Chapter 14, Problem 14.137QP

Phosphorus pentachloride, PCl₅, decomposes on heating to give phosphorus trichloride, PCl₅, and chlorine.

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

A closed 2.90-L vessel initially contains 0.0564 mol PCl₅. What is the total pressure at 250°C when equilibrium is achieved? The value of K_c at 250°C is 4.15 × 10⁻².

Interpretation Introduction

Interpretation:

The total pressure of the system at equilibrium has to be found.

Concept introduction:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$: A system is said to be in equilibrium when all the measurable properties of the system remains unchanged with the time.  Equilibrium constant is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where the reactant A is giving product B.

$\text{A}\rightleftharpoons \text{B}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}\left[\text{A}\right]{\text{=k}}_{\text{r}}\left[\text{B}\right]\end{array}$

On rearranging,

$\frac{\left[\text{A}\right]}{\left[\text{B}\right]}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}={\text{K}}_{\text{c}}$

Where,

${\text{k}}_f$ is the rate constant of the forward reaction.

${\text{k}}_r$ is the rate constant of the reverse reaction.

$K_c$ is the equilibrium constant.

Ideal gas equation is an equation that is describing the state of a imaginary ideal gas.

${\text{PV}}_{}\text{=n RT}$

Where,

P is the pressure of the gas

V is the volume

n is the number of moles of gas

R is the universal gas constant $\text{(R=0}{\text{.0821LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}})$

T is the temperature

Answer to Problem 14.137QP

The total pressure of  the system at equilibrium is found to be $\text{1}\text{.45atm}$

Explanation of Solution

Given,

The initial amount of ${\text{PCl}}_5\text{=}0.0564\text{mol}$

The volume of the vessel $\text{=}2.90\text{L}$

The equilibrium constant ${\text{K}}_{\text{c}}=4.15\times {10}^{-2}$

$\begin{array}{l}\text{Temperature}{\text{=250}}^{°}\text{C}\\ \text{=250+273}\text{.15}\\ \text{=}\text{523}\text{.15K}\end{array}$

To find initial concentration of reactant ${\mathrm{PCl}}_5$

The initial concentrations of the ${\text{H}}_2$ is found as given below.

$\begin{array}{l}\text{Initial concentration of}{\mathrm{PCl}}_5\text{=}\frac{\text{Num of moles}}{\text{Volume}}\\ =\frac{\text{0}\text{.0564mol}}{2.90\text{L}}\\ =0.01945\text{M}\end{array}$

To find the equilibrium composition.

Using the table approach, the equilibrium concentrations of the reactants and the products can be found.

$\text{Amount}\text{(M)}{\text{PCl}}_{\text{5(g)}}\rightleftharpoons {\text{PCl}}_3{}_{\text{(g)}}\text{+}{\text{Cl}}_2{}_{\text{(g)}}$

$\begin{array}{l}\text{Initial}0.01945\rightleftharpoons \text{0}\text{}0\\ \text{Change}\text{-x}\rightleftharpoons +x+x\\ \text{Equillibrium}\left(0.01945-x\right)\rightleftharpoons xx\end{array}$

The equilibrium concentration values are then substituted into the equilibrium expression to get the change in concentration x.

$\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\frac{\left[{\text{PCl}}_{\text{3}}\right]\left[{\text{Cl}}_{\text{2}}\right]}{\left[{\text{PCl}}_{\text{5}}\right]}\\ \text{4}\text{.15 }\times {\text{10}}^{-2}=\frac{x^2}{\text{(0}\text{.01945 )}}\end{array}$

On rearranging we get a quadratic equation.

$x^2+(4.15\times {10}^{-2})x-8.071\times {10}^{-4}=0$

On solving the quadratic equation the value of x obtained.

$x=\frac{-(4.15\times {\text{10}}^{-2})\pm \sqrt{{(4.15\times {\text{10}}^{-2})}^2-4(1)(\text{8}\text{.071}\times {\text{10}}^{-2})}}{2(1)}$

$\text{x}=-\text{5}\text{.59}\times {10}^{-2}\text{or}\text{1}\text{.443}\times {10}^{-2}$

On solving we get two values for x, the logical value for x is taken.

$\text{x}=\text{1}\text{.443}\times {10}^{-2}\text{M}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{PCl}}_5\text{=}0.01945\text{M-x}\\ \text{=}0.01945\text{M-1}\text{.443}\times {10}^{-2}\text{M}\\ \text{=}\text{0}\text{.00502M}\end{array}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{PCl}}_3\text{=}\text{x}\\ \text{=1}\text{.443}\times {10}^{-2}\text{M}\end{array}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{Cl}}_2\text{=}\text{x}\\ \text{=1}\text{.443}\times {10}^{-2}\text{M}\end{array}$

Total concentration of the reaction mixture at equilibrium is the sum of the concentration of products and reactants.

$\left[{\text{PCl}}_{\text{5}}\right]+\left[{\text{PCl}}_{\text{3}}\right]\text{+}\left[{\text{Cl}}_{\text{2}}\right]=\text{0}\text{.00502M}+\text{1}\text{.443}\times {10}^{-2}\text{M}+\text{1}\text{.443}\times {10}^{-2}\text{M}=\text{0}\text{.03388 M}$

To find the total pressure

At constant temperature,

$\begin{array}{l}{\text{PV}}_{}\text{=n RT}\\ \text{P=}\frac{\text{n RT}}{\text{V}}\end{array}$

Where $\frac{\text{n}}{\text{V}}$ is the concentration.

The total concentration is substituted in the ideal gas equation to get the total pressure.

$\begin{array}{l}\text{P}\text{=}\frac{\text{n RT}}{\text{V}}\\ \text{P}\text{=(0}\text{.03388 M)(0}{\text{.082057LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)(523}\text{.15K)}\\ \text{=}\text{1}\text{.45atm}\end{array}$

Conclusion

The equilibrium composition of the given reaction mixture was found.  The total concentration of the equilibrium mixture is then substituted in the ideal gas equation to get the value of total pressure.