Chapter 12, Problem 104CP

Interpretation Introduction

Interpretation: The equilibrium constant $\left(K_{\text{p}}\right)$ value at $1325\text{K}$ for the given reaction is $1.00\times {10}^{-1}$. The total pressure of the equilibrium mixture is given to be $1.00\text{atm}$. The equilibrium pressure of the species involved in the given reaction and the fraction (by moles) of ${\text{P}}_4\left(g\right)$ that has dissociated to reach equilibrium is 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 $\left(K_{\text{p}}\right)$

To determine: The equilibrium pressure of the species involved in the given reaction and the fraction (by moles) of ${\text{P}}_4\left(g\right)$.

The equilibrium constant expression for the given reaction is, $K_{\text{p}}=\frac{{\left({\text{P}}_{{\text{P}}_2}\right)}^2}{\left({\text{P}}_{{\text{P}}_4}\right)}$

Answer to Problem 104CP

Answer

The equilibrium pressure of the species involved in the given reaction and the fraction (by moles) of ${\text{P}}_4\left(g\right)$ is calculated as follows.

Explanation of Solution

Explanation

Given

The stated reaction is,

${\text{P}}_4\left(g\right)\rightleftharpoons {\text{2P}}_2\left(g\right)$

The total pressure of the equilibrium mixture is given to be $1.00\text{atm}$.

The equilibrium constant expression is,

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.

Substitute the value of the equilibrium pressures of ${\text{P}}_{\text{4}}\left(g\right)$ and ${\text{P}}_2\left(g\right)$ in the above expression.

$K_{\text{p}}=\frac{{\left({\text{P}}_{{\text{P}}_2}\right)}^2}{\left({\text{P}}_{{\text{P}}_4}\right)}$ (1)

The change in the partial pressure of ${\text{P}}_4\left(g\right)$ is $\underset{\_}{0.135atm}$.

The initial partial pressure of ${\text{P}}_4\left(g\right)$ is assumed to be $x$.

The change in the partial pressure of ${\text{P}}_4\left(g\right)$ is assumed to be $y$.

The equilibrium pressures are represented as,

$\begin{array}{cccc}& {\text{P}}_{\text{4}}\left(g\right)& \rightleftharpoons & {\text{2P}}_2\left(g\right)\\ \text{Initial}\left(\text{atm}\right)& x& & 0\\ \text{Change}& -y& & +2y\\ \text{Equilibrium}\left(\text{atm}\right)& x-y& & 2y\end{array}$

The equilibrium pressure of ${\text{P}}_{\text{4}}$ $\left({\text{P}}_{{\text{P}}_4}\right)$ is $\left(x-y\right)\text{atm}$.

The equilibrium pressure of ${\text{P}}_2$ $\left({\text{P}}_{{\text{P}}_2}\right)$ is $\left(2y\right)\text{atm}$.

Substitute the value of $\left({\text{P}}_{{\text{P}}_4}\right)$ and $\left({\text{P}}_{{\text{P}}_2}\right)$ in the equation (1).

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

The equilibrium constant value for this reaction is given to be $1.00\times {10}^{-1}$.

Substitute the value in the above expression.

$1.00\times {10}^{-1}=\frac{{\left(2y\right)}^2}{\left(x-y\right)}$ (2)

The total equilibrium pressure is calculated by the formula,

$\text{Total}\text{equilibrium}\text{pressure}=\left({\text{P}}_{{\text{P}}_4}\right)+\left({\text{P}}_{{\text{P}}_2}\right)$

Substitute the value of $\left({\text{P}}_{{\text{P}}_4}\right)$ and $\left({\text{P}}_{{\text{P}}_2}\right)$ in the above expression.

$\begin{array}{c}\text{Total}\text{equilibrium}\text{pressure}=\left(x-y\right)+\left(2y\right)\\ =\left(x+y\right)\end{array}$

The total equilibrium pressure is given to be $1.0\text{atm}$.

Therefore,

$\begin{array}{l}x+y=1\\ x=1-y\end{array}$

Substitute the value of $x$ in equation (2).

$\begin{array}{c}1.00\times {10}^{-1}=\frac{{\left(2y\right)}^2}{\left(\left(1-y\right)-y\right)}\\ 0.100=\frac{4y^2}{\left(1-2y\right)}\\ y=\underset{\_}{0.135}\end{array}$

Therefore, the change in the partial pressure of ${\text{P}}_4\left(g\right)$ is $\underset{\_}{0.135atm}$.

The initial partial pressure of ${\text{P}}_4\left(g\right)$ is $\underset{\_}{0.865atm}$.

The initial partial pressure of ${\text{P}}_4\left(g\right)$ is calculated by the formula,

$x=1-y$

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

$\begin{array}{c}x=\left(1-0.135\right)\text{atm}\\ =\underset{\_}{0.865atm}\end{array}$

Therefore, the initial partial pressure of ${\text{P}}_4\left(g\right)$ is $\underset{\_}{0.865atm}$.

The equilibrium pressure of ${\text{P}}_4\left(g\right)$ is $\underset{\_}{0.73atm}$.

The equilibrium pressure of ${\text{P}}_4\left(g\right)$ is calculated by the formula,

${\text{P}}_{{\text{P}}_{\text{4}}}=x-y$

Substitute the value of $x$ and  $y$ in the above expression.

$\begin{array}{c}{\text{P}}_{{\text{P}}_{\text{4}}}=x-y\\ =0.865-0.135\\ =\underset{\_}{0.73atm}\end{array}$

Therefore, the equilibrium pressure of ${\text{P}}_4\left(g\right)$ is $\underset{\_}{0.73atm}$.

The equilibrium pressure of ${\text{P}}_2\left(g\right)$ is $\underset{\_}{0.27atm}$.

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

${\text{P}}_{{\text{P}}_2}=2y$

Substitute the value of $x$ and  $y$ in the above expression.

$\begin{array}{c}{\text{P}}_{{\text{P}}_2}=2y\\ =2\left(0.135\right)\\ =\underset{\_}{0.27atm}\end{array}$

Therefore, the equilibrium pressure of ${\text{P}}_2\left(g\right)$ is $\underset{\_}{0.27atm}$.

The initial number of moles of ${\text{P}}_{\text{4}}\left(g\right)$ is $\underset{\_}{0.0079mol}$.

The initial pressure of ${\text{P}}_{\text{4}}\left(g\right)$ is $0.865\text{atm}$.

The volume of the container is assumed to be $1.0\text{L}$.

The given temperature is $1325\text{K}$.

The ideal gas equation is,

$PV=nRT$

Where,

• $P$ is the partial pressure.
• $V$ is the volume of the container.
• $n$ is the number of moles.
• $R$ is the universal gas constant $\left(0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)$
• $T$ is the temperature.

Rearrange the ideal gas equation.

$n=\frac{PV}{RT}$

Substitute the values of $P,V,R$ and $T$ for ${\text{P}}_{\text{4}}\left(g\right)$ in the above expression.

$\begin{array}{c}n=\frac{\left(0.865\text{atm}\right)\left(10.0\text{L}\right)}{\left(0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(1325\text{K}\right)}\\ =\underset{\_}{0.0079mol}\end{array}$

The number of moles of ${\text{P}}_{\text{4}}\left(g\right)$ that dissociated is $\underset{\_}{0.0012mol}$.

The fraction (by moles) of ${\text{P}}_{\text{4}}\left(g\right)$ dissociated is $\underset{\_}{0.15}$.

The volume of the container is assumed to be $1.0\text{L}$.

The given temperature is $1325\text{K}$.

The ideal gas equation is,

$PV=nRT$

Rearrange the ideal gas equation.

$n=\frac{PV}{RT}$

Substitute the values of $P,V,R$ and $T$ for ${\text{P}}_{\text{4}}\left(g\right)$ in the above expression.

$\begin{array}{c}n=\frac{\left(0.135\text{atm}\right)\left(10.0\text{L}\right)}{\left(0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(1325\text{K}\right)}\\ =\underset{\_}{0.0012mol}\end{array}$

The fraction (by moles) of ${\text{P}}_{\text{4}}\left(g\right)$ dissociated is $\underset{\_}{0.15}$.

The fraction (by moles) of ${\text{P}}_{\text{4}}\left(g\right)$ is calculated by the formula,

$\text{Fraction}\text{(by}\text{moles})=\frac{\text{Number}\text{of}\text{moles}\text{dissociated}}{\text{Initial}\text{number}\text{of}\text{moles}}$

Substitute the value of the number of moles of ${\text{P}}_{\text{4}}\left(g\right)$ dissociated and the initial number of moles of ${\text{P}}_{\text{4}}\left(g\right)$ in the above formula.

$\begin{array}{c}\text{Fraction}\text{(by}\text{moles})=\frac{\text{0}\text{.0012}\text{mol}}{\text{0}\text{.0079}\text{mol}}\\ =\underset{\_}{0.15}\end{array}$

Conclusion

Conclusion

The equilibrium pressure of ${\text{P}}_4\left(g\right)$ is $\underset{\_}{0.73atm}$ and the equilibrium pressure of ${\text{P}}_2\left(g\right)$ is $\underset{\_}{0.27atm}$. The fraction (by moles) of ${\text{P}}_{\text{4}}\left(g\right)$ dissociated is $\underset{\_}{0.15}$.