NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium

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.

Question 6.2 What is K c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂ ]= 0.60M, [O₂ ] = 0.82M and [SO₃ ] = 1.90M ?

$2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)\rightleftharpoons 2{\mathrm{SO}}_3(g)$

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;
$K_c=\frac{{\left[{\mathrm{SO}}_3\right]}^2}{{\left[{\mathrm{SO}}_2\right]}^2\left[{\mathrm{O}}_2\right]}$
$=\frac{[1.90{]}^2}{[0.60{]}^2[0.82]}=12.22M^{-1}($ approx $)$

Question 6.3 At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms

I(g)⇌2I(g)

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 $I_{(g)}\rightleftharpoons 2I_{(g)}$

$\begin{array}{rl}& K_p=\frac{{\left(p_I\right)}^2}{p_{l_2}}\\ & =\frac{{(4\times {10}^4)}^2}{6\times {10}^4}\mathrm{Pa}\\ & =2.67\times {10}^4\mathrm{Pa}\end{array}$

Question 6.4 Write the expression for the equilibrium constant, K_c, 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) ${\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5(aq)}+{\mathrm{H}}_2{\mathrm{O}}_{(l)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOH}}_{(aq)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{(aq)}$
(iv) ${\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5(aq)}+{\mathrm{H}}_2{\mathrm{O}}_{(l)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOH}}_{(aq)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{(aq)}$
(v) $I_{2(s)}+5F_2\rightleftharpoons 2I{F}_5$

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 K_c for each of the following equilibria from the value of K_p:

(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;

$K_p=K_c(RT{)}^{\mathrm{\Delta }n}$

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, K_c = 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:

$\begin{array}{rl}& K_c^{\prime }=\frac{1}{K_c}\\ & =\frac{1}{6.3\times {10}^{14}}\\ & =1.58\times {10}^{-15}\end{array}$

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 . $$\begin{array}{lcc}& 2{\mathrm{N}}_{2(g)}+{\mathrm{O}}_{2(g)}\rightleftharpoons 2{\mathrm{N}}_2{\mathrm{O}}_{(g)}& \\ \text{ initial conc }& 0.482& 0.933& 0& \text{ (in moles) }\\ \text{ at equilibrium }& 0.482-x& 0.933-x& x& \text{ (in moles) }\end{array}$$

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 K_c at this temperature?

Answer :

We have,
$K_p=2\times {10}^{10}/$ bar
We know that the relation between Kp and Kc ;

$K_p=K_c(RT{)}^{\mathrm{\Delta }n}$
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}}$

$\begin{array}{rl}& =74.79\times {10}^{10}{\mathrm{Lmol}}^{-1}\\ & =7.48\times {10}^{10}{\mathrm{Lmol}}^{-1}\end{array}$

Question 6.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

$2H{I}_{(g)}\rightleftharpoons H_{2(g)}+I_{2(g)}$

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:

$\begin{array}{c}2H{I}_{(g)}\rightleftharpoons H_{2(g)}+I_{2(g)}\\ H{I}_{(g)}\rightleftharpoons 1/2H_{2(g)}+1/2I_{2(g)}\end{array}$

At equilibrium,

$\begin{array}{rl}& \text{ pHI }=0.04p{H}_2=\frac{0.16}{2}\\ & =0.08p{I}_2=\frac{0.16}{2}=0.08\end{array}$

Therefore,

$\begin{array}{rl}& K_p=\frac{p{H}_2\times p{I}_2}{p^{2}HI}\\ & =\frac{0.08\times 0.08}{(0.04{)}^2}=4\end{array}$

Question 6.12 A mixture of 1.57 mol of N₂ , 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N_{2( g)}+3H_{2( g)}⇌2NH_{3( g)} is 1.7 × 10². Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer :

We have,${\mathrm{N}}_{2(g)}+3{\mathrm{H}}_{2(g)}\rightleftharpoons 2{\mathrm{NH}}_{3(g)}$

K_c=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

$\begin{array}{rl}& Q_c=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}\\ & =\frac{(8.13/20{)}^2}{(1.57/20)(1.92/20{)}^3}=2.4\times {10}^{-3}\end{array}$
The reaction is not in equilibrium. Since $Qc>K_c$, the equilibrium proceeds in reverse direction.

Question 6.14 One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,

${\mathrm{H}}_2{\mathrm{O}}_{(g)}+{\mathrm{CO}}_{(g)}\rightleftharpoons {\mathrm{H}}_{2(g)}+{\mathrm{CO}}_{2(g)}$

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;

$\begin{array}{rl}& K_C=\frac{\left[{\mathrm{H}}_2\right]\left[{\mathrm{CO}}_2\right]}{\left[{\mathrm{H}}_2\mathrm{O}\right][\mathrm{CO}]}\\ & =\frac{.04\times .04}{(.06{)}^2}\end{array}$

= 0.44 (approx)

Question 6.15 At 700 K, equilibrium constant for the reaction:

$H_{2(g)}+I_{2(g)}\rightleftharpoons 2H{I}_{(g)}$

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-

$H_{2(g)}+I_{2(g)}\rightleftharpoons 2H{I}_{(g)}$

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 $)\mathrm{mol}/L$

Question 6.16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?

$2IC{l}_{(g)}\rightleftharpoons I_{2(g)}+C{l}_{2(g)};K_c=0.14$

Answer :

The given reaction is:$$\begin{array}{lc}& 2\mathrm{ICl}(g)\rightleftharpoons I_2(g)+C{l}_2(g)\\ \text{ Initial conc. }& 0.78\mathrm{M}00\\ \text{ At equilibrium }& (0.78-2x)\mathrm{M}x\mathrm{M}x\mathrm{M}\end{array}$$

The value of Kc = 0.14

Now we can write,$\begin{array}{rl}& K_c=\frac{\left[I_2\right]\left[{\mathrm{Cl}}_2\right]}{[\mathrm{ICl}{]}^2}\\ & 0.14=\frac{x^2}{(.78-x{)}^2}0.374\end{array}$

$=\frac{x}{(.78-x)}$

By solving this we can get the value of $x=0.167$

Question 6.17 Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

$C_2{H}_{6(g)}\rightleftharpoons C_2{H}_{4(g)}+H_{2(g)}$

Answer :

Suppose the pressure exerted by the hydrogen and ethene gas be p at equilibrium.

the given reaction is-

$\begin{array}{cllll}{\mathrm{C}}_2{\mathrm{H}}_{6(\mathrm{g})}\rightleftharpoons {\mathrm{C}}_2{\mathrm{H}}_{4(\mathrm{g})}+{\mathrm{H}}_{2(\mathrm{g})}& & & & \\ \text{ initial pressure }& 4atm& 0& 0\\ \text{ At equilibrium }& 4-p& p& p\end{array}$

Now,

$\begin{array}{rl}& K_p=p_{C_2{H}_4}\times p_{H_2}/p_{C_2{H}_6}\\ & 0.04=\frac{p^2}{4-p}\end{array}$

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.

$\begin{array}{c}{\mathrm{CH}}_3{\mathrm{COOH}}_{(l)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{(l)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOO}}_2{\mathrm{H}}_{5(l)}+{\mathrm{H}}_2{\mathrm{O}}_{(l)}\\ \text{ initial conc. }& 1/\mathrm{V}& 0.18/\mathrm{V}& 0\\ \text{ At equilibrium }& \frac{1-.171}{V}(=0.829/\mathrm{V})\frac{1-.171}{V}& 0.171& 0.171\end{array}$

So equilibrium constant for the reaction can be calculated as;

$\begin{array}{c}{K}_c=\frac{(0.171{)}^2}{(0.829)(0.009)}\\ =3.92\text{ (approx) }\end{array}$

Question 6.18(iii) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

$$\begin{array}{lc}& {\mathrm{CH}}_3{\mathrm{COOH}}_{(l)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{(l)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5(l)}+{\mathrm{H}}_2{\mathrm{O}}_{(l)}\\ \text{ initial conc. }& 1/\mathrm{V}& 0.5/\mathrm{V}& 0\\ \text{ At equilibrium }& \frac{1-.214}{V}(=0.786/\mathrm{V})\frac{1-.214}{V}& 0.214& 0.214\end{array}$$

(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-

$\begin{array}{rl}& Q_c=\frac{(0.214{)}^2}{(0.786)(0.286)}\\ & =0.2037\text{ (approx) }\end{array}$

Since $Q_c<K_c$ equilibrium has not been reached.

Question 6.19 A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^-1 mol L ^-1 . If value of K_c is 8.3 × 10 ^-3 , what are the concentrations of PCl₃ and Cl₂ at equilibrium?

$PC{l}_{5(\mathrm{g})}\rightleftharpoons PC{l}_{3(\mathrm{g})}+C{l}_{2(\mathrm{g})}$

Answer :

We have,

concentration of$PC{l}_5=0.05\mathrm{mol}/\mathrm{L}$
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:

${\mathrm{PCl}}_5(g)\rightleftharpoons {\mathrm{PCl}}_3(g)+{\mathrm{Cl}}_2(g)$

$\begin{array}{llll}\text{ at equilibrium }& 0.05& x& x\end{array}$

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:

$\begin{array}{rl}& K_c=\frac{x^2}{0.05}=8.3\times {10}^{-3}\\ & \Rightarrow x=\sqrt{4.15\times {10}^{-4}}\\ & =0.0204\text{ (approx) }\end{array}$

Hence the concentration of${\mathrm{PCl}}_3$ and ${\mathrm{Cl}}_2$ is $0.0204\mathrm{mol}/\mathrm{L}$

Question 6.20 One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO₂ . $Fe{O}_{(s)}+C{O}_{(g)}\rightleftharpoons F{e}_{(s)}+C{O}_{2(g)}$; Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and ${\mathrm{CO}}_2$ at 1050 K if the initial partial pressures are: pCO = 1.4 atm and $p{\mathrm{CO}}_2$ = 0.80 atm?

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

$Q_p=\frac{pC{O}_2}{pCO}=\frac{0.80}{1.4}=0.571$
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

$\therefore K_p=\frac{pC{O}_2}{pCO}$

$\begin{array}{rl}& =\frac{0.80-p}{1.4+p}=0.265\\ & =0.371+.265p=0.80-p\end{array}$

By solving the above equation we get the value of p = 0.339 atm

Question 6.21 Equilibrium constant, K_c 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 N_{2 ,} 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:

$${\mathrm{N}}_{2(g)}+3{\mathrm{H}}_{2(g)}\rightleftharpoons 2{\mathrm{NH}}_{3(g)}$$ $$3.0{\mathrm{molL}}^{-1}2.0{\mathrm{molL}}^{-1}0.5{\mathrm{molL}}^{-1}$$

Now, we know that,

$\begin{array}{rl}& Qc=\left[{\mathrm{NH}}_3\right]2^{/}\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3\\ & =(0.5{)}^2/(3.0)(2.0{)}^3=0.0104\end{array}$

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 K_c = 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:

$$\begin{array}{rl}& 2{\mathrm{BrCl}}_{(g)}\rightleftharpoons {\mathrm{Br}}_{2(g)}+{\mathrm{Cl}}_{2(g)}\\ & \text{ Initial Conc. }3.3\times {10}^{-3}00\\ & \text{ at equilibrium }3.3\times {10}^{-3.2x}xx\end{array}$$

Now, we can write,

$\begin{array}{rl}& K_c=\frac{\left[{\mathrm{BR}}_2\right]\left[{\mathrm{Cl}}_2\right]}{{\left[{\mathrm{BrCl}}_2\right]}^2}\\ & 32=\frac{x^2}{{(3.3\times {10}^{-3}-2x)}^2}\\ & 5.66=\frac{x}{(3.3\times {10}^{-3}-2x)}\\ & x+11.32x=18.678\times {10}^{-3}\end{array}$

By solving the above equation we get,
$x=1.51\times {10}^{-3}$

Hence, at equilibrium$\left[{\mathrm{BrCl}}_2\right]=3.3\times {10}^{-3}-(2\times 1.51\times {10}^{-3})$
=0.3×10−3 M

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

$\begin{array}{rl}& {\mathrm{\Delta }}_f{G}^{+}\left[{\mathrm{NO}}_2\right]=52.0\mathrm{kJ}/\mathrm{mol}\\ & {\mathrm{\Delta }}_f{G}^{+}[\mathrm{NO}]=87.0\mathrm{kJ}/\mathrm{mol}\\ & {\mathrm{\Delta }}_f{G}^{+}\left[{\mathrm{O}}_2\right]=0\mathrm{kJ}/\mathrm{mol}\end{array}$

Answer :

Given data,

$\begin{array}{rl}& {\mathrm{\Delta }}_f{G}^{+}\left[{\mathrm{NO}}_2\right]=52.0\mathrm{kJ}/\mathrm{mol}\\ & {\mathrm{\Delta }}_f{G}^{+}[\mathrm{NO}]=87.0\mathrm{kJ}/\mathrm{mol}\\ & {\mathrm{\Delta }}_f{G}^{+}\left[{\mathrm{O}}_2\right]=0\mathrm{kJ}/\mathrm{mol}\end{array}$

Given chemical reaction

-$\mathrm{NO}(g)+\frac{1}{2}{\mathrm{O}}_2(g)\rightleftharpoons {\mathrm{NO}}_2(g)$

for the reaction,$\begin{array}{rl}& \mathrm{\Delta }{G}^0=\mathrm{\Delta }{G}^0\text{ (products) }-\mathrm{\Delta }{G}^0\text{ (reactants) }\\ & =(52-87-0)\\ & =-35\mathrm{kJ}/\mathrm{mol}\end{array}$

Question 6.25 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(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,

• (a)The number of moles of reaction products will increase.
• (b)The number of moles of reaction products will decrease
• (c)The number of moles of reaction products remains the same.

Question 6.26 Which of the following reactions will get affected by increasing the pressure? Also, mention whether the change will cause the reaction to go into forward or backward direction.

(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

H_2(g)+Br_2(g)⇌2HBr_(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

Answer :

Given that,

$\begin{array}{rl}& K_p\text{ for the reaction }=1.6\times {10}^5\\ & K_p^{\prime }=1/K_p=\frac{1}{1.6\times {10}^5}=6.25\times {10}^{-6}\end{array}$

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,
$K_p^{\prime }=\frac{p^{B{r}_2}\cdot p^{H_2}}{p_{HBr}^2}$
$\begin{array}{rl}& 6.25\times {10}^{-6}=\frac{p^2}{(10-2p{)}^2}\\ & 5\times {10}^{-3}=\frac{p}{(5-p)}\end{array}$
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

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

Question 6.30 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl₅ is 8.3 ×10 ^-3 . If decomposition is depicted as,

${\mathrm{PCl}}_5(g)\rightleftharpoons {\mathrm{PCl}}_3(g)+{\mathrm{Cl}}_2(g),{\mathrm{\Delta }}_r{\mathrm{H}}^{+}=124.0{\mathrm{kJmol}}^{-1}$

(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,
${\mathrm{PCl}}_5(g)\rightleftharpoons {\mathrm{PCl}}_3(g)+{\mathrm{Cl}}_2(g),{\mathrm{\Delta }}_r{\mathrm{H}}^{+}=124.0{\mathrm{kJmol}}^{-1}$
Equilibrium constant for the above reaction = 8.3×10⁻³

(a) Expression of Kc for this reaction-$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{PC{l}_5}$

(b) The value of the reverse equilibrium constant can be calculated as;

$\begin{array}{rl}& K_c^{\prime }=\frac{1}{K_c}\\ & =\frac{1}{8.3\times {10}^{-3}}=1.20\times {10}^2\end{array}$

(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.

Question 6.31 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂ . In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

$\mathrm{CO}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})\rightleftharpoons {\mathrm{CO}}_2+{\mathrm{H}}_2(\mathrm{g})$

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-
$$\begin{array}{lcc}& \mathrm{CO}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})& \rightleftharpoons {\mathrm{CO}}_2+{\mathrm{H}}_2(\mathrm{g})\\ \text{ Initial concentration }& 4\text{ bar }& 4\text{ bar }& 0& 0\\ \text{ At equilibrium }& 4-p& 4-p& p& p\end{array}$$

Therefore, we can write,$\begin{array}{rl}& 10.1=\frac{p_{{\mathrm{CO}}_2}\cdot p_{{\mathrm{H}}_2}}{p_{\mathrm{CO}}\cdot p_{{\mathrm{H}}_2\mathrm{O}}}\\ & 10.1=\frac{p^2}{(4-p{)}^2}3.178\\ & =\frac{p}{(4-p)}p=(4-p)\times 3.178\end{array}$
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) (K_c=1.8) will have an appreciable concentration of reactants and products.

Question 6.33 The value of K c for the reaction 3O_2(g)⇌2O_3(g) is 2.0 ×10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 ×10^-2 , what is the concentration of O₂ ?

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-

$3O_2(g)\rightleftharpoons 2O_3(g)$
Then we have,
(equilibrium constant)
$K_c={[O_3(g)]}^2/{[O_2(g)]}^3$
$\begin{array}{rl}& 2\times {10}^{-50}={\left[{\mathrm{O}}_3\right]}^2/{(1.6\times {10}^{-2})}^3{\left[{\mathrm{O}}_3\right]}^2\\ & =2\times {10}^{-50}\times {(1.6\times {10}^{-2})}^3\\ & =8.192\times {10}^{-56}\\ & \left[{\mathrm{O}}_3\right]=\sqrt{8.192\times {10}^{-56}}=2.86\times {10}^{-28}\end{array}$

Thus the concentration of dioxygen is$2.86\times {10}^{-28}$

Question 6.34(a) The reaction,

$\mathrm{CO}(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})\rightleftharpoons {\mathrm{CH}}_4(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$

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, $0.3\mathrm{mol}/\mathrm{L}0.1\mathrm{mol}/\mathrm{L}x0.02\mathrm{mol}/\mathrm{L}$