Chapter 14, Problem 14.75QP

(a)

Interpretation Introduction

Interpretation: The equilibrium partial pressure of the reactants and products at $500\text{K}$ is to be calculated for the given reaction and the change in concentration of ${\text{PCl}}_{\text{5}}$ and ${\text{PCl}}_{\text{3}}$ on addition of more chlorine after reaching the equilibrium is to be predicted.

Concept introduction: The equilibrium constant $\left(K_{\text{p}}\right)$ is expressed as,

$K_{\text{p}}=\frac{{\left(P_{\text{C}}\right)}^c{\left(P_{\text{D}}\right)}^d}{{\left(P_{\text{A}}\right)}^a{\left(P_{\text{B}}\right)}^b}$

To determine: The equilibrium partial pressure of the reactants and products at $500\text{K}$.

Answer to Problem 14.75QP

Solution

The equilibrium Partial pressure of ${\text{PCl}}_{\text{5}}$ is $\underset{\_}{0.025atm}$, the equilibrium Partial pressure of ${\text{PCl}}_3$ is $\underset{\_}{1.035atm}$ and the equilibrium Partial pressure of ${\text{Cl}}_2$ is $\underset{\_}{0.535atm}$.

Explanation of Solution

Explanation

Given

The balanced chemical equation is,

${\text{PCl}}_5\left(g\right)\rightleftharpoons {\text{PCl}}_3\left(g\right)+{\text{Cl}}_2\left(g\right)$

The equilibrium constant at $500\text{K}$ is $23.6$.

The initial partial pressure of ${\text{PCl}}_5$ is $0.560\text{atm}$.

The initial partial pressure of ${\text{PCl}}_3$ is $0.500\text{atm}$.

The equilibrium constant is calculated by the formula,

$K_{\text{p}}=\frac{\left(P_{{\text{PCl}}_3\left(g\right)}\right)\left(P_{{\text{Cl}}_2\left(g\right)}\right)}{\left(P_{{\text{PCl}}_5\left(g\right)}\right)}$ (1)

Where,

• $\left(P_{{\text{PCl}}_3\left(g\right)}\right)$ is the equilibrium partial pressure of ${\text{PCl}}_3$.
• $\left(P_{{\text{Cl}}_2\left(g\right)}\right)$ is the equilibrium partial pressure of ${\text{Cl}}_2$.
• $\left(P_{{\text{PCl}}_5\left(g\right)}\right)$ is the equilibrium partial pressure of ${\text{PCl}}_5$.

Consider the following table showing the values of partial pressure at initial and at equilibrium stage,

$\begin{array}{cccccc}& {\text{PCl}}_5\left(g\right)& \rightleftharpoons & {\text{PCl}}_3\left(g\right)& +& {\text{Cl}}_2\left(g\right)\\ \text{Initial}& 0.560\text{atm}& & 0.500\text{atm}& & 0\\ \text{At}\text{equilibrium}& 0.560-x& & 0.500+x& & x\end{array}$ (2)

Where,

• $x$ is the change in partial pressure occurred during the reaction.

Substitute the values of equilibrium partial pressure of reactants and product and equilibrium constant in equation (1).

$\begin{array}{c}23.6=\frac{\left(0.500+x\right)\left(x\right)}{\left(0.560-x\right)}\\ 13.22-23.6x=0.500x+x^2\\ x^2+24.1x-13.22=0\end{array}$

Solve this quadratic equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.
• $b$ is the coefficient of $x$.
• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{c}x=\frac{-24.1\pm \sqrt{{\left(24.1\right)}^2\pm 4\times 1\times \left(-13.216\right)}}{2\times 1}\\ =\frac{-24.1\pm 25.17}{2}\\ =\frac{1.07}{2}\\ =0.535\end{array}$

The value if $x$ is $0.535$.

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

The equilibrium Partial pressure of ${\text{PCl}}_{\text{5}}$ is $\begin{array}{l}=0.560-x\\ =0.560-0.535\\ =\underset{\_}{0.025atm}\end{array}$

The equilibrium Partial pressure of ${\text{PCl}}_3$ is $\begin{array}{l}=0.500+x\\ =0.560+0.535\\ =\underset{\_}{1.035atm}\end{array}$

The equilibrium Partial pressure of ${\text{Cl}}_2$ is $\begin{array}{l}=x\\ =\underset{\_}{0.535atm}\end{array}$

Therefore, The equilibrium Partial pressure of ${\text{PCl}}_{\text{5}}$ is $\underset{\_}{0.025atm}$, the equilibrium Partial pressure of ${\text{PCl}}_3$ is $\underset{\_}{1.035atm}$ and the equilibrium Partial pressure of ${\text{Cl}}_2$ is $\underset{\_}{0.535atm}$.

(b)

Interpretation Introduction

To determine: The change in concentration of ${\text{PCl}}_{\text{5}}$ and ${\text{PCl}}_{\text{3}}$ on addition of more chlorine after reaching the equilibrium.

Answer to Problem 14.75QP

Solution

The concentration of ${\text{PCl}}_{\text{5}}$ and ${\text{PCl}}_{\text{3}}$ increases and decreases respectively on addition of more chlorine after reaching the equilibrium.

Explanation of Solution

Explanation

On addition of ${\text{Cl}}_{\text{2}}$ in the reaction mixture, the partial pressure of ${\text{Cl}}_{\text{2}}$ increases. The concentration of product is more than the concentration of reactant. To maintain the equilibrium value, the reaction will shift toward the backward direction.

Therefore, on addition of ${\text{Cl}}_{\text{2}}$, the concentration of ${\text{PCl}}_{\text{5}}$ increases and the concentration of ${\text{PCl}}_{\text{3}}$ decreases.

Conclusion

1. a. The equilibrium Partial pressure of ${\text{PCl}}_{\text{5}}$ is $\underset{\_}{0.025atm}$, the equilibrium Partial pressure of ${\text{PCl}}_3$ is $\underset{\_}{1.035atm}$ and the equilibrium Partial pressure of ${\text{Cl}}_2$ is $\underset{\_}{0.535atm}$.
2. b. The concentration of ${\text{PCl}}_{\text{5}}$ and ${\text{PCl}}_{\text{3}}$ increases and decreases respectively on addition of more chlorine after reaching the equilibrium.