Chapter 14, Problem 14.58QE

Interpretation Introduction

Interpretation:

The concentration of all species after the attainment of equilibrium in the following reaction has to be determined.

  ${\text{PCl}}_3\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\rightleftharpoons {\text{PCl}}_{\text{5}}\left(\text{g}\right)$

Concept Introduction:

The condition of equilibrium is a state of balance of processes that runs in opposite directions. At equilibrium, the formation of a product from the reactant balances the formation of a reactant from the product. Also, the change in concentration of reaction and product seems to be negligible at equilibrium state.

A reaction quotient $Q$ is an algebraic form of equilibrium constant $\left(K_{\text{eq}}\right)$ for all concentrations even at reaction concentrations at equilibrium. The reaction quotient $Q$ can also help in prediction of direction of reaction when compared with the equilibrium constant $\left(K_{\text{eq}}\right)$.

The general equilibrium reaction is as follows:

  $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

Here,

$\text{A}$ and $\text{B}$ are the reactants.

$\text{C}$ and $\text{D}$ are products.

$a$ and $b$ are the stoichiometric coefficients of reactants.

$c$ and $d$ are the stoichiometric coefficients of products.

The expression of the equilibrium constant for the above reaction is as follows:

  $Q=\frac{P_{\text{C}}^c{P}_{\text{D}}^d}{P_{\text{A}}^a{P}_{\text{B}}^b}$

Here,

$Q$ is the reaction quotient.

$P_{\text{C}}$ is the partial pressure of $\text{C}$.

$P_{\text{D}}$ is the partial pressure of $\text{D}$.

$P_{\text{A}}$ is the partial pressure of $\text{A}$.

$P_{\text{B}}$ is the partial pressure of $\text{B}$.

The concentration of reactants and products changes in order to bring reaction quotient and equilibrium constant closer. Therefore, the direction of reaction can be predicted as follows:

(1) If $Q$ is less than $K_{\text{eq}}$ then the reaction moves to increase the concentration of products, which means reaction moves towards right direction.

(2) If $Q$ is greater than $K_{\text{eq}}$ then the reaction moves to decrease the concentration of products, that means reaction moves towards the left direction.

(3) If $Q$ is equal to $K_{\text{eq}}$ then the reaction is in equilibrium.

Answer to Problem 14.58QE

The pressure of ${\text{PCl}}_{\text{5}}$ is $1.361\text{ atm}$, that of ${\text{PCl}}_3$ is $\text{1}\text{.624 atm}$ and that of ${\text{Cl}}_2$ is $\text{0}\text{.873 atm}$.

Explanation of Solution

The given reaction is as follows:

  ${\text{PCl}}_3\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\rightleftharpoons {\text{PCl}}_{\text{5}}\left(\text{g}\right)$

The mathematical expression for the ideal gas equation is as follows:

  $PV=nRT$        (1)

Here,

$P$ is the pressure of the gas.

$V$ is the volume of the gas.

$n$ is the moles of the gas.

$R$ is the universal gas constant.

$T$ is the absolute temperature of the gas.

Rearrange equation (1) to calculate the pressure of gas.

  $P=\frac{nRT}{V}$        (2)

The formula to convert degree Celsius to Kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\text{ K}$        (3)

Substitute $\text{332 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (3) to calculate $T$.

  $\begin{array}{c}T\left(\text{K}\right)=332\text{ °C}+273.15\text{ K}\\ =605.15\text{ K}\end{array}$

Substitute $0.15\text{ mol}$ for $n$, $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $605.15\text{ K}$ for $T$ and $10\text{ L}$ for $V$ in equation (2) to calculate the pressure of ${\text{PCl}}_{\text{3}}$.

  $\begin{array}{c}{P}_{{\text{PCl}}_{\text{3}}}=\frac{\left(0.15\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(605.15\text{ K}\right)}{10\text{ L}}\\ =0.744\text{ atm}\end{array}$

Substitute $0.20\text{ mol}$ for $n$, $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $605.15\text{ K}$ for $T$ and $10\text{ L}$ for $V$ in equation (2) to calculate the pressure of ${\text{Cl}}_2$.

  $\begin{array}{c}{P}_{{\text{Cl}}_2}=\frac{\left(0.20\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(605.15\text{ K}\right)}{10\text{ L}}\\ =0.993\text{ atm}\end{array}$

Substitute $0.25\text{ mol}$ for $n$, $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $605.15\text{ K}$ for $T$ and $10\text{ L}$ for $V$ in equation (2) to calculate the pressure of ${\text{PCl}}_{\text{5}}$.

  $\begin{array}{c}{P}_{{\text{PCl}}_{\text{5}}}=\frac{\left(0.25\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(605.15\text{ K}\right)}{10\text{ L}}\\ =1.241\text{ atm}\end{array}$

The formula to calculate $Q$ for the given reaction is as follows:

  $Q=\frac{P_{{\text{PCl}}_{\text{5}}}}{P_{{\text{PCl}}_{\text{3}}}{P}_{{\text{Cl}}_2}}$        (4)

Substitute $1.241\text{ atm}$ for $P_{{\text{PCl}}_{\text{5}}}$, $0.993\text{ atm}$ for $P_{{\text{Cl}}_2}$ and $0.744\text{ atm}$ for $P_{{\text{PCl}}_{\text{3}}}$ in equation (4).

  $\begin{array}{c}Q=\frac{1.241\text{ atm}}{\left(0.744\text{ atm}\right)\left(\text{0}\text{.993 atm}\right)}\\ =1.679\end{array}$

The value of $Q$ is less than $K_{\text{p}}$ so the reaction is not in equilibrium but proceeds in forward direction. So the pressures calculated are not equilibrium pressures.

The ICE table for the above reaction is as follows:

$\begin{array}{cccccc}\text{Equation}& {\text{PCl}}_3& +& {\text{Cl}}_{\text{2}}& \rightleftharpoons & {\text{PCl}}_{\text{5}}\\ \text{Initial}\left(\text{atm}\right)& 0.744& & 0.993& & 1.241\\ \text{Change}\left(\text{atm}\right)& -x& & -x& & +x\\ \text{Equilibrium}\left(\text{atm}\right)& 0.744-x& & 0.993-x& & 1.241+x\end{array}$

The expression to calculate $K_{\text{p}}$ for the given reaction is as follows:

  $K_{\text{p}}=\frac{P_{{\text{PCl}}_{\text{5}}}}{P_{{\text{PCl}}_{\text{3}}}{P}_{{\text{Cl}}_2}}$        (5)

Substitute $1.241+x$ for $P_{{\text{PCl}}_{\text{5}}}$, $0.744-x$ for $P_{{\text{PCl}}_3}$ and $0.993-x$ for $P_{{\text{Cl}}_2}$ and 2.5 for $K_{\text{p}}$ in equation (5).

  $\begin{array}{l}2.5=\frac{\left(1.241+x\right)}{\left(0.744-x\right)\left(0.993-x\right)}\\ x=\left[\left(2.5\right)\left(0.744-x\right)\left(0.993-x\right)\right]-1.241\end{array}$

Rearrange the above equation as follows:

  $x^2-2.137x+0.2424=0$

Solve the quadratic equation for $x$.

  $x=0.12\text{ atm}$

Or,

  $x=2.01\text{ atm}$

But $x=2.01\text{ atm}$ is rejected as it will result in negative pressures of both ${\text{PCl}}_3$ and ${\text{Cl}}_{\text{2}}$.

The pressure of ${\text{PCl}}_{\text{5}}$ can be calculated as follows:

  $\begin{array}{c}{P}_{{\text{PCl}}_5}=1.241\text{ atm}+0.12\text{ atm}\\ =\text{1}\text{.361 atm}\end{array}$

The pressure of ${\text{PCl}}_3$ can be calculated as follows:

  $\begin{array}{c}{P}_{{\text{PCl}}_3}=0.744\text{ atm}-0.12\text{ atm}\\ =\text{1}\text{.624 atm}\end{array}$

The pressure of ${\text{Cl}}_{\text{2}}$ can be calculated as follows:

  $\begin{array}{c}{P}_{{\text{Cl}}_{\text{2}}}=0.993\text{ atm}-0.12\text{ atm}\\ =0.873\text{ atm}\end{array}$