O}\right][\mathrm{CO}]} \\ & =\frac{.04 \times .04}{(.06)^2}\end{aligned}$
= 0.44 (approx)
Question 6.15 At 700 K, equilibrium constant for the reaction:
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2 H I_{(g)}$
is 54.8. If 0.5 mol L -1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2 (g) and I2 (g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
Answer :
We have,
The equilibrium constant of the reaction = 54.8
moles of HI = 0.5 mol/L
The given reaction is-
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2 H I_{(g)}$
So, the reverse equilibrium constant is$K_c^{\prime}=1 / K_c$
Suppose the concentration of hydrogen and iodine at equilibrium be x
$\left[I_2\right]=\left[H_2\right]=x$
Therefore,
$K_c^{\prime}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]}=\frac{x^2}{(.5)^2}$
So, the value of$x=\sqrt{\frac{0.25}{54.8}}=0.0675($ approx $) \mathrm{mol} / L$
$2 I C l_{(g)} \rightleftharpoons I_{2(g)}+C l_{2(g)} ; K_c=0.14$
Answer :
The given reaction is:$$
\begin{array}{lcl}
& 2 \operatorname{ICl}(g) \rightleftharpoons I_2(g)+C l_2(g) \\
\text { Initial conc. } & 0.78 \mathrm{M} \quad 0 \quad 0 \\
\text { At equilibrium } & (0.78-2 x) \mathrm{M} \quad x \mathrm{M} \quad x \mathrm{M}
\end{array}
$$
The value of Kc = 0.14
Now we can write,$\begin{aligned} & K_c=\frac{\left[I_2\right]\left[\mathrm{Cl}_2\right]}{[\mathrm{ICl}]^2} \\ & 0.14=\frac{x^2}{(.78-x)^2} 0.374\end{aligned}$
$=\frac{x}{(.78-x)}$
By solving this we can get the value of $x=0.167$
$C_2 H_{6(g)} \rightleftharpoons C_2 H_{4(g)}+H_{2(g)}$
Answer :
Suppose the pressure exerted by the hydrogen and ethene gas be p at equilibrium.
the given reaction is-
$\begin{array}{cllll}\mathrm{C}_2 \mathrm{H}_{6(\mathrm{~g})} \rightleftharpoons \mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} & & & & \\ \text { initial pressure } & 4 a t m & 0 & 0 \\ \text { At equilibrium } & 4-p & p & p\end{array}$
Now,
$\begin{aligned} & K_p=p_{C_2 H_4} \times p_{H_2} / p_{C_2 H_6} \\ & 0.04=\frac{p^2}{4-p}\end{aligned}$
By solving the quadratic equation we can get the value of p = 0.38
Hence,at equilibrium,
$p_{C_2 H_6}=4-\mathrm{p}=4-.038$
= 3.62 atm
Question 6.18(i) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
$\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)}$
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
Answer :
The given reaction is-
$\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)}$ the concentration ratio (reaction quotient) of the given chemical reaction is-
$Q_c=\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}$
Question 6.18(ii) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
$\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)}$
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
Answer :
Let the volume of the mixture will be V.
$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOO}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)} \\ \text { initial conc. } & 1 / \mathrm{V} & 0.18 / \mathrm{V} & 0 \\ \text { At equilibrium } & \frac{1-.171}{V}(=0.829 / \mathrm{V}) \frac{1-.171}{V} & 0.171 & 0.171\end{array}$
So equilibrium constant for the reaction can be calculated as;
$\begin{gathered}K_c=\frac{(0.171)^2}{(0.829)(0.009)} \\ =3.92 \text { (approx) }\end{gathered}$
Question 6.18(iii) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
$$
\begin{array}{lccc}
& \mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)} \\
\text { initial conc. } & 1 / \mathrm{V} & 0.5 / \mathrm{V} & 0 \\
\text { At equilibrium } & \frac{1-.214}{V}(=0.786 / \mathrm{V}) \frac{1-.214}{V} & 0.214 & 0.214
\end{array}
$$
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Answer :
Let the volume of the mixture will be V.
$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOO}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)} & & & \\ \text { initial conc. } & 1 / \mathrm{V} & 0.18 / \mathrm{V} & 0 \\ \text { At equilibrium } & \frac{1-.171}{V}(=0.829 / \mathrm{V}) \frac{1-.171}{V} & 0.171 & 0.171\end{array}$
Therefore the reaction quotient of the reaction-
$\begin{aligned} & Q_c=\frac{(0.214)^2}{(0.786)(0.286)} \\ & =0.2037 \text { (approx) }\end{aligned}$
Since $Q_c<K_c$ equilibrium has not been reached.
$P C l_{5(\mathrm{~g})} \rightleftharpoons P C l_{3(\mathrm{~g})}+C l_{2(\mathrm{~g})}$
Answer :
We have,
concentration of$P C l_5=0.05 \mathrm{~mol} / \mathrm{L}$
and $K_c=8.3 \times 10^{-3}$
Suppose the concentrations of both PCl3 and Cl2 at equilibrium be x mol/L. The given reaction is:
$
\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)
$
$\begin{array}{llll}\text { at equilibrium } & \quad0.05 & \quad x & \quad x\end{array}$
it is given that the value of the equilibrium constant,$K_c=8.3 \times 10^{-3}$
Now we can write the expression for equilibrium as:
$\begin{aligned} & K_c=\frac{x^2}{0.05}=8.3 \times 10^{-3} \\ & \Rightarrow x=\sqrt{4.15 \times 10^{-4}} \\ & =0.0204 \text { (approx) }\end{aligned}$
Hence the concentration of$\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ is $0.0204 \mathrm{~mol} / \mathrm{L}$
Answer :
We have,
$K_p=0.265$
the initial pressure of CO and CO2 are 1.4 atm and 0.80atm resp.
The given reaction is-
$\mathrm{FeO}_{(s)}+\mathrm{CO}_{(g)} \rightleftharpoons \mathrm{Fe}_{(s)}+\mathrm{CO}_{2(g)}$
initially, 1.4 atm 0.80 atm
$Q_p=\frac{p C O_2}{p C O}=\frac{0.80}{1.4}=0.571$
Since Qp>Kc the reaction will proceed in the backward direction to attain equilibrium. The partial pressure of CO2 will increase = decrease in the partial pressure of CO2 = p
$\therefore K_p=\frac{p C O_2}{p C O}$
$\begin{aligned} & =\frac{0.80-p}{1.4+p}=0.265 \\ & =0.371+.265 p=0.80-p\end{aligned}$
By solving the above equation we get the value of p = 0.339 atm
Question 6.21 Equilibrium constant, Kc for the reaction
$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}$
At 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L-1 N2 , 2.0 mol L-1 H2 and 0.5 mol L-1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
Answer :
The given reaction is:
$$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)} \\$$ $$\quad 3.0 \mathrm{molL}^{-1} 2.0 \mathrm{molL}^{-1} 0.5 \mathrm{molL}^{-1}$$
Now, we know that,
$\begin{aligned} & Q c=\left[\mathrm{NH}_3\right] 2^{/}\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3 \\ & =(0.5)^2 /(3.0)(2.0)^3=0.0104\end{aligned}$
It is given that Kc = 0.061
Since, Qc ≠ Kc, the reaction mixture is not at equilibrium.
Again, Qc<Kc , the reaction will proceed in the forward direction to attain the equilibrium.
Question 6.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
$2 \mathrm{BrCl}_{(g)} \rightleftharpoons \mathrm{Br}_{2(g)}+\mathrm{Cl}_{2(g}$
For which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10-3 mol L-1 , what is its molar concentration in the mixture at equilibrium?
Answer :
Suppose the x amount of bromine and chlorine formed at equilibrium. The given reaction is:
$$
\begin{aligned}
& \qquad 2 \mathrm{BrCl}_{(g)} \rightleftharpoons \mathrm{Br}_{2(g)}+\mathrm{Cl}_{2(g)} \\
& \text { Initial Conc. } 3.3 \times 10^{-3} \quad 0 \quad 0 \\
& \text { at equilibrium } 3.3 \times 10^{-3.2 x} \quad x \quad x
\end{aligned}
$$
Now, we can write,
$\begin{aligned} & K_c=\frac{\left[\mathrm{BR}_2\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{BrCl}_2\right]^2} \\ & 32=\frac{x^2}{\left(3.3 \times 10^{-3}-2 x\right)^2} \\ & 5.66=\frac{x}{\left(3.3 \times 10^{-3}-2 x\right)} \\ & x+11.32 x=18.678 \times 10^{-3}\end{aligned}$
By solving the above equation we get,
$x=1.51 \times 10^{-3}$
Hence, at equilibrium$\left[\mathrm{BrCl}_2\right]=3.3 \times 10^{-3}-\left(2 \times 1.51 \times 10^{-3}\right)$
=0.3×10−3 M
Question 6.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass
$2 \mathrm{BrCl}_{(g)} \rightleftharpoons \mathrm{Br}_{2(g)}+\mathrm{Cl}_{2(g}$
Calculate Kc for this reaction at the above temperature .
Answer :
Suppose the total mass of the gaseous mixture is 100 g.
Total pressure is 1 atm
Mass of CO = 90.55 g
And, mass of CO2 = (100 - 90.55) = 9.45 g
Now, number of moles of CO = 90.55/28 = 3.234 mol (mol. wt of CO = 28)
Number of moles of CO2 = 9.45/44 = 0.215 mol (mol. wt of CO2 = 44)
Partial pressure of CO ,
$\begin{aligned} & p C O=\frac{n_{\mathrm{CO}}}{n_{\mathrm{CO}}+n_{\mathrm{CO}_2}} p_T \\ & =\frac{3.234}{3.234+0.215} \times 1 \\ & =0.938 \mathrm{~atm}\end{aligned}$
Similarly partial pressure of CO2,
$\begin{aligned} & p C O_2=\frac{n_{\mathrm{CO}_2}}{n_{\mathrm{CO}}+n_{\mathrm{CO}_2}} p_T \\ & =0.062 \mathrm{~atm}\end{aligned}$
Thus, $K_p=\frac{(0.938)^2}{(0.062)} \approx 14.19$
By using the relation $K_p=K_c(R T)^{\Delta n}$
$\begin{aligned} & K_c=\frac{14.19}{(0.083 \times 1127)^1} \\ & =0.159(\text { approx })\end{aligned}$
Question 6.24 Calculate a) ΔG0 for the formation of NO2 from NO and O2 at 298K
$\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)$
where
$\begin{aligned} & \Delta_f G^{+}\left[\mathrm{NO}_2\right]=52.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}[\mathrm{NO}]=87.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}\left[\mathrm{O}_2\right]=0 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$
Answer :
Given data,
$\begin{aligned} & \Delta_f G^{+}\left[\mathrm{NO}_2\right]=52.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}[\mathrm{NO}]=87.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}\left[\mathrm{O}_2\right]=0 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$
Given chemical reaction
-$\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)$
for the reaction,$\begin{aligned} & \Delta G^0=\Delta G^0 \text { (products) }-\Delta G^0 \text { (reactants) } \\ & =(52-87-0) \\ & =-35 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$
(a) $\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(g)}$
(b) $\mathrm{CaO}(s)+\mathrm{CO}_2(s) \rightleftharpoons \mathrm{CaCO}_3(s)$
(c) $3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{H}_2(\mathrm{~g})$
Answer :
According to Le Chatellier's principle, if the pressure is decreased, then the equilibrium will shift in the direction in which more number of moles of gases is present.
So,
(i) $\mathrm{COCl}_2(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_2(g)$
(ii) $\mathrm{CH}_4(g)+2 \mathrm{~S}_2(g) \rightleftharpoons \mathrm{CS}_2(g)+2 \mathrm{H}_2 \mathrm{~S}(g)$
(iii) $\mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$
(iv) $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
(v) $\mathrm{CaCO}_3(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2(g)}$
(vi)$4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_2 \mathrm{O}(g)$
Answer :
According to Le Chatellier's principle, if the pressure is increased, then the equilibrium will shift in the direction in which less number of moles of gases is present. So, as per this rule following given reactions are affected by the increasing pressure-
The reaction (i), (iii), and (vi)- all proceeds in the backward direction
Reaction(iv) will shift in the forward direction because the number of moles of gaseous reactants is more than that of products.
Question 6.27 The equilibrium constant for the following reaction is 1.6 ×105 at 1024K
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Answer :
Given that,
$\begin{aligned} & K_p \text { for the reaction }=1.6 \times 10^5 \\ & K_p^{\prime}=1 / K_p=\frac{1}{1.6 \times 10^5}=6.25 \times 10^{-6}\end{aligned}$
Let the pressure of both H2 and Br2 at equilibrium be p.
$\begin{array}{cccc}2 \mathrm{HBr} & \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) & & \\ \text { initial conc. } & 10 & 0 & 0 \\ \text { at eq } & 10-2 \mathrm{p} & \mathrm{p} & \mathrm{p}\end{array}$
Now,
$K_p^{\prime}=\frac{p^{B r_2} \cdot p^{H_2}}{p_{H B r}^2}$
$\begin{aligned} & 6.25 \times 10^{-6}=\frac{p^2}{(10-2 p)^2} \\ & 5 \times 10^{-3}=\frac{p}{(5-p)}\end{aligned}$
By solving the above equation we get,
p = 0.00248 bar
Hence the pressure of H2 and Br2 is 0.00248 bar and pressure of HBr is 0.00496 bar
Write an expression for Kp for the above reaction.
Answer :
$\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g})$
the expression of ionisation constant ( Kp) for the reaction can be defined as the ratio of the product of concentration to the product of reactants.$K_p=\frac{p_{C O} \cdot p_{H_2}^3}{p_{C H_4} \cdot p_{H_2 O}}$
Question 6.28(b) Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
$\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g})$
How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst?
Answer :
(i) According to Le Chatellier's principle, if pressure is increased, then the reaction will shift towards the less number of moles of gases. So, here the direction of equilibrium is backward and the value of Kp decreases.
(ii) According to Le Chatellier's principle, as the reaction is endothermic, the equilibrium will shift in the forward direction. The value of Kp is increases.
(iii) The equilibrium of the reaction is not affected by the presence of the catalyst. It only increases the rate of reaction.
Question 6.29(a) Describe the effect of :
(a) addition of H2
$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
Answer :
$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
(a)According to Le Chatelliers principle, on the addition of dihydrogen, the number of mole of H2 increases on the reactant side. Thus to attain the equilibrium again the reaction will move in the forward direction.
(b) According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.
(c) If we remove the CO from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction
(d) On removal of CH3OH the equilibrium will shift in the forward direction.
Question 6.29(b) Describe the effect of :
addition of CH3OH
$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
Answer :
$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.
Question 6.29(c) Describe the effect of :
removal of CO
$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
Answer :
$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
If we remove the CO from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction
(a) write an expression for Kc for the reaction.
(b) what is the value of Kc for the reverse reaction at the same temperature ?
(c) what would be the effect on Kc if
(i) more PCl5 is added
(ii) pressure is increased
(iii) the temperature is increased ?
Answer :
We have,
$\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g), \Delta_r \mathrm{H}^{+}=124.0 \mathrm{kJmol}^{-1}$
Equilibrium constant for the above reaction = 8.3×10−3
(a) Expression of Kc for this reaction-$K_c=\frac{\left[P C l_3\right]\left[C l_2\right]}{P C l_5}$
(b) The value of the reverse equilibrium constant can be calculated as;
$\begin{aligned} & K_c^{\prime}=\frac{1}{K_c} \\ & =\frac{1}{8.3 \times 10^{-3}}=1.20 \times 10^2\end{aligned}$
(c). i Kc would remain the same because the temperature is constant in this case.
(c). ii If we increase the pressure, there is no change in Kc because the temperature is constant in this case also.
(c). iii In an endothermic reaction, the value of Kc increases with an increase in temperature.
If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that $p_{\mathrm{CO}}=p_{\mathrm{H}_2 \mathrm{O}}=4.0 \mathrm{bar}$, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C
Answer :
We have,
The partial pressure of CO and H2O is 4 bar and the Kp=10.1
Let p be the partial pressure of CO2 and H2 at equilibrium. The given reaction is-
$$
\begin{array}{lcccc}
& \mathrm{CO}(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) & \rightleftharpoons \mathrm{CO}_2+\mathrm{H}_2(\mathrm{~g}) \\
\text { Initial concentration } & 4 \text { bar } & 4 \text { bar } & 0 & 0 \\
\text { At equilibrium } & 4-p & 4-p & p & p
\end{array}
$$
Therefore, we can write,$
\begin{aligned}
& 10.1=\frac{p_{\mathrm{CO}_2} \cdot p_{\mathrm{H}_2}}{p_{\mathrm{CO}} \cdot p_{\mathrm{H}_2 \mathrm{O}}} \\
& 10.1=\frac{p^2}{(4-p)^2} 3.178 \\
& =\frac{p}{(4-p)} p=(4-p) \times 3.178
\end{aligned}
$
By solving the above equation we get, p = 3.04
Hence, the partial pressure of dihydrogen at equilibrium is 3.04 bar
Question 6.32 Predict which of the following reaction will have an appreciable concentration of reactants and products:
(a) $\mathrm{Cl}_2(g) \rightleftharpoons 2 \mathrm{Cl}(g) ; K_c=5 \times 10^{-39}$
(b) $\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}) ; K_c=3.7 \times 10^8$
(c) $\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2 \mathrm{Cl}(\mathrm{g}) ; \mathrm{K}_c=1.8$
Answer :
If the value of Kc is in the range of 10−3 to 103, then the reaction has appreciable concentrations of both reactants and products.
Therefore, the third reaction (c) (Kc=1.8) will have an appreciable concentration of reactants and products.
Answer :
We have,
Equilibrium constant of the reaction = 2×10−50
the concentration of dioxygen$\left[O_2\right]=1.6 \times 10^{-2}$
The given reaction is-
$3 O_2(g) \rightleftharpoons 2 O_3(g)$
Then we have,
(equilibrium constant)
$K_c=\left[O_3(g)\right]^2 /\left[O_2(g)\right]^3$
$\begin{aligned} & 2 \times 10^{-50}=\left[\mathrm{O}_3\right]^2 /\left(1.6 \times 10^{-2}\right)^3\left[\mathrm{O}_3\right]^2 \\ & =2 \times 10^{-50} \times\left(1.6 \times 10^{-2}\right)^3 \\ & =8.192 \times 10^{-56} \\ & {\left[\mathrm{O}_3\right]=\sqrt{8.192 \times 10^{-56}}=2.86 \times 10^{-28}}\end{aligned}$
Thus the concentration of dioxygen is$2.86 \times 10^{-28}$
Question 6.34(a) The reaction,
is at equilibrium at 1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O, and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant Kc, for the reaction at the given temperature is 3.90.
Answer :
Given that,
Total volume = 1L
0.3 mol of CO , 0.10 mol of dihydrogen( H2 )and the 0.02 mol of water( H2O )
the equilibrium constant = 3.90
Let x be the concentration of methane at equilibrium. The given reaction is-
$
\mathrm{CO}(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g})
$
At equilibrium, $\quad 0.3 \mathrm{~mol} / \mathrm{L} \quad 0.1 \mathrm{~mol} / \mathrm{L} \quad x \quad 0.02 \mathrm{~mol} / \mathrm{L}$
Therefore,
$\begin{aligned} & K_c=\frac{0.02 \times x}{(0.3)(0.1)}=3.90 \\ & \Rightarrow x=\frac{(3.9)(0.3)(0.1)^3}{(0.02)} \\ & =5.85 \times 10^{-2}\end{aligned}$
Thus the concentration of methane at equilibrium is$=5.85 \times 10^{-2}$
Question 6.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:
Answer :
A conjugate acid-base pair means that the species are differed by only one proton. for example; HCl is an acid because it donates a proton to water. So $\mathrm{HCl}-\mathrm{Cl}^{-}$, and $\mathrm{H}_3 \mathrm{O}^{+}-\mathrm{H}_2 \mathrm{O}$ these pairs are called conjugate acid-base pair.
Species
|
Conjugate acid-base
|
HNO2
|
NO2− (Base)
|
CN−
|
HCN (Acid)
|
HClO4
|
ClO4− (base)
|
F−
|
HF (acid)
|
OH−
|
H2O (acid)
|
CO32−
|
HCO3− (acid)
|
S2−
|
HS− (acid)
|
Question 6.36 Which of the followings are Lewis acids?$\mathrm{H}_2 \mathrm{O}, \mathrm{BF}_3, \mathrm{H}^{+}$and $\mathrm{NH}_4^{+}$
Answer :
Lewis acid-
Those species that can accept the pair of electrons are called Lewis acids. For example Boron trifluoride ( BF3 ), ammonium ion( NH4+) and the hydrogen ion( H+). Among them water molecule is a Lewis base, it can donate pair of electrons.
Question 6.37 What will be the conjugate bases for the Brönsted acids: $\mathrm{HF}, \mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HCO}^{-3}$?
Answer :
When Brönsted acids lose their proton then they become a conjugate base of that corresponding acids.
Followings are the conjugate base of Brönsted acid-
$\begin{aligned} & \mathrm{HF}_2-F^{-} \\ & \mathrm{H}_2 \mathrm{SO}_4^{-}-\mathrm{HSO}_4^{-} \\ & \mathrm{HCO}_3^{-}-\mathrm{CO}_3^{2-}\end{aligned}$
Question 6.38 Write the conjugate acids for the following Brönsted bases: NH2−,NH3 and HCOO− .
Answer :
When Brönsted base accepts a proton then they become a conjugate acid of that corresponding base.
Followings are the conjugate acid of Brönsted base-
Answer :
When acid or base accept or lose a proton, they form conjugate acid or base of that corresponding species.
Lists of the conjugate acid and conjugate base of the given species-
Species
|
Conjugate acid
|
Conjugate base
|
H2O
|
H3O+
|
OH−
|
HCO3−
|
H2CO3
|
CO32−
|
HSO4−
|
H2SO4
|
SO42−
|
NH3
|
NH4+
|
NH2−
|
Answer :
Species which donate pair of an electron are called Lewis base and which accepts pair of electrons are called acid.
(a) OH− is a Lewis base since it can donate its lone pair of electrons.
(b) F− is a Lewis base since it can donate a pair of electrons.
(c) H+ is a Lewis acid since it can accept a pair of electrons.
(d) BCl3 is a Lewis acid since it can accept a pair of electrons.
Question 6.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3 M. what is its pH?
Answer :
We have,
the concentration of Hydrogen ion sample is 3.8×10−3 M
So, $p H=-\log [H]^{+}$
$
=-\log \left(3.8 \times 10^{-3}\right)=3-\log 3.8=2.42
$
Question 6.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer :
We have,
The pH of a sample of vinegar is 3.76$p H=?$
Therefore,
$p H=-\log \left[H^{+}\right] \log \left[H^{+}\right]=-p H$
Taking antilog on both sides we get,
$\begin{aligned} & {\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-3.76)} \\ & {\left[\mathrm{H}^{+}\right]=1.74 \times 10^{-4}}\end{aligned}$
Hence the concentration of hydrogen ion$\left[H^{+}\right]=1.74 \times 10^{-4}$
Answer :
We have,
IOnization constant of hydrogen fluoride, methanoic acid and hydrogen cyanide are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$respectively.
It is known that,
$
K_b=\frac{K_w}{K_a}
$..........................(i)
$K_b$ of the conjugate base $F^{-}$
$
\begin{aligned}
& =\frac{10^{-14}}{6.8 \times 10^{-4}} \\
& =1.5 \times 10^{-11}
\end{aligned}
$
Similarly,
By using the equation (i)
$K_b$ of the conjugate base $\mathrm{HCOO}^{-}$
$
\begin{aligned}
& =\frac{10^{-14}}{1.8 \times 10^{-4}} \\
& =5.6 \times 10^{-11}
\end{aligned}
$
$K_b$ of the conjugate base $\mathrm{CN}^{-}$
$
\begin{aligned}
& =\frac{10^{-14}}{4.8 \times 10^{-9}} \\
& =2.8 \times 10^{-6}
\end{aligned}
$
Answer :
We have,
The ionization constant of phenol is $1.0 \times 10^{-10}$ , and
the concentration of phenol is 0.05 M
degree of ionisation = ?
Ionization of phenol;
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}+\mathrm{H}_3 \mathrm{O}^{+}$
At equilibrium,
the concentration of various species are-
$\begin{aligned} & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right]=0.05-x} \\ & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=x}\end{aligned}$
As we see, the value of ionisation is very less. Also x will be very small. Thus we can ignore x .
$\begin{aligned} & K_a=\frac{\left[C_6 \mathrm{H}_5 \mathrm{O}^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[C_6 \mathrm{H}_5 \mathrm{OH}\right]} 10^{-10} \\ & =\frac{x^2}{0.05} x=\sqrt{10^{-10} \times 0.05} \\ & x=2.2 \times 10^{-6}\end{aligned}$
Hence the concentration of phenolate ion is$\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]=2.2 \times 10^{-6}$
Let α be the degree of dissociation of phenol in the presence of 0.01 M of phenolate ion.
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}+\mathrm{H}^{+}$
Concentration (1 - α ) 0.05 0.05 α 0.05 α
So,$\begin{aligned} & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right]=0.05(1-\alpha)=0.05} \\ & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]=0.05 \alpha+0.01 \approx 0.01 \mathrm{M}} \\ & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=0.05 \alpha}\end{aligned}$
therefore,
$K_a=\frac{(0.01)(0.05 \alpha)}{0.05} 10^{-10}=0.01 \alpha$
$\alpha=10^{-8}$
The degree of dissociation is 10−8
Answer :
We have,
1st ionisation constant of hydrogen sulphide is 9.1×10−8 and the 2nd dissociation constant is 1.2×10−13
Case 1st-(absence of hydrochloric acid)
To calculate the concentration of HS−
Let x be the concentration of HS− and the ionisation of hydrogen sulphide is;
$\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}$
0.1 M
At equilibrium, the concentration of various species is,
Since the dissociation constant is very small. So, x can be neglected.
the concentration of $H_2 S=1-x \mathrm{M}$
the concentration of$H S^{-}$and $H^{+}$is $x \mathrm{M}$
So,
$K_a=\frac{x^2}{0.1}=9.1 \times 10^{-8}$
from here x can be calculated and we get, $x=9.54 \times 10^{-5} M$
Case 2nd (In presence of 0.1 M, HCl)
Suppose H2S is dissociated is x .Then at equilibrium,
$\left[\mathrm{H}_2 \mathrm{~S}\right]=0.1-x \simeq 0.1,\left[\mathrm{H}^{+}\right]=0.1+x \simeq 0.1$ and the $\left[\mathrm{HS}^{-}\right]=x \mathrm{M}$
So,
$
K_a=\frac{x(0.1)}{0.1}=9.1 \times 10^{-8}
$
Thus the concentration of $\left[\mathrm{HS}^{-}\right]$is $9.1 \times 10^{-8}$
Answer :
It is given,
The ionisation constant of acetic acid is 1.74×10−5 and concentration is 0.05 M
The ionisation of acetic acid is;
$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
Therefore,
$\begin{aligned} & \mathrm{K}_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & \therefore\left[\mathrm{H}^{+}\right]=\sqrt{\left(1.74 \times 10^{-5}\right)\left(5 \times 10^{-2}\right)} \\ & =9.33 \times 10^{-4}=\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\end{aligned}$
$=9.33 \times 10^{-4}=\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]$
$
\begin{aligned}
& =4-\log (9.33) \\
& =3.03
\end{aligned}
$
We know that,
$
\begin{aligned}
\alpha & =\sqrt{\frac{K_a}{C}} \\
\alpha & =\sqrt{\frac{1.74 \times 10^{-5}}{0.05}} \\
& =1.86 \times 10^{-2}
\end{aligned}
$
Answer :
We have,
pH of organic acid is 4.15 and its concentration is 0.01M
Suppose the organic acid be HA. The dissociation of organic acid can be written as;
$\begin{aligned} & H A \rightleftharpoons H^{+}+A^{-} \\ & p H=4.15\end{aligned}$$-\log \left[H^{+}\right]=4.15\left[H^{+}\right]=7.08 \times 10^{-5}$
Now,
$\begin{aligned} & K_a=\left[H^{+}\right]\left[A^{-}\right] /[H A] \\ & {\left[H^{+}\right]=\left[A^{-}\right]=7.08 \times 10^{-5}}\end{aligned}$
[HA] = 0.01
Then,
$K_a=\left(7.08 \times 10^{-5}\right)^2 / 0.01 K_a=5.01 \times 10^{-7}$
Thus
$p K_a=-\log \left(5.01 \times 10^{-7}\right)=6.30$
Question 6.48 Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH
Answer :
Assuming the complete dissociation. So, α=1
(a) The ionisation of hydrochloric acid is
$\mathrm{HCl} \rightleftharpoons \mathrm{H}^{+}+\mathrm{Cl}^{-}$
Since it is fully ionised then$\left[\mathrm{H}^{+}\right]=\left[\mathrm{Cl}^{-}\right]=0.003 \mathrm{M}$
Therefore,$\begin{aligned} & p H \text { of the solution }=-\log (0.003) \\ & =3-\log (3) \\ & =2.52\end{aligned}$
(b) The ionisation of 0.005M NaOH
$\begin{aligned} & \mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{OH}^{-} \\ & {\left[\mathrm{Na}^{+}\right]=\left[\mathrm{OH}^{-}\right]=0.005 \mathrm{M}}\end{aligned}$
Therefore,
pOH of the solution $
=-\log (0.005)=3-\log 5=2.301
$
pH of the solution is equal to (14 - 2.301 =11.70)
(c) The ionisation of 0.002MHBr
$\begin{aligned} & \mathrm{HBr} \rightleftharpoons \mathrm{H}^{+}+\mathrm{Br}^{-} \\ & {\left[\mathrm{Br}^{-}\right]=\left[\mathrm{H}^{+}\right]=0.002 \mathrm{M}}\end{aligned}$
Therefore,
$p H$ of the solution $
=-\log (0.002)=3-\log 2=2.69
$
pH of the solution is equal to (2.69)
(d) The ionisation of 0.002M KOH
$\begin{aligned} & \mathrm{KOH} \rightleftharpoons \mathrm{K}^{+}+\mathrm{OH}^{-} \\ & {\left[\mathrm{OH}^{-}\right]=\left[\mathrm{K}^{+}\right]=0.002 \mathrm{M}}\end{aligned}$
Therefore,
pOH of the solution $
=-\log (0.002)=3-\log 2=2.69
$
pH of the solution is equal to (14 - 2.69 = 11.31)
Question 6.49(a) Calculate the pH of the following solutions:
2 g of TlOH dissolved in water to give 2 litre of solution.
Answer :
Here, 2 g of TlOH dissolves in water to give 2 litres of solution
So, the concentration of TlOH =$[$ TlOH $($ aq $)]=\frac{2}{2} g / L=\frac{1}{221} M$ (the molar mass of TlOH is 221)
TlOH can be dissociated as $\mathrm{TlOH}(\mathrm{aq}) \rightarrow \mathrm{Tl}^{+}+\mathrm{OH}^{-}$
$
O H^{-}(a q)=\operatorname{TlOH}(a q)=\frac{1}{221} M
$
Therefore, $K_w=\left[H^{+}\right]\left[O H^{-}\right]\left(\right.$since $\left.\mathrm{K} w=10^{-14}\right)$
So, the concentration of $\left[H^{+}\right]=221 \times 10^{-14}$
Thus $P^H=-\log \left[H^{+}\right]=-\log \left(221 \times 10^{-14}\right)$
= 11.65(approx)
Question 6.49(b) Calculate the pH of the following solutions:
0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.
Answer :
The calcium hydroxide ion dissociates into-
$
\mathrm{Ca}(\mathrm{OH})_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-}
$
Molecular weight of $\mathrm{Ca}(\mathrm{OH})_2=74$
the concentration of $\left[\mathrm{Ca}(\mathrm{OH})_2\right]=\frac{0.3 \times 1000}{74 \times 500}=0.0081 \mathrm{M}$
$\therefore\left[\mathrm{OH}^{-}\right]=\left[\mathrm{Ca}(\mathrm{OH})_2\right]=0.0081 \mathrm{M}$
We know that,
$\begin{aligned} & {\left[H^{+}\right]=\frac{K_w}{\left[O H^{-}\right]}=\frac{10^{-14}}{0.0162}} \\ & =61.7 \times 10^{-14}\end{aligned}$
Thus $pH=-\log \left[H^{+}\right]=-\log \left(61.7 \times 10^{-14}\right)$
$=14-1.79=12.21$
Question 6.49(c) Calculate the pH of the following solutions:
0.3 g of NaOH dissolved in water to give 200 mL of solution.
Answer :
NaOH dissociates into$\mathrm{NaOH} \rightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}$
So, the concentration of $[\mathrm{NaOH}]=\frac{0.3 \times 1000}{200 \times 40} \mathrm{M}=0.0375 \mathrm{M}$
$\therefore\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=0.0375 \mathrm{M}$
We know that ,
$
\left[H^{+}\right]=\frac{K_W}{\left[O H^{-}\right]}=\frac{10^{-14}}{0.0375}=2.66 \times 10^{-13}
$
Now, $pH=-\log \left[H^{+}\right]$
$
=-\log \left(2.66 \times 10^{-13}\right)=12.57
$
Question 6.49(d) Calculate the pH of the following solutions:
1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.
Answer :
We know that,
M1V1 (before dilution) = M2V2 (after dilution)
initially V1 = 1mL and M1 = 13.6 M
and V2 = 1L and M1 = ?
By putting all these values we get,
$
\begin{aligned}
& M_2=\frac{13.6 \times 10^{-3}}{1}=1.36 \times 10^{-2} \\
& \therefore\left[H^{+}\right]=1.36 \times 10^{-2}
\end{aligned}
$
Thus $pH=-\log \left[H^{+}\right]=-\log \left(1.36 \times 10^{-2}\right)$
= 1.86 (approx)
Answer :
We have,
Degree of ionization(a) = 0.132
Concentration of bromoacetic acid (C) = 0.1 M
Thus the concentration of H3O+=C.a
= 0.0132
Therefore
$\begin{aligned} & pH=-\log \left[H^{+}\right]=-\log (0.0132) \\ & =1.879\end{aligned}$
Now, we know that,
$\begin{aligned} & K_a=C . a^2 \\ & \text { So, } p K_a=-\log \left(C . a^2\right) \\ & =-\log \left(0.1 \times(0.0132)^2\right) \\ & =-\log (0.0017)=2.76 \text { (approx) }\end{aligned}$
Answer :
We have,
$\begin{aligned} & \mathrm{C}=0.005 \mathrm{M} \\ & pH=9.95 \text { and } pOH=14-9.95=4.05\end{aligned}$
we know that pOH=-$\log \left[O H^{-}\right]$$4.05=-\log \left[\mathrm{OH}^{-}\right]$
By taking antilog on both sides we get.
concentration of$\left[\mathrm{OH}^{-}\right]=8.91 \times 10^{-5}$
C.a $=8.91 \times 10^{-5}$
So, $\mathrm{a}=1.782 \times 10^{-2}$
We know that,
$
\begin{aligned}
& K_b=C . a^2 \\
& =0.005 \times\left(1.782 \times 10^{-2}\right)^2=0.0158 \times 10^{-4}
\end{aligned}
$
Thus
$
\begin{aligned}
& p K_b=-\log \left(K_b\right) \\
& =-\log \left(0.0158 \times 10^{-4}\right)=5.80
\end{aligned}
$
Answer :
We have,
C = 0.001 M
Kb = 4.27×10−10
Degree of inozation of aniline (a) = ?
Ionization constant of the conjugate acid ( Ka ) = ?
We know that
$\begin{aligned} & K_b=C . a^2 \\ & 4.27 \times 10^{-10}=(0.001) a^2 \\ & \text { Thus } a=\sqrt{4270 \times 10^{-10}} \\ & =65.34 \times 10^{-4}\end{aligned}$
$\begin{aligned} & \text { Then [Base] }=\text { C.a }=\left(65.34 \times 10^{-4}\right)(0.001) \\ & =0.653 \times 10^{-5}\end{aligned}$
$\begin{aligned} & \text { Now, } pOH=-\log \left(0.65 \times 10^{-5}\right)=6.187 \\ & P^H=14-PO H \\ & =14-6.187 \\ & =7.813\end{aligned}$
It is known that,
$
K_a \times K_b=K_w
$
So, $K_a=\frac{10^{-14}}{4.27 \times 10^{-10}}$ $=2.34 \times 10^{-5}$ This is ionization constant.
0.01M
Answer :
We have,
C = 0.05 M
$p K_a=4.74=-\log \left(K_a\right)$
By taking antilog on both sides we get,
$
\left(K_a\right)=1.82 \times 10^{-5}=C \cdot(\alpha)^2
$
from here we get the value of $\alpha=\sqrt{\frac{1.82 \times 10^{-5}}{5 \times 10^{-2}}}$ $=1.908 \times 10^{-2}$
After adding hydrochloric acid, the concentration of H+ ions increases and due to that the equilibrium shifts towards the backward direction. It means dissociation will decrease.
(i) when 0.01 HCl is taken
$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} & & \\ \text {Initial conc. } & 0.05 & 0 & 0 \\ \text { after dissociation } & 0.05-x & 0.001+x & x\end{array}$
As the dissociation is very small.
So we can write $0.001+x \approx 0.001$ and $0.05-x \approx 0.05$
$\begin{aligned} & \text { Now, } K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & =\frac{(0.001)(x)}{(0.05)}=\frac{x}{50}\end{aligned}$
So, the value of $x=\frac{\left(1.82 \times 10^{-5}\right)(0.05)}{0.01}$
Now degree of dissociation $=($ amount dissociated $) /($ amount taken $)$
$
\begin{aligned}
& =\frac{1.82 \times 10^{-3}(0.05)}{0.05} \\
& =1.82 \times 10^{-3}
\end{aligned}
$
0.1M in HCl?
Answer :
Let the x amount of acetic acid is dissociated in this case
$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} & & \\ \text {Initial conc. } & 0.05 & 0 & 0 \\ \text { after dissociation } & 0.05-x & 0.001+x & x\end{array}$
As the dissociation is very small.
So we can write $0.1+x \approx 0.1$ and $0.05-x \approx 0.05$
$\begin{gathered}K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ =\frac{(0.1)(x)}{(0.05)}=2 x\end{gathered}$
So, the value of$x=\frac{\left(1.82 \times 10^{-4}\right)(0.05)}{0.1}$
Now the degree of dissociation = (amount dissociated) / (amount is taken)
$\begin{aligned} & =\frac{1.82 \times 10^{-4}(0.05)}{0.05} \\ & =1.82 \times 10^{-4}\end{aligned}$
Answer :
We have,
(Degree of ionization) $K_b=5.4 \times 10^{-4}$
Concentration of dimethylamine = 0.02 M
$\therefore \alpha=\sqrt{\frac{K_b}{C}}=\sqrt{\frac{5.4 \times 10^{-4}}{0.02}}$$\alpha=0.1643$
If we add 0.1 M of sodium hydroxide. It is a strong base so, it goes complete ionization
$\mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{OH}^{-}$
(0.1 M) (0.1 M)
and also,
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons\left(\mathrm{CH}_3\right)_2 \mathrm{NH}_2^{+}+\mathrm{OH}^{-}$
0.02- x x x
$\left[\mathrm{OH}^{-}\right]=x+0.1 \approx 0.1$ (since the dissociation is very small)
Therefore,
$\begin{aligned} & K_b=\frac{x(0.1)}{0.02} \\ & x=5.4 \times 10^{-4}\left(\frac{0.02}{0.1}\right) \\ & =0.0054\end{aligned}$
Hence in the presece of 0.1 M of sodium hydroxide , 0.54% of dimethylamine get dissociated.
Question 6.55(a) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
Human muscle-fluid, 6.83
Answer :
We have pH 6.83
It is known that $pH=-\log \left[H^{+}\right]$
Therefore,
$6.83=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,
$\left[H^{+}\right]=1.48 \times 10^{-7} M$
Question 6.55(b) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
Human stomach fluid, 1.2
Answer :
We have $pH=1.2$
It is known that $pH=-\log \left[H^{+}\right]$
Therefore,
$1.2=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,
$\left[H^{+}\right]=0.063 M$
Question 6.55(c) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
Human blood, 7.38
Answer :
we have $pH=7.38$
It is known that pH=−log[H+]
Therefore,
$7.38=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,
$\left[H^{+}\right]=4.17 \times 10^{-8} M$
Question 6.55(d) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
Human saliva, 6.4.
Answer :
we have $pH=6.4$
It is known that pH=−log[H+]
Therefore,
$6.4=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,
$\left[H^{+}\right]=3.98 \times 10^{-7} M$
Answer :
We already know that pH can be calculated as- $-\log \left[H^{+}\right]$
to calculate the concentration of $\left[H^{+}\right]=\operatorname{antilog}\left(-pH\right)$
Thus, the hydrogen ion concentration of followings pH values are-
(i) pH of milk = 6.8
Since, pH=$-\log \left[H^{+}\right]$
6.8 = −log[H+]
$\log \left[H^{+}\right]$ = -6.8
[H+] = anitlog(-6.8)
$=1.5 \times 10^{-7} M$
(ii) pH of black coffee = 5.0
Since,pH=$-\log \left[H^{+}\right]$
5.0 = $-\log \left[H^{+}\right]$
$\log \left[H^{+}\right]$= -5.0
$\left[H^{+}\right]$ = anitlog(-5.0)
$=10^{-5} M$
(iii) pH of tomato juice = 4.2
Since, pH=$-\log \left[\mathrm{H}^{+}\right]$
4.2 = $-\log \left[\mathrm{H}^{+}\right]$
$\begin{aligned} & \log \left[H^{+}\right]=-4.2 \\ & {\left[H^{+}\right]=\operatorname{anitlog}(-4.2)} \\ & =6.31 \times 10^{-5} \mathrm{M}\end{aligned}$
(iv) pH of lemon juice = 2.2
Since, pH=$-\log \left[H^{+}\right]$
2.2 = $-\log \left[H^{+}\right]$
$\begin{aligned} & 2.2=-\log \left[H^{+}\right] \\ & \log \left[H^{+}\right]=-2.2 \\ & {\left[H^{+}\right]=\operatorname{anitlog}(-2.2)} \\ & =6.31 \times 10^{-3} \mathrm{M}\end{aligned}$
(v) pH of egg white = 7.8
Since, pH=$-\log \left[\mathrm{H}^{+}\right]$
7.8 = $-\log \left[\mathrm{H}^{+}\right]$
$\begin{aligned} & \log \left[H^{+}\right]=-7.8 \\ & {\left[H^{+}\right]=\operatorname{anitlog}(-7.8)} \\ & =1.58 \times 10^{-8} M\end{aligned}$
Answer :
We have 0.562 g of potassium hydroxide ( KOH ). On dissolving in water gives 200 mL of solution.
Therefore, concentration of
$[K O H(a q)]=\frac{0.561 \times 1000}{200} g / L$
= 2.805 g/L
$=2.805 \times \frac{1}{56.11} M=0.05 M$
$\mathrm{KOH}(a q) \rightarrow \mathrm{K}^{+}(a q)+\mathrm{OH}^{-}(a q)$
It is a strong base. So, that it goes complete dissociation.
$\left[O H^{-}\right]=0.05 M=\left[K^{+}\right]$
It is known that,
$\begin{aligned} & K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]} \\ & =\frac{10^{-14}}{0.05}=2 \times 10^{-13} \mathrm{M}\end{aligned}$
Therefore,$pH=-\log \left(2 \times 10^{-13}\right)=12.69$
Answer :
By given abova data, we know the solubility of Sr(OH)2 at 298 K = 19.23 g/L
So, concentration of [Sr(OH)2]
= 19.23/121.63M (Molecular weight of Sr(OH)2 = 121.63 u)
= 0.1581 M
$\mathrm{Sr}(\mathrm{OH})_2(\mathrm{aq}) \rightarrow \mathrm{Sr}^{2+}(\mathrm{aq})+2\left(\mathrm{OH}^{-}\right)(\mathrm{aq})$
$\therefore S r^{2+}=0.1581 M$
and the concentration of [OH−] =2×0.1581M=0.3162M
Now
It is known that,
$
\begin{aligned}
& K_w=\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right] \\
& {\left[\mathrm{H}^{+}\right]=\frac{10^{-14}}{0.3162}} \\
& =3.16 \times 10^{-14}
\end{aligned}
$
Therefore $pH=-\log \left(3.16 \times 10^{-14}\right)=13.5$
Answer :
Let the degree of ionization of propanoic acid be α . Then Let suppose we can write propanoic acid to be HA,
It is known that,
We have
ionization constant of propanoic acid $\left(K_a\right)=1.32 \times 10^{-5}$ and the concentration is 0.005 M
$\alpha=\sqrt{\frac{K_a}{C}}$
By putting the values in above formula we get,
$\alpha=\sqrt{\frac{1.32 \times 10^{-5}}{0.05}}=1.62 \times 10^{-2}$
$
{ [Acid] }=\left[\mathrm{H}_3 \mathrm{O}\right]^{+}=\text {C. } \alpha $
$ =8.15 \times 10^{-4}$
Therefore, $p^H=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]$
$
\begin{aligned}
& =-\log \left[8.15 \times 10^{-4}\right] \\
& =3.08
\end{aligned}
$
If we add 0.01M hydrochloric acid then,
$\begin{array}{ccc}\mathrm{AH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons & \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-} & \\ \text {initial con } & \mathrm{C} & 0 \\ \text { at equi. } & \mathrm{C}-x \approx \mathrm{C} & 0.01+x\end{array}$
Now, by using the formula of $K_a=\frac{(x)(0.01)}{c-x}$$=\frac{(x)(0.01)}{C}$
The value of x is calculated as ;
$\Rightarrow 1.32 \times 10^{-5} \times \frac{0.01}{0.01}=1.32 \times 10^{-5}$ (Degree of ionisation)
Answer :
We have,
Concentration of cyanic acid = 0.1 M
pH=$-\log \left[H^{+}\right]=2.34$
Therefore, the concentration of [H+] = antilog (-2.34)
$=4.5 \times 10^{-3}$
It is known that,
$\begin{aligned} & {\left[H^{+}\right]=C . \alpha=4.5 \times 10^{-3}} \\ & \alpha=\frac{4.5 \times 10^{-3}}{0.1}=4.5 \times 10^{-2}\end{aligned}$
Then Ionization constant
$\left(K_a\right)=C \cdot \alpha^2$$\begin{aligned} & =(0.1)\left(4.5 \times 10^{-2}\right)^2 \\ & =2.02 \times 10^{-4}\end{aligned}$
Answer :
We have,
Ionization constant of nitrous acid =4.5×10−5
Concentration of sodium nitrite ( NaNO2 ) = 0.04 M
Degree of hydrolysis can be calculated as;
$K_h=\frac{K_w}{K_a}=\frac{10^{-14}}{4.5 \times 10^{-4}}=0.22 \times 10^{-10}$
Sodium nitrite is a salt of sodium hydroxide (strong base) and the weak acid ( $\mathrm{HNO}_2$ )
$\mathrm{NO}_2^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{HNO}_2+\mathrm{OH}^{-}$
Suppose x moles of salt undergoes hydrolysis, then the concentration of-
$
\left[\mathrm{NO}_2^{-}\right]=0.04-x \approx 0.04
$
$\left[\mathrm{HNO}_2^{-}\right]=x$, and
$
\left[O H^{-}\right]=x
$
Therefore
$k_h=\frac{x^2}{0.04}=0.22 \times 10^{-10}$
from here we can calculate the value of x ;
$\begin{aligned} & \Rightarrow x=\sqrt{0.0088 \times 10^{-10}} \\ & =0.093 \times 10^{-5}=\left[\mathrm{OH}^{-}\right] \\ & \Rightarrow\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]}=\frac{10^{-14}}{0.093 \times 10^{-5}} \\ & =10.75 \times 10^{-9} \mathrm{M}\end{aligned}$
Therefore the degree of hydrolysis
Question 6.62 A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer :
Given,
pH = 3.44
We know that
pH=$-\log \left[H^{+}\right]$
By taking antilog on both sides we get,
[H+] = antilog (- 3.44)
$\therefore\left[H^{+}\right]=3.63 \times 10^{-4}$
pyridinium hydrochloride completely ionised.
Then Kh = (conc. of products)/ (conc, of reactants)
$=\frac{\left(3.63 \times 10^{-2}\right)^2}{0.02}$ (? Concentration is 0.02M)
$\Rightarrow K_h=6.58 \times 10^{-6} $
Now,$\begin{aligned} & K_h=K_w / K_a \\ & \Rightarrow K a=K_w / K_h\end{aligned}$
$\begin{aligned} & =10^{-14} / 6.58 \times 10^{-6} \\ & =1.51 \times 10^{-9} \text { (approx) }\end{aligned}$
Answer :
Salts of strong acid and strong base are neutral in nature for example-
Salts of a strong base and weak acid are basic in nature for example-
Salts of strong acid and a weak base are acidic in nature for example-
Answer :
We have,
Ionisation constant of chloroacetic acid( Ka ) is 1.35×10−3
The concentration of acid = 0.1 M
Ionisation if acid, =$\mathrm{ClCH}_2 \mathrm{COOH} \rightleftharpoons \mathrm{ClCH}_2 \mathrm{COO}^{-}+\mathrm{H}^{+}$
We know that,
$
\Rightarrow K_a=\frac{\left[\mathrm{ClCH}_2 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{ClCH}_2 \mathrm{COOH}\right]}
$....................(i)
As it completely ionised$\left[\mathrm{ClCH}_2 \mathrm{COO}^{-}\right]=\left[\mathrm{H}^{+}\right]$
Putting the values in eq (i)
$\begin{aligned} & 1.35 \times 10^{-3}=\frac{\left[H^{+}\right]^2}{0.02} \\ & {\left[H^{+}\right]=\sqrt{1.35 \times 10^{-3} \times 0.02}=1.16 \times 10^{-2}}\end{aligned}$
Therefore, pH of the solution$=-\log \left(1.16 \times 10^{-2}\right)$
= 2−log(1.16)
= 1.94
Now,
$0.1 \mathrm{M} \mathrm{ClCH}_2 \mathrm{COONa}^{-}$ (sod. chloroacetate) is basic due to hydrolysis-
$\mathrm{ClCH}_2 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{CH}_2 \mathrm{ClCOOH}+\mathrm{OH}^{-}$
For a salt of strong base+strong acid
$p H=7+\frac{p K_a+\log C}{2}=7+\frac{2.87+\log 0.1}{2}$
pH = 7.94
Question 6.65 Ionic product of water at 310 K is 2.7 × 10 -14 . What is the pH of neutral water at this temperature?
Answer :
We have the ionic product of water at 310 K is $2.7 \times 10^{-14}$
It is known that,
ionic product $K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]$
Since $\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]$, therefore $K_w=\left[\mathrm{H}^{+}\right]^2$
$\begin{aligned} & \Rightarrow K_w \text { at } 310 \mathrm{~K} \text { is } 2.7 \times 10^{-14} \\ & \therefore K_w=2.7 \times 10^{-14}=\left[H^{+}\right]^2\end{aligned}$
here we can calculate the value of [H+] concentration.
$
\left[H^{+}\right]=\sqrt{2.7 \times 10^{-14}}=1.64 \times 10^{-7}
$
Thus, $pH=-\log \left[H^{+}\right]$
$
\begin{aligned}
& =-\log \left(1.64 \times 10^{-7}\right) \\
& =6.78
\end{aligned}
$
Hence the pH of neutral water is 6.78
Question 6.66(a) Calculate the pH of the resultant mixtures:
10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
Answer :
Given that,
Vol. of 0.2 M Ca(OH)2 = 10 mL
Vol. of 0.1 M HCl = 25 mL
therefore, by using the formula,
$M\left(O H^{-}\right)=\frac{M_1 V_1(\text { base })-M_2 V_2(\text { acid })}{V_1+V_2}$
By substituting the value in these equations, we get;
$\Rightarrow \frac{(0.2 \times 2)-(0.1 \times 2)}{10+25}=\frac{1.5}{25}=0.06$
Now, pOH=$-\log \left[\mathrm{OH}^{-}\right]$
$=-\log (0.06)=1.221$
since pH+pOH=14
pH=14−pOH
= 14-1.221
= 12.78
Question 6.66(b) Calculate the pH of the resultant mixtures:
10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
Answer :
In this case, both the solutions have the same number of moles of H+ and OH− , therefore they both can get completely neutralised. Hence the pH = 7.0
Question 6.66(c) Calculate the pH of the resultant mixtures:
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH
Answer :
Given that,
Volume of 0.1 M KOH = 10 mL, and
Volume of 0.1 M H2SO4 = 10 mL
So, by using the formula of,
$M\left(H^{+}\right)=\frac{M_1 V_1(\text { acid })-M_2 V_2(\text { base })}{V_1+V_2}$
By putting the values we get,
$\Rightarrow \frac{2(0.1 \times 10)-0.1 \times 10}{10+10}=\frac{1}{20}=5 \times 10^{-2}$
Hence, $p H=-\log \left[H^{+}\right]=-\log \left(5 \times 10^{-2}\right)=1.30$
Answer :
Solubility product is the product of ionic concentrations in a saturated solution.
$K_{s p}=\left[A^{+}\right]\left[B^{-}\right]$
(i) silver chromate ( Ag2CrO4 )
Ionization of silver chromate
$\mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_4^{2-}$
Let " s " be the solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$
$\begin{aligned} & {\left[\mathrm{Ag}^{+}\right]=2 s} \\ & {\left[\mathrm{CrO}_4^{2-}\right]=s}\end{aligned}$
According to the table $K_{s p}$ of $\mathrm{Ag}_2 \mathrm{CrO}_4=1.1 \times 10^{-12}$
$\begin{aligned} & \Rightarrow 1.1 \times 10^{-12}=(2 s)^2 . s=1.1 \times 10^{-12}=2 s^3 \\ & s=\sqrt[3]{\frac{1.1 \times 10^{-12}}{4}} \\ & =0.65 \times 10^{-4}\end{aligned}$
(ii) Barium chromate ( BaCrO4 )
Ionization of silver chromate
$\mathrm{BaCrO}_4 \rightleftharpoons \mathrm{Ba}^{2+}+\mathrm{CrO}_4^{2-}$
Let " s " be the solubility of BaCrO4
$\begin{aligned} & {\left[\mathrm{Ba}^{2+}\right]=s} \\ & {\left[\mathrm{CrO}_4^{2-}\right]=s}\end{aligned}$
According to the table $K_{s p}$ of $\mathrm{BaCrO}_4=1.2 \times 10^{-12}$
$\begin{aligned} & \Rightarrow 1.2 \times 10^{-10}=s . s=1.1 \times 10^{-10}=s^2 \\ & s=\sqrt{\frac{1.2 \times 10^{-10}}{1}} \\ & =1.09 \times 10^{-5}\end{aligned}$
(iii) Ferric hydroxide Fe(OH)3)
Ionization of Ferric hydroxide
$\mathrm{Fe}(\mathrm{OH})_3 \rightleftharpoons \mathrm{Fe}^{3+}+3 \mathrm{OH}^{-}$
Let " s " be the solubility of Fe(OH)3
$\begin{aligned} & {\left[F e^{3+}\right]=s} \\ & {\left[O H^{-}\right]=3 s}\end{aligned}$
According to the table Ksp of Fe(OH)3 = 1.0×10−38
$\begin{aligned} & \Rightarrow 1.0 \times 10^{-38}=s .(3 s)^3=1.0 \times 10^{-38}=27 s^4 \\ & s=\sqrt[4]{\frac{1.0 \times 10^{-38}}{27}} \\ & =1.39 \times 10^{-10}\end{aligned}$
(iv)
Lead chloride ( PbCl2 )
Ionization of Lead chloride
$
\mathrm{PbCl}_2 \rightleftharpoons \mathrm{~Pb}^{2+}+2 \mathrm{Cl}^{-}
$
Let " $s$ " be the solubility of $\mathrm{PbCl}_2$
$
\begin{aligned}
& {\left[\mathrm{Pb}^{2+}\right]=s} \\
& {\left[\mathrm{Cl}^{-}\right]=2 s}
\end{aligned}
$
According to the table $K_{s p}$ of $\mathrm{PbCl}_2=1.6 \times 10^{-5}$
$\begin{aligned} & \Rightarrow 1.6 \times 10^{-5}=s \cdot(2 s)^2 \\ & =1.6 \times 10^{-5}=4 s^3 \\ & s=\sqrt[3]{\frac{1.6 \times 10^{-5}}{4}} \\ & =1.58 \times 10^{-2}\end{aligned}$
So molarity of $\mathrm{Pb}^{2+}=1.58 \times 10^{-2} \mathrm{M}$ and molarity of $\mathrm{Cl}^{-}=3.16 \times 10^{-2} \mathrm{M}$
Answer :
silver chromate $\left(\mathrm{Ag}_2 \mathrm{CrO}_4\right)$
Ionization of silver chromate
$\mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_4^{2-}$
Let " s " be the solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$
$\begin{aligned} & {\left[\mathrm{Ag}^{+}\right]=2 s} \\ & {\left[\mathrm{CrO}_4^{2-}\right]=s} \\ & K_{s p} \text { of } \mathrm{Ag}_2 \mathrm{CrO}_4=1.1 \times 10^{-12} \\ & \Rightarrow 1.1 \times 10^{-12} \\ & =(2 s)^2 \cdot s=1.1 \times 10^{-12}=2 s^3 \\ & s=\sqrt[3]{\frac{1.1 \times 10^{-12}}{4}} \\ & =0.65 \times 10^{-4}\end{aligned}$
Ionization of Silver bromide ( AgBr )
$\begin{aligned} & A g B r \rightleftharpoons A g^{+}+B r^{-} \\ & K_{s p} \text { of } A g B r=5 \times 10^{-13} \\ & {\left[\mathrm{Ag}^{+}\right]=s^{\prime}} \\ & {\left[\mathrm{Br}^{-}\right]=s^{\prime}} \\ & \Rightarrow 5 \times 10^{-13}=s^{\prime} . s^{\prime} \\ & =5 \times 10^{-13}=s^{\prime 2} \\ & s^{\prime}=\sqrt{\frac{5 \times 10^{-13}}{1}}=\sqrt{0.5 \times 10^{-12}} \\ & =7.07 \times 10^{-7}\end{aligned}$
Now, the ratio of solubilities
$\begin{aligned} & \Rightarrow \frac{s}{s^{\prime}}=\frac{6.5 \times 10^{-5}}{1.1 \times 10^{-12}} \\ & =9.91\end{aligned}$
Answer :
We have,
solubility product $\left(K_{s p}\right)$ of cupric iodate $=7.4 \times 10^{-8}$
When equal volumes of sodium iodate and cupric chlorate are mixed together the molar concentration of both the solution becomes half (= 0.001)
Ionization of cupric iodate is;
$\underset{0.001 \mathrm{M}}{\mathrm{Cu}\left(\mathrm{IO}_3\right)_2} \rightleftharpoons \mathrm{Cu}^{2+}+\underset{0.001 \mathrm{M}}{2 \mathrm{IO}_3^{-}}$
So, Ksp can be calculated as;
$\begin{aligned} & K_{s p}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{IO}_3^{-}\right]^2 \\ & =(0.001)^3=10^{-9}\end{aligned}$
Sinc eionic product is less than the Ksp so no precipitation occurs.
Answer :
Suppose S is the molar solubility of silver benzoate in water, then
$C_6 \mathrm{H}_5 \mathrm{COOAg}_s \rightleftharpoons C_6 \mathrm{H}_5 \mathrm{COO}_{a q}^{-}+\mathrm{Ag}_{a q}^{+}$
so S
$=\sqrt{2.5 \times 10^{-13}}=5.0 \times 10^{-7} \mathrm{M}$
If the solubility of salt of weak acid of ionization constant Ka is S , then Ksp,Ka and S′ are related to each other at pH=3.19.
So$\left[H^{+}\right]=6.46 \times 10^{-4} M(\because \mathrm{pH}=3.19)$
$K_{s p}={S^{\prime}}^2\left[\frac{K_a}{K_a+\left[H^{+}\right]}\right]$
$S^{\prime}=\left\{\frac{2.5 \times 10^{-13}}{\left[\frac{6.46 \times 10^{-5}}{6.46 \times 10^{-5}+6.46 \times 10^{-4}}\right]}\right\}^{\frac{1}{2}}$
$S^{\prime}=\left\{\frac{2.5 \times 10^{-13} \times 7.106 \times 10^{-4}}{6.46 \times 10^{-5}}\right\}^{\frac{1}{2}}$
$=\left(2.75 \times 10^{-12}\right)^{\frac{1}{2}}$
$=1.658 \times 10^{-6} M$
So the ratio of
$\frac{S I}{S}=\frac{1.658 \times 10^{-6}}{5.0 \times 10^{-7}}=3.32$
Silver benzoate is 3.32 times more soluble in buffer of $p H=3.19$ than in pure water.
Answer :
We have,
The solubility product of the $F e S=6.3 \times 10^{-18}$
Equals number of moles of ferrous sulphate and sodium sulphide are mixed in an equal volume.
Let s be the concentration of ferrous sulphate and sodium sulphide. On mixing the equimolar solution, the volume of the concentration becomes half.
So, $\left[\mathrm{FeSO}_4\right]=\left[\mathrm{Na}_2 \mathrm{~S}\right]=\frac{s}{2} M$
The ionisation of ferrous sulphide;
$F e S \rightleftharpoons F e^{2+}+S^{2-}$
Therefore, for no precipitation, ionic product = solubility product
$K_{s p}=\left(\frac{s}{2}\right)\left(\frac{s}{2}\right) 6.3 \times 10^{-18}=\frac{s^2}{4}$
By solving the above equation, we get
$s=5.02 \times 10^{-9}$
The maximum concentration of both the solution is$s=5.02 \times 10^{-9}$M
Answer :
We have,
The solubility product of calcium sulphate is 9.1×10−6 .
given mass of calcium sulphate = 1g
Ionization of calcium sulphate;
$\mathrm{CaSO}_4 \rightleftharpoons \mathrm{Ca}^{2+}+\mathrm{SO}_4^{2-}$
Therefore,$\mathrm{K}_{s p}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]$
Let the solubility of calcium sulphate be s .
Then,$\begin{aligned} & K_{s p}=s^2 9.1 \times 10^{-6} \\ & =s^2 s=\sqrt{9.1 \times 10^{-6}} \\ & =3.02 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\end{aligned}$
Thus,
mass/ (mol. wt) × volume =3.02×10−3 Molarity
mass$=3.02 \times 136 \times 10^{-3}=0.41 \mathrm{~g}$
So, that to dissolve 1 g of calcium sulphate we need $=1 / 0.41 L=2.44 L$of water.
Answer :
We have,
the concentration of [S2−]=1×10−19 and the volume of the solution containing sulphur ion = 10 mL.
Volume of metal salts solution added = 5mL
Before mixing,
$\begin{aligned} & {\left[S^{2-}\right]=1 \times 10^{-19} \mathrm{M}} \\ & {\left[M^{2+}\right]=0.04 M}\end{aligned}$
After mixing,
Volume = 15 mL
So, the concentration of$\left[S^{2-}\right]=1 \times 10^{-19}$
$=6.67 \times 10^{-20} M$
$\begin{aligned} & \text { concentration of }\left[M^{2+}\right]=0.04 M \times \frac{5}{15} \\ & =1.33 \times 10^{-2}\end{aligned}$
$\begin{aligned} & \text { So, the ionic product }=\left[M^{2+}\right]\left[S^{2-}\right] \\ & =\left(6.67 \times 10^{-20}\right)\left(1.33 \times 10^{-2}\right) \\ & =8.87 \times 10^{-22}\end{aligned}$
For the precipitation of the solution, the ionic product should be greater than the corresponding Ksp values.
$\mathrm{K}_{s p}$ of FeS, MnS, ZnS, CdS are $6.3 \times 10^{-18}, 2.5 \times 10^{-13}, 1.6 \times 10^{-24}$ and $8 \times 10^{-27}$ respectively.
Hence precipitation will take place in CdCl2 and ZnCl2 metal salts.
To enhance conceptual understanding of Chapter 6 Equilibrium, practising HOTS questions is essential. These questions from NCERT Class 11 Chemistry Chapter 6 help you prepare effectively for competitive exams like NEET and JEE.
Question: Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows.$\mathrm{NH}_2 \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{CO}_{2(\mathrm{~g})}$ At equilibrium, ammonia is added such that partial pressures of NH3 now equals the original total pressure. Calculate the ratio of the total pressure now to the original total pressure.
1)$\frac{31}{27}$
2) $\frac{60}{40}$
3) $\frac{31}{9}$
4) $\frac{62}{27}$
Answer:
The reaction:-
$\mathrm{NH}_2 \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons \underset{2 \mathrm{P}}{2 \mathrm{NH}_{3(\mathrm{~g})}}+\underset{2 \mathrm{P}}{\mathrm{CO}_{2(\mathrm{~g})}}$
$\begin{gathered}\mathrm{K}_{\mathrm{p}}=\left(\mathrm{P}_{\mathrm{NH}_3}\right)^2\left(\mathrm{P}_{\mathrm{CO}_2}\right) \\ \mathrm{K}_{\mathrm{p}}=(2 \mathrm{P})^2(\mathrm{P}) \\ \mathrm{P}_{\mathrm{T}}(\text { initial })=3 \mathrm{P}\end{gathered}$
$\mathrm{NH}_2 \mathrm{COONH}_{4(\mathrm{~s})} \rightleftharpoons \underset{3 \mathrm{P}}{2 \mathrm{NH}_{3(\mathrm{~g})}}+\underset{\mathrm{P}^{\prime}}{\mathrm{CO}_{2(\mathrm{~g})}}$
$\mathrm{K}_{\mathrm{P}}=(3 \mathrm{P})^2\left(\mathrm{P}^{\prime}\right)$
From eq (I) and (ii)
$\begin{aligned}(2 \mathrm{P})^2(\mathrm{P}) & =(3 \mathrm{P})^2\left(\mathrm{P}^{\prime}\right) \\ \mathrm{P}^{\prime} & =\frac{4 \mathrm{P}}{9}\end{aligned}$
$\frac{\mathrm{P}_{\mathrm{T}}(\text { new })}{\mathrm{P}_{\mathrm{T}}(\text { old })}=\frac{3 \mathrm{P}+\mathrm{P}^{\prime}}{3 \mathrm{P}}=\frac{3 \mathrm{P}+\frac{4 \mathrm{P}}{9}}{3 \mathrm{P}}=\frac{31}{27}$
Hence, the correct answer is option (1).
Question: Consider the following chemical equilibrium of the gas phase reaction at a constant temperature : $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
If p being the total pressure, Kp is the pressure equilibrium constant and α is the degree of dissociation, then which of the following is true at equilibrium?
1) If p value is extremely high compared to Kp,α≈1
2) When p increases α decreases
3) If kp value is extremely high compared to p,α becomes much less than unity
4) When p increases α increases
Answer:
$\begin{array}{rlrll}\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}) & & & \\ \mathrm{t} & =0 & \mathrm{a} & 0 & 0 \\ \mathrm{t} & =\mathrm{t} & \mathrm{a}(1-\alpha) & \mathrm{a} \alpha & \mathrm{a} \alpha\end{array}$
a moles of A(g) taken initially and at time Now moles fraction of A(g),B(g) and C(g) are
$\begin{aligned} & X_A=\frac{a-a \alpha}{a+a \alpha}=\frac{1-\alpha}{1+\alpha} \\ & X_B=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha} \\ & X_C=\frac{a \alpha}{a+a \alpha}=\frac{\alpha}{1+\alpha}\end{aligned}$
Now if P is total pressure then partial pressure of A(g),B(g) and C(g) are
$\begin{gathered}\mathrm{P}_{\mathrm{A}}=\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P} \\ \mathrm{P}_{\mathrm{B}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\ \mathrm{P}_{\mathrm{C}}=\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P} \\ \mathrm{K}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}\left(\frac{\alpha}{1+\alpha}\right) \mathrm{P}}{\left(\frac{1-\alpha}{1+\alpha}\right) \mathrm{P}}\end{gathered}$
As KP is only function of temperature.
So as $\mathrm{P} \uparrow \alpha \downarrow$
Hence, the correct answer is option (2).
Question: Consider the equilibrium
$\mathrm{CO}(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g})$
If the pressure applied over the system increases by two fold at constant temperature then
(A) Concentration of reactants and products increases.
(B) Equilibrium will shift in forward direction.
(C) Equilibrium constant increases since concentration of products increases.
(D) Equilibrium constant remains unchanged as concentration of reactants and products remain same.
Choose the correct answer from the options given below :
(1) (A) and (B) only
(2) (A), (B) and (D) only
(3) (B) and (C) only
(4) (A), (B) and (C) only
Answer:
$$ \mathrm{CO}(g)+3 \mathrm{H}_2(g) \rightleftharpoons \mathrm{CH}_4(g)+\mathrm{H}_2 \mathrm{O}(g) $$
Statement 1. With the increase in pressure, at constant temperature, the concentration of both reactants and products will increase because the same amount of gas is now in a smaller volume.
Hence, statement 1 is correct
Statement 2. An increase in pressure causes product formation as the equilibrium shifts in the forward direction.
Hence, statement 2 is correct
Statement 3. The expression for equilibrium constant $\left(\mathrm{K}_c\right)$ for the given equilibrium can be written as
$K_c=\frac{\left[\mathrm{CH}_4\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_2\right]^3}$
Equilibrium constant is not changed with increase in pressure and increase in concentration of reactants or products.
Hence, Statement 3 is incorrect.
Statement 4. Equilibrium constant remains unchanged as concentration of reactants and products remain same.
Hence, Statement 4 is incorrect.
Hence, the correct answer is option (1).
A clear understanding of the right approach is essential before solving questions from NCERT Class 11 Chemistry Chapter 6 Equilibrium. Given below are simple and effective methods that will help you solve each question accurately and with a structured approach
1. The first step is to understand the key concepts
Try to focus on the basics of equilibrium - dynamic nature, reversible reactions, equilibrium constant (Kc \& Kp), Le Chatelier's Principle, etc. Learning the concepts will help you attempt the Equilibrium questions with ease.
2. Learn formulas thoroughly
Chapter 6 Equilibrium has a lot of numerical questions so try to memorize important formulas like
1. Equilibrium Constant
$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$
2. Relationship Between $K_p$ and $K_c$
$K_p=K_c(R T)^{\Delta n}$
3. Reaction Quotient
$Q=\frac{[C]^c[D]^d}{[A]^a[B]^b}$
4. pH and pOH
$\mathrm{pH}=-\log \left[H^{+}\right]$
$\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]$
$\mathrm{pH}+\mathrm{pOH}=14$
5. Ostwald’s Dilution Law
$K=\frac{C \alpha^2}{1-\alpha}$
3. Use ICE Tables (Initial, Change, Equilibrium)
For Class 11 Chemistry Equilibrium problems, you can create an ICE table to track concentrations and simplify calculations. This is required while solving the questions relating the change in equilibrium.
4. Logarithms and pH calculations
It is important to learn to calculate pH and pOH using $-\log \left[H^{+}\right]$and using Ka or Kb values for weak acids/bases.
5. Solve conceptual and numerical questions
Students can start practising with theory-based questions to develop a clear understanding of the fundamental concepts of Equilibrium Class 11 Chemistry then proceed to numerical problems to strengthen their problem solving skills. Students can refer to NCERT Exemplar as it offers a variety of both conceptual and numerical questions.
Topics covered in NCERT Chapter 6 Equilibrium is essential for learning the concepts of chemical and ionic balance.
6.1 Equilibrium in physical processes
6.2 Equilibrium in chemical processes
6.3 Law of chemical equilibrium and equilibrium constant
6.4 Homogeneous equilibrium
6.5 Heterogeneous equilibrium
6.6 Applications of equilibrium constants
6.7 Relationship between equilibrium constant K reaction quotient and Q and Gibbs energy G
6.8 Factors affecting equilibria
6.9 Ionic equilibrium in solutions
6.10 Acids, bases and salts
6.11 Ionisation of acids and bases
6.12 Buffer solutions
6.13 Solubility equilibria of sparingly soluble salts
NCERT Chapter 6 Equilibrium provides a strong base for JEE and NEET Chemistry. The table below will help you distinguish the topics based on the board and competitive exams.
Class 11 NCERT chapter-wise solutions are given below:
You can also access the NCERT solutions for other subjects as well. Click on the link below
Follow the links below to get the syllabus and the recommended books.
Frequently Asked Questions (FAQs)
Chapter 6 Equilibrium, are very helpful for annual exam preparation, as they provide clear, step-by-step explanations of textbook questions. These solutions strengthen conceptual understanding, improve problem-solving accuracy.
Equilibrium refers to a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction.
Temperature changes affect equilibrium depending on whether the reaction is exothermic or endothermic. If the temperature increases, the equilibrium will shift in the direction that absorbs heat, but if the temperature decreases, the equilibrium will shift towards the exothermic direction, releasing heat.
The reaction quotient is a measure of the relative concentrations of products and reactants at any given moment. By comparing Q to K, we can determine which direction the reaction must shift to reach Equilibrium.
A reversible reaction is a reaction that can proceed in both the forward and reverse directions. Reversible reactions are essential for establishing equilibrium. If a reaction is not reversible, it will proceed to completion, and there will be no equilibrium.
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