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: [SO 2 ]= 0.60M, [O 2 ] = 0.82M and [SO 3 ] = 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-
$$\begin{gathered} [S{O}_2]=0.60M \\ [O_2]=0.82M \\ [S{O}_3]=1.90M \end{gathered}$$
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;
$K_c=\frac{[S{O}_3{]}^2}{[S{O}_2{]}^2[O_2]}$

$$\begin{gathered} =\frac{[1.90{]}^2}{[0.60{]}^2[0.82]} \\ =12.22M^{-1} \end{gathered}$$ (approx)

Question 6.3 At a certain temperature and total pressure of 10 5 Pa, iodine vapour contains 40% by volume of I atoms

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

Calculate Kp for the equilibrium.

Answer :

It is given that total pressure (P T )is ${10}^{5}Pa$ and partial pressure of $I$ atom = 40 % of P T
So, the partial pressure of $I$ atom = $\frac{40}{100}\times {10}^5=4\times {10}^4$ Pascal

The partial pressure of $I_2$ = 60% of P T

So, the partial pressure of $I_2$ = $\frac{60}{100}\times {10}^5=6\times {10}^4$ Pascal

Now, for the reaction $I_{(g)}\rightleftharpoons 2I_{(g)}$
$K_p=\frac{(p_I{)}^2}{p_{I_2}}=\frac{(4\times {10}^4{)}^2}{6\times {10}^4}Pa$
$=2.67\times {10}^4\mathrm{Pa}$

Question 6.4 Write the expression for the equilibrium constant, K c for each of the following reactions:

(i) $2{\mathrm{NOCl}}_{(g)}\rightleftharpoons 2{\mathrm{NO}}_{(g)}+{\mathrm{Cl}2}_{(g)}$

(ii) $2\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_{2(\mathrm{s})}\rightleftharpoons 2\mathrm{CuO}{\mathrm{S}}_{(\mathrm{s})}+4{\mathrm{NO}}_{2(\mathrm{g})}+{\mathrm{O}}_{2(\mathrm{g})}$

(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)
$K_c=\frac{[NO{]}^2\left[C{l}_2\right]}{[NOCl{]}^2}$

(ii)
$K_c=\frac{[CuO{]}^2[N{O}_2{]}^4[O_2]}{[Cu(N{O}_3{)}_2{]}^2}$
(iii)

$K_c=\frac{[C{H}_{3}COOH][C_2{H}_{5}OH]}{[C{H}_{3}COO{C}_{2}H5]}$

(iv)
$K_c=\frac{1}{[F{e}^{3+}][O{H}^{-}]}$

(v)
$K_c=\frac{[I{F}_5{]}^2}{[F_2{]}^5}$

Question 6.5 Find out the value of K c for each of the following equilibria from the value of K p :

(i) $2{\mathrm{NOCl}}_{(g)}\rightleftharpoons 2{\mathrm{NO}}_{(g)}+{\mathrm{Cl}}_{2(g)};K_p=1.8\times {10}^{-2}at500\mathrm{K}$

(ii)${\mathrm{CaCO}}_{3(s)}\rightleftharpoons {\mathrm{CaO}}_{(s)}+{\mathrm{CO}}_{2(g)};{\mathrm{K}}_p=167at1073\mathrm{K}$

Answer :

We know that the relation between $K_p$ and $K_c$ is expressed as;

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

$\mathrm{\Delta }n$ = 3 - 2 = 1
By putting the all values in eq (i) we get

$K_c=\frac{1.8\times {10}^{-2}}{0.0831\times 500}=4.33\times {10}^{-4}$

For (ii)
$K_p$ = $167$ and temp(T) = 1073 K
$\mathrm{\Delta }n$ = 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 × 10 14 at 1000 K

$N{O}_{(g)}+O_{3(g)}\rightleftharpoons N{O}_{2(g)}+O_{2(g)}$

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K c , for the reverse reaction?

Answer :

It is given that,
$K_c=6.3\times {10}^{14}$
we know that $K_c^{\prime }$ for the reverse reaction is the inverse of the forward equilibrium constant. Thus it can be calculated as:

$K_c^{\prime }=\frac{1}{K_c}$
$=\frac{1}{6.3\times {10}^{14}}$
$=1.58\times {10}^{-15}$

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 N 2 and O 2- takes place as follows:

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

If a mixture of 0.482 mol N 2 and 0.933 mol of O 2 is placed in a 10 L reaction vessel and allowed to form N 2 O at a temperature for which K c = 2.0 ×10 -37 , determine the composition of equilibrium mixture.

Answer :

It is given that,
$K_c=2.0\times {10}^{-37}$
Let the concentration of $N_{2}O$ at equilibrium be $x$ . So,
$2{\mathrm{N}}_{2(g)}+{\mathrm{O}}_{2(\mathrm{g})}\rightleftharpoons 2{\mathrm{N}}_2{\mathrm{O}}_{(\mathrm{g})}$

initial conc 0.482 0.933 0 (in moles)
at equilibrium 0.482- $x$ 0.933 - $x$ $x$ (in moles)

The equilibrium constant is very small. So, we can assume 0.482- $x$ = 0.482 and 0.933 - $x$ = 0.933

We know that,

$K_c=\frac{[N_{2}O{]}^2}{[N_2{]}^2[O_2]}$
$2\times {10}^{-37}=\frac{(x/10{)}^2}{(0.0482{)}^2(0.0933)}$ {dividing the moles by 10 to get concentration of ions)

$$\begin{gathered} \frac{x^2}{100}=2\times {10}^{-37}\times (0.0482{)}^2\times (0.0933) \\ {x}^2=43.35\times {10}^{-40} \\ x=\sqrt{43.35\times {10}^{-40}} \\ x=6.6\times {10}^{-20} \end{gathered}$$

So, the concentration of $[N_{2}O]=\frac{x}{10}=6.6\times {10}^{-20}$

Question 6.9 Nitric oxide reacts with Br 2 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 Br 2 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 Br 2 .

Answer :

The initial concentration of $NO$ and $B{r}_2$ is 0.087 mol and 0.0437 mol resp.

The given chemical reaction is-
$2N{O}_{(g)}+{\mathrm{Br}}_{2(g)}\rightleftharpoons 2{\mathrm{NOBr}}_{(g)}$

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 $B{r}_2$ two mol of $NOBr$ produced. So, to produce 0.518 mol of $NOBr$ we need $\frac{0.518}{2}=0.0259$ mol of $B{r}_2$ .

Thus, the amount of $NO$ present at equilibrium = 0.087 - 0.0518 = 0.0352 mol

and the amount of $B{r}_2$ present at the equilibrium = 0.0437-0.0259 = 0.0178 mol

Question 6.10 At 450K, K p = 2.0 × 10 10 /bar for the given reaction at equilibrium.

$2S{O}_{2(g)}+O_{2(g)}\rightleftharpoons 2S{O}_{3(g)}$

What is K c at this temperature?

Answer :

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

$K_p=K_c(RT{)}^{\mathrm{\Delta }n}$
Here $\mathrm{\Delta }n$ = ( moles of product) - (moles of reactants)

$2{\mathrm{SO}}_{2(g)}+O_{2(g)}\rightleftharpoons 2{\mathrm{SO}}_{3(g)}$

So. here $\mathrm{\Delta }n$ = 2-3 = -1

By applying the formula we get;

$2\times {10}^{10}=K_c(0.0831Lbar/K/mol)\times 450K$
$K_c=\frac{2\times {10}^{10}}{(0.0831Lbar/K/mol)\times 450K}$

$=74.79\times {10}^{10}Lmo{l}^{-1}$
= 7.48 $\times {10}^{10}Lmo{l}^{-1}$

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 K p 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: $\begin{array}{rl}& 2H{I}_{(g)}\rightleftharpoons H_{2(g)}+I_{2(g)}\\ & H{I}_{(g)}\rightleftharpoons 1/2H_{2(g)}+1/2I_{2(g)}\end{array}$

At equilibrium,
$$\begin{gathered} pHI=0.04 \\ p{H}_2=\frac{0.16}{2}=0.08 \\ p{I}_2=\frac{0.16}{2}=0.08 \end{gathered}$$

Therefore,
$K_p=\frac{p{H}_2\times p{I}_2}{p^{2}HI}$
$$\begin{gathered} =\frac{0.08\times 0.08}{(0.04{)}^2} \\ =4 \end{gathered}$$

Question 6.12 A mixture of 1.57 mol of N 2 , 1.92 mol of H 2 and 8.13 mol of NH 3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K c for the reaction ${\mathrm{N}}_{2(\mathrm{g})}+3{\mathrm{H}}_{2(\mathrm{g})}\rightleftharpoons 2{\mathrm{NH}}_{3(\mathrm{g})}$ is 1.7 × 10 2 . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer :

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

$K_c=1.7\times {10}^2$

The concentration of species are-
$$\begin{gathered} [N_2]=1.57/20mol{L}^{-1} \\ [H_2]=1.92/20mol{L}^{-1} \\ [N{H}_3]=8.13/20mol{L}^{-1} \end{gathered}$$

We know the formula of

$Q_c=\frac{[N{H}_3{]}^2}{[N_2][H_2{]}^3}$
$$\begin{gathered} =\frac{(8.13/20{)}^2}{(1.57/20)(1.92/20{)}^3} \\ =2.4\times {10}^{-3} \end{gathered}$$
The reaction is not in equilibrium. Since $Qc>K_c$ , the equilibrium proceeds in reverse direction.

Question 6.13 The equilibrium constant expression for a gas reaction is,

$K_c=\frac{{\left[N{H}_3\right]}^4{\left[O_2\right]}^5}{{\left[NO\right]}^4{\left[H_{2}O\right]}^5}$

Write the balanced chemical equation corresponding to this expression.

Answer :

The balanced chemical equation corresponding to the given expression can be written as:

$4NO+6H_{2}O\rightleftharpoons 4N{H}_3+5O_2$

Question 6.14 One mole of H 2 O 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-

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

initial conc 1/10 1/10 0 0
At equilibrium 0.6/10 0.6/10 0.04 0.04

Now, the equilibrium constant for the reaction can be calculated as;

$K_c=\frac{[H_2][C{O}_2]}{[H_{2}O][CO]}$
$=\frac{.04\times .04}{(.06{)}^2}$

= 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 H 2 (g) and I 2 (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 $x$

$[I_2]=[H_2]=x$

Therefore,

$K_c^{\prime }=\frac{[H_2][I_2]}{[HI]}=\frac{x^2}{(.5{)}^2}$
So, the value of $x$ = $\sqrt{\frac{0.25}{54.8}}=0.0675(approx)mol/L$

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:
$2ICl(g)\rightleftharpoons I_2(g)+C{l}_2(g)$
Initial conc. 0.78 M 0 0
At equilibrium (0.78 - 2 $x$ ) M $x$ M $x$ M

The value of $K_c=0.14$

Now we can write,
$K_c=\frac{[I_2][C{l}_2]}{[ICl{]}^2}$
$$\begin{gathered} 0.14=\frac{x^2}{(.78-x{)}^2} \\ 0.374=\frac{x}{(.78-x)} \end{gathered}$$

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

Question 6.17 K p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C 2 H 6 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-

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

initial pressure 4 atm 0 0
At equilibrium 4 - $p$ $p$ $p$

Now,
$K_p=p_{C_2{H}_4}\times p_{H_2}/p_{C_2{H}_6}$
$0.04=\frac{p^2}{4-p}$

By solving the quadratic equation we can get the value of $p$ = 0.38

Hence,at equilibrium,

$p_{C_2{H}_6}$ = 4 - 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), Q c , 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-

$Q_c=\frac{[C{H}_{3}COO{C}_2{H}_5][H_{2}O]}{[C{H}_{3}COOH][C_2{H}_{5}OH]}$

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.${\mathrm{CH}}_3{\mathrm{COOH}}_{(l)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{(l)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5(l)}+{\mathrm{H}}_2{\mathrm{O}}_{(l)}$

initial conc. 1/V 0.18/V 0 0

At equilibrium $\frac{1-.171}{V}$ (= 0.829/V) $\frac{1-.171}{V}$ $0.171$ $0.171$

So the equilibrium constant for the reaction can be calculated as;

$K_c=\frac{(0.171{)}^2}{(0.829)(0.009)}$

$=3.92$ (approx)

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

${\mathrm{CH}}_3{\mathrm{COOH}}_{(l)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{(l)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5(l)}+{\mathrm{H}}_2{\mathrm{O}}_{(l)}$

(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.${\mathrm{CH}}_3{\mathrm{COOH}}_{(l)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{(l)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5(l)}+{\mathrm{H}}_2{\mathrm{O}}_{(l)}$

initial conc. 1/V 0.5/V 0 0

At equilibrium $\frac{1-.214}{V}$ (= 0.786/V) $\frac{1-.214}{V}$ $0.214$ $0.214$

Therefore the reaction quotient of the reaction-

$Q_c=\frac{(0.214{)}^2}{(0.786)(0.286)}$

$=0.2037(approx)$

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

Question 6.19 A sample of pure PCl 5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl 5 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 PCl 3 and Cl 2 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 mol/L
and $K_c=8.3\times {10}^{-3}$

Suppose the concentrations of both $PC{l}_3$ and $C{l}_2$ at equilibrium be $x$ mol/L. The given reaction is:

$PC{l}_5(g)\rightleftharpoons PC{l}_3(g)+C{l}_2(g)$

at equilibrium 0.05 $x$ $x$

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:

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

Hence the concentration of $PC{l}_3$ and $C{l}_2$ is 0.0204 mol / 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 CO 2 . ${\mathrm{FeO}}_{(s)}+{\mathrm{CO}}_{(\mathrm{g})}\rightleftharpoons {\mathrm{Fe}}_{(\mathrm{s})}+{\mathrm{CO}}_{2(\mathrm{g}}$ ; K p = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO 2 at 1050 K if the initial partial pressures are: $p_{CO}$ = 1.4 atm and $pC{O}_2$ = 0.80 atm?

Answer :

We have,

$K_p=0.265$
the initial pressure of $CO$ and $C{O}_2$ 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 $Q_p>K_c$ the reaction will proceed in the backward direction to attain equilibrium. The partial pressure of $C{O}_2$ will increase = decrease in the partial pressure of $C{O}_2$ = $p$

$\therefore K_p=\frac{pC{O}_2}{pCO}=\frac{0.80-p}{1.4+p}=0.265$
$=0.371+.265p=0.80-p$

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 H 2 and 0.5 mol L -1 NH 3 . Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Answer :

The given reaction is:

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

at a particular time: 3.0molL -1 2.0 molL -1 0.5molL -1

Now, we know that,

$Qc=[N{H}_3]2^{/}[N_2][H_2{]}^3$
$$\begin{gathered} =(0.5{)}^2/(3.0)(2.0{)}^3 \\ =0.0104 \end{gathered}$$

It is given that K c = 0.061

Since, Qc $\ne$ Kc, the reaction mixture is not at equilibrium.

Again, $Q_c<K_c$ , 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:

$2BrC{l}_{(g)}\rightleftharpoons B{r}_{2(g)}+C{l}_{2(g)}$

Initial Conc. $3.3\times {10}^{-3}$ 0 0

at equilibrium $3.3\times {10}^{-3}$ - 2 $x$ $x$ $x$

Now, we can write,

$K_c=\frac{[B{R}_2][C{l}_2]}{[BrC{l}_2{]}^2}$
$32=\frac{x^2}{(3.3\times {10}^{-3}-2x{)}^2}$
$5.66=\frac{x}{(3.3\times {10}^{-3}-2x)}$
$x+11.32x=18.678\times {10}^{-3}$

By solving the above equation we get,

$x$ = $1.51\times {10}^{-3}$

Hence, at equilibrium $[BrC{l}_2]=3.3\times {10}^{-3}-(2\times 1.51\times {10}^{-3})$
$=0.3\times {10}^{-3}$ M

Question 6.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO 2 in equilibrium with soild carbon has 90.55% CO by mass

$C_{(s)}+{\mathrm{CO}}_{2(g)}\rightleftharpoons 2{\mathrm{CO}}_{(g)}$

Calculate K c 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 $C{O}_2$ = (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 $C{O}_2$ = 9.45/44 = 0.215 mol (mol. wt of $C{O}_2$ = 44)

Partial pressure of $CO$ ,

$pCO=\frac{n_{CO}}{n_{CO}+n_{C{O}_2}}{p}_T$

$=\frac{3.234}{3.234+0.215}\times 1$
= 0.938 atm

Similarly partial pressure of $C{O}_2$ ,

$pC{O}_2=\frac{n_{C{O}_2}}{n_{CO}+n_{C{O}_2}}{p}_T$
= 0.062 atm

Thus, $K_p=\frac{(0.938{)}^2}{(0.062)}\approx 14.19$

By using the relation $K_p=K_c(RT{)}^{\mathrm{\Delta }n}$
$K_c=\frac{14.19}{(0.083\times 1127{)}^1}$
= 0.159 (approx)

Question 6.24 Calculate a) $\mathrm{\Delta }{G}^0$ for the formation of NO2 from NO and O2 at 298K

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

where

${\mathrm{\Delta }}_f{G}^{+}\left[N{O}_2\right]=52.0kJ/mol$

${\mathrm{\Delta }}_f{G}^{+}\left[NO\right]=87.0kJ/mol$

${\mathrm{\Delta }}_f{G}^{+}\left[O_2\right]=0kJ/mol$

Answer :

Given data,

${\mathrm{\Delta }}_f{G}^{+}\left[N{O}_2\right]=52.0kJ/mol$
${\mathrm{\Delta }}_f{G}^{+}\left[NO\right]=87.0kJ/mol$

${\mathrm{\Delta }}_f{G}^{+}\left[O_2\right]=0kJ/mol$

given chemical reaction-
$NO(g)+\frac{1}{2}{O}_2(g)\rightleftharpoons N{O}_2(g)$

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

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

(b) $CaO(s)+C{O}_2(s)\rightleftharpoons CaC{O}_3(s)$

(c) $3Fe(s)+4H_{2}O(g)\rightleftharpoons F{e}_3{O}_4(s)+H_2(g)$

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) $COC{l}_2(g)\rightleftharpoons CO(g)+C{l}_2(g)$

(ii) $C{H}_4(g)+2S_2(g)\rightleftharpoons C{S}_2(g)+2H_{2}S(g)$

(iii) $C{O}_2(g)+C(s)\rightleftharpoons 2CO(g)$

(iv) $2H_2(g)+CO(g)\rightleftharpoons C{H}_{3}OH(g)$

(v) $CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2(g)$

(vi) $4N{H}_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_{2}O(g)$

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 ×10 5 at 1024K

$H_2(g)+B{r}_2(g)\rightleftharpoons 2HBr(g)$

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

Answer :

Given that,
$K_p$ for the reaction = $1.6\times {10}^5$

$K_p^{\prime }=1/K_p=\frac{1}{1.6\times {10}^5}=6.25\times {10}^{-6}$

Let the pressure of both $H_2$ and $B{r}_2$ at equilibrium be $p$ .

$2HBr\rightleftharpoons H_2(g)+B{r}_2(g)$
initial conc. 10 0 0
at eq 10-2p p p

Now,
$K_p^{\prime }=\frac{p^{B{r}_2}.p^{H_2}}{p_{HBr}^2}$
$6.25\times {10}^{-6}=\frac{p^2}{(10-2p{)}^2}$
$5\times {10}^{-3}=\frac{p}{(5-p)}$
By solving the above equation we get,

$p$ = 0.00248 bar

Hence the pressure of $H_2$ and $B{r}_2$ is 0.00248 bar and pressure of $HBr$ is 0.00496 bar

Question 6.28(a) Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

$C{H}_4(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_2(g)$

Write an expression for Kp for the above reaction.

Answer :

$C{H}_4(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_2(g)$
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.
$K_p=\frac{p_{CO}.p_{H_2}^3}{p_{C{H}_4}.p_{H_{2}O}}$

Question 6.28(b) Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

$C{H}_4(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_2(g)$

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 $K_p$ decreases.

(ii) According to Le Chatellier's principle, as the reaction is endothermic, the equilibrium will shift in the forward direction. The value of $K_p$ 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 H 2

$2H_2(g)+CO(g)\rightleftharpoons C{H}_{3}OH(g)$

Answer :

$2H_2(g)+CO(g)\rightleftharpoons C{H}_{3}OH(g)$
(a)According to Le Chatelliers principle, on the addition of dihydrogen, the number of mole of $H_2$ 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 $C{H}_{3}OH$ the equilibrium will shift in the forward direction.

Question 6.29(b) Describe the effect of :

addition of CH 3 OH

$2H_2(g)+CO(g)\rightleftharpoons C{H}_{3}OH(g)$

Answer :

$2H_2(g)+CO(g)\rightleftharpoons C{H}_{3}OH(g)$
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

$2H_2(g)+CO(g)\rightleftharpoons C{H}_{3}OH(g)$

Answer :

$2H_2(g)+CO(g)\rightleftharpoons C{H}_{3}OH(g)$
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 K c for decomposition of phosphorus pentachloride, PCl 5 is 8.3 ×10 -3 . If decomposition is depicted as,

$PC{l}_5(g)\rightleftharpoons PC{l}_3(g)+C{l}_2(g),{\mathrm{\Delta }}_r{H}^{+}=124.0kJmo{l}^{-1}$

(a) write an expression for Kc for the reaction.
(b) what is the value of K c for the reverse reaction at the same temperature ?
(c) what would be the effect on K c if

(i) more PCl 5 is added
(ii) pressure is increased
(iii) the temperature is increased ?

Answer :

We have,
$PC{l}_5(g)\rightleftharpoons PC{l}_3(g)+C{l}_2(g),{\mathrm{\Delta }}_r{H}^{+}=124.0kJmo{l}^{-1}$
Equilibrium constant for the above reaction = $8.3\times {10}^{-3}$

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

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

$K_c^{\prime }=\frac{1}{K_c}$
$=\frac{1}{8.3\times {10}^{-3}}=1.20\times {10}^2$

(c).i $K_c$ would remain the same because the temperature is constant in this case.

(c). ii If we increase the pressure, there is no change in $K_c$ because the temperature is constant in this case also.

(c). iii In an endothermic reaction, the value of $K_c$ increases with 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 H 2 . In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

$CO(g)+H_{2}O(g)\rightleftharpoons C{O}_2+H_2(g)$

If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that $p_{CO}=p_{H_{2}O}$ = 4.0 bar, what will be the partial pressure of H 2 at equilibrium? Kp= 10.1 at 400°C

Answer :

We have,
The partial pressure of $CO$ and $H_{2}O$ is 4 bar and the $K_p=10.1$
Let $p$ be the partial pressure of $C{O}_2$ and $H_2$ at equilibrium. The given reaction is-

$CO(g)+H_{2}O(g)\rightleftharpoons C{O}_2+H_2(g)$
Initial concentration 4 bar 4 bar 0 0
At equilibrium 4 - $p$ 4 - $p$ $p$ $p$

Therefore, we can write,
$10.1=\frac{p_{C{O}_2}.p_{H_2}}{p_{CO}.p_{H_{2}O}}$
$$\begin{gathered} 10.1=\frac{p^2}{(4-p{)}^2} \\ 3.178=\frac{p}{(4-p)} \\ p=(4-p)\times 3.178 \end{gathered}$$
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) $C{l}_2(g)\rightleftharpoons 2Cl(g);K_c=5\times {10}^{-39}$
(b) $C{l}_2(g)+2NO(g)\rightleftharpoons 2NOCl(g);K_c=3.7\times {10}^8$
(c) $C{l}_2(g)+2N{O}_2(g)\rightleftharpoons 2N{O}_{2}Cl(g);K_c=1.8$

Answer :

If the value of $K_c$ is in the range of ${10}^{-3}$ to ${10}^3$ , 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 $3O_2(g)\rightleftharpoons 2O_3(g)$ is 2.0 ×10 -50 at 25°C. If the equilibrium concentration of O 2 in air at 25°C is 1.6 ×10 -2 , what is the concentration of O 3 ?

Answer :

We have,

equilibrium constant of the reaction = $2\times {10}^{-50}$
the concentration of dioxygen $[O_2]$ = $1.6\times {10}^{-2}$

the given reaction is-

$3O_2(g)\rightleftharpoons 2O_3(g)$
Then we have,
(equilibrium constant)
$K_c=[O_3(g){]}^2/[O_2(g){]}^3$

$$\begin{gathered} 2\times {10}^{-50}=[O_3{]}^2/(1.6\times {10}^{-2}{)}^3 \\ [O_3{]}^2=2\times {10}^{-50}\times (1.6\times {10}^{-2}{)}^3 \end{gathered}$$

$=8.192\times {10}^{-56}$

$[O_3]=\sqrt{8.192\times {10}^{-56}}=2.86\times {10}^{-28}$

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

Question 6.34(a) The reaction,

$CO(g)+3H_2(g)\rightleftharpoons C{H}_4(g)+H_{2}O(g)$

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of $CO$ , 0.10 mol of $H_2$ and 0.02 mol of $H_{2}O$ and an unknown amount of CH 4 in the flask. Determine the concentration of CH 4 in the mixture. The equilibrium constant, K c 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( $H_2$ )and the 0.02 mol of water( $H_{2}O$ )
the equilibrium constant = 3.90

Let $x$ be the concentration of methane at equilibrium. The given reaction is-

$CO(g)+3H_2(g)\rightleftharpoons C{H}_4(g)+H_{2}O(g)$
At equilibrium, 0.3 mol/L 0.1 mol/L $x$ 0.02 mol/L