Chapter 13, Problem 56E

At a particular temperature, K_p = 0.25 for the reaction

${\text{N}}_2{\text{O}}_4\left(g\right)\rightleftharpoons 2{\text{NO}}_2\left(g\right)$

a. A flask containing only N₂O₄ at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.

b. A flask containing only NO₂ at an initial pressure of 9.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases.

c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K_{\text{p}}\right)$ value for the decomposition reaction of ${\text{N}}_2{\text{O}}_4$ is given. The equilibrium partial pressures at the given initial pressure values 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 pressure, it is represented as $K_{\text{p}}$.

To determine: The equilibrium partial pressures of the species involved.

Answer to Problem 56E

The equilibrium partial pressures of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ and ${\text{NO}}_2{}_{\left(g\right)}$ are $\underset{\_}{4.0atm}$ and $\underset{\_}{1.0atm}$ respectively.

Explanation of Solution

Given

The initial partial pressure of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ $\left({\text{P}}_{{\text{N}}_2{\text{O}}_4}\right)$ is $4.5\text{atm}$.

The equilibrium constant $\left(K_{\text{p}}\right)$ is $0.25$.

The reaction that takes place between hydrogen gas and nitrogen gas is,

${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\left(\text{g}\right)}\rightleftharpoons {\text{2NO}}_{\text{2}}{}_{\left(\text{g}\right)}$

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

$K_{\text{p}}=\frac{\text{Partial}\text{pressure}\text{of}\text{products}}{\text{Partial}\text{pressure}\text{of}\text{reactants}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant in terms of partial pressure.

The equilibrium ratio for the given reaction is,

$K_{\text{p}}=\frac{{\left({\text{P}}_{{\text{NO}}_2}\right)}^2}{\left({\text{P}}_{{\text{N}}_2{\text{O}}_4}\right)}$ (1)

The equilibrium partial pressure values are represented as,

$\begin{array}{cccc}& {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\left(\text{g}\right)}& \rightleftharpoons & {\text{2NO}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \text{Initial}\left(\text{atm}\right)& \text{4}\text{.5}& & \text{0}\\ \text{Change}\left(\text{atm}\right)& \text{-x}& & \text{+2x}\\ \text{Equilibrium}\left(\text{atm}\right)& \text{4}\text{.5-x}& & \text{2x}\end{array}$

According to the ICE table formed,

The equilibrium partial pressure of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ $\left({\text{P}}_{{\text{N}}_2{\text{O}}_4}\right)$ is $4.5-x$.

The equilibrium partial pressure of ${\text{NO}}_2{}_{\left(g\right)}$ $\left({\text{P}}_{{\text{NO}}_{\text{2}}}\right)$ is $2x$.

Substitute the values of ${\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_4}$ and ${\text{P}}_{{\text{NO}}_2}$ in equation (1).

$K_{\text{p}}=\frac{{\left(2x\right)}^2}{\left(4.5-x\right)}$

Substitute the given value of $K_{\text{p}}$ in the above expression.

$\begin{array}{c}0.25=\frac{{\left(2x\right)}^2}{\left(4.5-x\right)}\\ {\left(2x\right)}^2=0.25\left(4.5-x\right)\end{array}$

Simplifying the above expression, the value of $x$ is obtained.

$x=0.5$

The equilibrium partial pressure of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{N}}_2{\text{O}}_4{}_{\left(g\right)}=\left(4.5-x\right)$

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

$\begin{array}{c}\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{N}}_2{\text{O}}_4{}_{\left(g\right)}=\left(4.5-x\right)\\ =\left(4.5-0.5\right)\\ =\underset{\_}{4.0atm}\end{array}$

The equilibrium partial pressure of ${\text{NO}}_2{}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{NO}}_2{}_{\left(g\right)}=2x$

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

$\begin{array}{c}\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{NO}}_2{}_{\left(g\right)}=2\left(0.5\right)\text{atm}\\ =\underset{\_}{1.0atm}\end{array}$

Conclusion

The equilibrium partial pressures of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ and ${\text{NO}}_2{}_{\left(g\right)}$ are $\underset{\_}{4.0atm}$ and $\underset{\_}{1.0atm}$ respectively.

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K_{\text{p}}\right)$ value for the decomposition reaction of ${\text{N}}_2{\text{O}}_4$ is given. The equilibrium partial pressures at the given initial pressure values 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 pressure, it is represented as $K_{\text{p}}$.

To determine: The equilibrium partial pressures of the species involved.

Answer to Problem 56E

The equilibrium partial pressures of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ and ${\text{NO}}_2{}_{\left(g\right)}$ are $\underset{\_}{4.0atm}$ and $\underset{\_}{1.0atm}$ respectively.

Explanation of Solution

Given

The initial partial pressure of ${\text{NO}}_2{}_{\left(g\right)}$ $\left({\text{P}}_{{\text{NO}}_2}\right)$ is $9.0\text{atm}$.

The given equilibrium constant value is $0.25$.

The reaction that takes place between hydrogen gas and nitrogen gas is,

$2{\text{NO}}_2{}_{\left(g\right)}\rightleftharpoons {\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$

The given equilibrium constant $\left(K_{\text{p}}\right)$ value for this reaction is $\frac{1}{0.25}$.

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

$K_{\text{p}}=\frac{\text{Partial}\text{pressure}\text{of}\text{products}}{\text{Partial}\text{pressure}\text{of}\text{reactants}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant in terms of partial pressure.

The equilibrium ratio for the given reaction is,

$K_{\text{p}}=\frac{\left({\text{P}}_{{\text{N}}_2{\text{O}}_4}\right)}{{\left({\text{P}}_{{\text{NO}}_2}\right)}^2}$ (1)

The equilibrium partial pressure values are represented as,

$\begin{array}{cccc}& {\text{2NO}}_2{}_{\left(g\right)}& \rightleftharpoons & {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{}_{\left(g\right)}\\ \text{Initial}\left(\text{atm}\right)& 9.0& & 0\\ \text{Change}\left(\text{atm}\right)& -2x& & +x\\ \text{Equilibrium}\left(\text{atm}\right)& 9.0-2x& & +x\end{array}$

According to the ICE table formed,

The equilibrium partial pressure of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ $\left({\text{P}}_{{\text{N}}_2{\text{O}}_4}\right)$ is $x$.

The equilibrium partial pressure of ${\text{NO}}_2{}_{\left(g\right)}$ $\left({\text{P}}_{{\text{NO}}_{\text{2}}}\right)$ is $9.0-2x$.

Substitute the values of ${\text{P}}_{{\text{N}}_{\text{2}}{\text{O}}_4}$ and ${\text{P}}_{{\text{NO}}_2}$ in equation (1).

$K_{\text{p}}=\frac{\left(x\right)}{{\left(9.0-2x\right)}^2}$

Substitute the given value of $K_{\text{p}}$ in the above expression.

$\begin{array}{c}\frac{1}{0.25}=\frac{\left(x\right)}{{\left(9.0-2x\right)}^2}\\ {\left(9.0-2x\right)}^2=0.25\left(x\right)\end{array}$

Simplifying the above expression, the value of $x$ is obtained.

$x=4.0$

The equilibrium partial pressure of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{N}}_2{\text{O}}_4{}_{\left(g\right)}=\left(x\right)$

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

$\begin{array}{c}\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{N}}_2{\text{O}}_4{}_{\left(g\right)}=\left(x\right)\\ =\underset{\_}{4.0atm}\end{array}$

The equilibrium partial pressure of ${\text{NO}}_2{}_{\left(g\right)}$ is calculated by the formula,

$\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{NO}}_2{}_{\left(g\right)}=\left(9.0-2x\right)$

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

$\begin{array}{c}\text{Equilibrium}\text{partial}\text{pressure}\text{of}{\text{NO}}_2{}_{\left(g\right)}=\left(9.0-2x\right)\\ =\left(9.0-2\left(4\right)\right)\text{atm}\\ =\underset{\_}{1.0atm}\end{array}$

Conclusion

The equilibrium partial pressures of ${\text{N}}_2{\text{O}}_4{}_{\left(g\right)}$ and ${\text{NO}}_2{}_{\left(g\right)}$ are $\underset{\_}{4.0atm}$ and $\underset{\_}{1.0atm}$ respectively.

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K_{\text{p}}\right)$ value for the decomposition reaction of ${\text{N}}_2{\text{O}}_4$ is given. The equilibrium partial pressures at the given initial pressure values 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 pressure, it is represented as $K_{\text{p}}$.

To determine: If it matters that from which direction an equilibrium position is reached.

Answer to Problem 56E

The values obtained in part (a) and (b) are same. Therefore, it is does not matter.

Explanation of Solution

The equilibrium partial pressure values obtained in both (a) and (b) are same. Hence, it does not matter from which direction the equilibrium position is obtained in a chemical reaction.

Conclusion

It does not matter from which direction the equilibrium position is reached in a chemical reaction