Chapter 12, Problem 57E

(a)

Interpretation Introduction

Interpretation: The equilibrium constant value for a decomposition reaction of $\text{NOCl}$ is given. The equilibrium concentrations of the species involved in the given reaction are to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented as $K$.

To determine: The equilibrium concentrations of the species involved in the given reaction.

Answer to Problem 57E

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\underset{\_}{1.0M}$, of ${\text{NO}}_{\left(g\right)}$ is $\underset{\_}{0.032M}$ and of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\underset{\_}{0.016M}$.

Explanation of Solution

Given

The stated reaction is,

${\text{2NOCl}}_{\left(g\right)}\rightleftharpoons 2{\text{NO}}_{\left(g\right)}+{\text{Cl}}_2{}_{\left(g\right)}$

The initial number of moles $\text{NOCl}$ is $2.0\text{moles}$.

The volume of the flask is $2.0\text{L}$.

The equilibrium constant $\left(K\right)$ is $1.6\times {10}^{-5}$.

The concentration of a reactant is calculated by the formula,

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}\left(\text{L}\right)}$

The initial concentration of $\text{NOCl}$ is calculated by the formula,

$\text{Concentration}\text{of}\text{NOCl}=\frac{\text{Moles}\text{of}\text{NOCl}}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of $\text{NOCl}$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}\text{NOCl}=\frac{2.0\text{mole}}{\text{2}\text{.0}\text{L}}\\ =1.0M\end{array}$

The concentration of $\text{NOCl}$ consumed is assumed to be $2x$.

The equilibrium concentrations are represented as,

$\begin{array}{cccccc}& 2{\text{NOCl}}_{\left(g\right)}& \rightleftharpoons & 2{\text{NO}}_{\left(g\right)}& +& {\text{Cl}}_2{}_{\left(g\right)}\\ \text{Initial}\text{concentration}& 1.0& & 0& & 0\\ \text{Change}& -2x& & +2x& & +x\\ \text{Equilibrium}\text{concentration}& 1.0-2x& & 2x& & x\end{array}$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K=\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$ (1)

According to the formulated ICE table,

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\left(1.0-2x\right)\text{M}$.

The equilibrium initial concentration of ${\text{NO}}_{\left(g\right)}$ is $\left(2x\right)\text{M}$.

The equilibrium initial concentration of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\left(x\right)\text{M}$.

Substitute these values in equation (1).

$\begin{array}{l}K=\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}\\ K=\frac{{\left[2x\right]}^2\left[x\right]}{{\left[1.0-2x\right]}^2}\end{array}$

The given value of $K$ is $1.6\times {10}^{-5}$.

Substitute the value of $K$ in the above expression.

$\begin{array}{c}1.6\times {10}^{-5}=\frac{{\left[2x\right]}^2\left[x\right]}{{\left[1.0-2x\right]}^2}\\ x=0.0155\\ \approx 0.016\end{array}$

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{NOCl}}_{\left(g\right)}=\left(1.0-2x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{NOCl}}_{\left(g\right)}=\left(1.0-2x\right)\\ =\left(1.0-2\left(0.016\right)\right)\text{M}\\ \approx \underset{\_}{1.0M}\end{array}$

The equilibrium concentration of ${\text{NO}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{NO}}_{\left(g\right)}=\left(2x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{NO}}_{\left(g\right)}=\left(2x\right)\\ =2\left(0.016\right)\text{M}\\ =\underset{\_}{0.032M}\end{array}$

The equilibrium concentration of ${\text{Cl}}_2{}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{}_{\left(g\right)}=\left(x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{}_{\left(g\right)}=\left(x\right)\\ =\underset{\_}{0.016M}\end{array}$

Conclusion

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\underset{\_}{1.0M}$, of ${\text{NO}}_{\left(g\right)}$ is $\underset{\_}{0.032M}$ and of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\underset{\_}{0.016M}$.

(b)

Interpretation Introduction

Interpretation: The equilibrium constant value for a decomposition reaction of $\text{NOCl}$ is given. The equilibrium concentrations of the species involved in the given reaction are to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented as $K$.

To determine: The equilibrium concentrations of the species involved in the given reaction.

Answer to Problem 57E

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\underset{\_}{1.0M}$, of ${\text{NO}}_{\left(g\right)}$ is $\underset{\_}{1.0M}$ and of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\underset{\_}{1.6\times 10^{-5}M}$.

Explanation of Solution

Given

The stated reaction is,

${\text{2NOCl}}_{\left(g\right)}\rightleftharpoons 2{\text{NO}}_{\left(g\right)}+{\text{Cl}}_2{}_{\left(g\right)}$

The initial number of moles $\text{NOCl}$ is $1.0\text{mole}$.

The initial number of moles $\text{NO}$ is $1.0\text{mole}$.

The volume of the flask is $1.0\text{L}$.

The equilibrium constant $\left(K\right)$ is $1.6\times {10}^{-5}$.

The concentration of a reactant is calculated by the formula,

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}\left(\text{L}\right)}$

The initial concentration of $\text{NOCl}$ is calculated by the formula,

$\text{Concentration}\text{of}\text{NOCl}=\frac{\text{Moles}\text{of}\text{NOCl}}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of $\text{NOCl}$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}\text{NOCl}=\frac{1.0\text{mole}}{\text{1}\text{.0}\text{L}}\\ =1.0M\end{array}$

The initial concentration of $\text{NO}$ is calculated by the formula,

$\text{Concentration}\text{of}\text{NO}=\frac{\text{Moles}\text{of}\text{NO}}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of $\text{NO}$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}\text{NO}=\frac{1.0\text{mole}}{\text{1}\text{.0}\text{L}}\\ =1.0M\end{array}$

The concentration of $\text{NOCl}$ consumed is assumed to be $2x$.

The equilibrium concentrations are represented as,

$\begin{array}{cccccc}& 2{\text{NOCl}}_{\left(g\right)}& \rightleftharpoons & 2{\text{NO}}_{\left(g\right)}& +& {\text{Cl}}_2{}_{\left(g\right)}\\ \text{Initial}\text{concentration}& 1.0& & 1.0& & 0\\ \text{Change}& -2x& & +2x& & +x\\ \text{Equilibrium}\text{concentration}& 1.0-2x& & 1.0+2x& & x\end{array}$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K=\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$ (1)

According to the formulated ICE table,

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\left(1.0-2x\right)\text{M}$.

The equilibrium initial concentration of ${\text{NO}}_{\left(g\right)}$ is $\left(1.0+2x\right)\text{M}$.

The equilibrium initial concentration of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\left(x\right)\text{M}$.

Substitute these values in equation (1).

$\begin{array}{l}K=\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}\\ K=\frac{{\left[1+2x\right]}^2\left[x\right]}{{\left[1.0-2x\right]}^2}\end{array}$

The given value of $K$ is $1.6\times {10}^{-5}$.

Substitute the value of $K$ in the above expression.

$\begin{array}{c}1.6\times {10}^{-5}=\frac{{\left[1+2x\right]}^2\left[x\right]}{{\left[1.0-2x\right]}^2}\\ x=0.0001599\\ \approx 1.6\times 10^{-5}\end{array}$

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{NOCl}}_{\left(g\right)}=\left(1.0-2x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{NOCl}}_{\left(g\right)}=\left(1.0-2x\right)\\ =\left(1.0-2\left(1.6\times {10}^{-5}\right)\right)\text{M}\\ \approx \underset{\_}{1.0M}\end{array}$

The equilibrium concentration of ${\text{NO}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{NO}}_{\left(g\right)}=\left(1.0+2x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}\text{NO}\left(g\right)=\left(1.0+2x\right)\\ =\left(1.0+2\left(0.016\right)\right)\text{M}\\ \approx \underset{\_}{1.0M}\end{array}$

The equilibrium concentration of ${\text{Cl}}_2{}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{}_{\left(g\right)}=\left(x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{}_{\left(g\right)}=\left(x\right)\\ =\underset{\_}{1.6\times 10^{-5}M}\end{array}$

Conclusion

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\underset{\_}{1.0M}$, of ${\text{NO}}_{\left(g\right)}$ is $\underset{\_}{1.0M}$ and of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\underset{\_}{1.6\times 10^{-5}M}$.

(c)

Interpretation Introduction

Interpretation: The equilibrium constant value for a decomposition reaction of $\text{NOCl}$ is given. The equilibrium concentrations of the species involved in the given reaction are to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented as $K$.

To determine: The equilibrium concentrations of the species involved in the given reaction.

Answer to Problem 57E

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\underset{\_}{2.0M}$, of ${\text{NO}}_{\left(g\right)}$ is $\underset{\_}{8.0\times 10^{-3}M}$ and of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\underset{\_}{1M}$.

Explanation of Solution

Given

The stated reaction is,

${\text{2NOCl}}_{\left(g\right)}\rightleftharpoons {\text{NO}}_{\left(g\right)}+{\text{Cl}}_2{}_{\left(g\right)}$

The initial number of moles $\text{NOCl}$ is $2.0\text{moles}$.

The initial number of moles ${\text{Cl}}_2$ is $1.0\text{mole}$.

The volume of the flask is $1.0\text{L}$.

The equilibrium constant $\left(K\right)$ is $1.6\times {10}^{-5}$.

The concentration of a reactant is calculated by the formula,

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}\left(\text{L}\right)}$

The initial concentration of $\text{NOCl}$ is calculated by the formula,

$\text{Concentration}\text{of}\text{NOCl}=\frac{\text{Moles}\text{of}\text{NOCl}}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of $\text{NOCl}$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}\text{NOCl}=\frac{2.0\text{mole}}{\text{1}\text{.0}\text{L}}\\ =2.0M\end{array}$

The initial concentration of ${\text{Cl}}_2$ is calculated by the formula,

$\text{Concentration}\text{of}{\text{Cl}}_2=\frac{\text{Moles}\text{of}{\text{Cl}}_2}{\text{Volume}\text{of}\text{the}\text{flask}\left(\text{L}\right)}$

Substitute the given values of the number of moles of ${\text{Cl}}_2$ and the volume of the flask in the above expression.

$\begin{array}{c}\text{Concentration}\text{of}{\text{Cl}}_2=\frac{1.0\text{mole}}{\text{1}\text{.0}\text{L}}\\ =1.0M\end{array}$

The concentration of $\text{NOCl}$ consumed is assumed to be $2x$.

The equilibrium concentrations are represented as,

$\begin{array}{cccccc}& 2{\text{NOCl}}_{\left(g\right)}& \rightleftharpoons & 2{\text{NO}}_{\left(g\right)}& +& {\text{Cl}}_2{}_{\left(g\right)}\\ \text{Initial}\text{concentration}& 2.0& & 0& & 1.0\\ \text{Change}& -2x& & +2x& & +x\\ \text{Equilibrium}\text{concentration}& 2.0-2x& & +2x& & 1.0+x\end{array}$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K$ is the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K=\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}$ (1)

According to the formulated ICE table,

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\left(2.0-2x\right)\text{M}$.

The equilibrium initial concentration of ${\text{NO}}_{\left(g\right)}$ is $\left(2x\right)\text{M}$.

The equilibrium initial concentration of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\left(1.0+x\right)\text{M}$.

Substitute these values in equation (1).

$\begin{array}{l}K=\frac{{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]}{{\left[\text{NOCl}\right]}^2}\\ K=\frac{{\left[2x\right]}^2\left[1.0+x\right]}{{\left[2.0-2x\right]}^2}\end{array}$

The given value of $K$ is $1.6\times {10}^{-5}$.

The value of $x$ is assumed to very less than $1$.

Substitute the value of $K$ in the above expression.

$\begin{array}{c}1.6\times {10}^{-5}=\frac{{\left[2x\right]}^2\left[1.0\right]}{{\left[2.0\right]}^2}\\ x=4\times 10^{-3}\end{array}$

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is calculated by the formula,

${\text{NOCl}}_{\left(g\right)}$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{NOCl}}_{\left(g\right)}=\left(2.0-2x\right)\\ =\left(2.0-2\left(4\times {10}^{-3}\right)\right)\text{M}\\ \approx \underset{\_}{2.0M}\end{array}$

The equilibrium concentration of ${\text{NO}}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{NO}}_{\left(g\right)}=\left(2x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{NO}}_{\left(g\right)}=\left(2x\right)\\ =\left(2\left(4\times {10}^{-3}\right)\right)\text{M}\\ =\underset{\_}{8.0\times 10^{-3}M}\end{array}$

The equilibrium concentration of ${\text{Cl}}_2{}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{}_{\left(g\right)}=\left(1.0+x\right)$

Substitute the calculated value of $x$ in the above expression.

$\begin{array}{c}\text{Equilibrium}\text{concentration}\text{of}{\text{Cl}}_2{}_{\left(g\right)}=\left(1.0+x\right)\\ =\left(1.0+\left(4\times {10}^{-3}\right)\right)\\ \approx \underset{\_}{1M}\end{array}$

Conclusion

The equilibrium concentration of ${\text{NOCl}}_{\left(g\right)}$ is $\underset{\_}{2.0M}$, of ${\text{NO}}_{\left(g\right)}$ is $\underset{\_}{8.0\times 10^{-3}M}$ and of ${\text{Cl}}_2{}_{\left(g\right)}$ is $\underset{\_}{1M}$