Chapter 14, Problem 14.103QP

The equilibrium constant K_c for the reaction

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

equals 4.1 at 300°C.

1. a A sample of 35.8 g of PCl₅ is placed in a 5.0-L reaction vessel and heated to 300°C. What are the equilibrium concentrations of all of the species?
2. b What fraction of PCl₅ has decomposed?
3. c If 35.8 g of PCl₅ were placed in a 1.0-L vessel, what qualitative effect would this have on the fraction of PCl₅ that has decomposed (give a qualitative answer only; do not do the calculation)? Why?

Interpretation Introduction

Interpretation:

The equilibrium composition of the given reaction mixture has to be found and the fraction of decomposition of ${\text{PCl}}_5$ also has to be found. The effect of addition of more amount of ${\text{PCl}}_5$ to the vessel with less volume has to be explained.

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.

Answer to Problem 14.103QP

The equilibrium mixture contains $0.031M$ of ${\text{PCl}}_{\text{3}}$, $0.031M$ of ${\text{Cl}}_2$ and $\text{0}\text{.004M}$ of ${\text{PCl}}_5$

Explanation of Solution

Given,

The weight of ${\text{PCl}}_{\text{3}}=35.8g$

Molecular weight of ${\text{PCl}}_{\text{3}}=208.22{\text{gmol}}^{\text{-1}}$

Volume of the vessel $\text{=1 L}$

To find the initial concentration of ${\text{PCl}}_{\text{5}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles=}\frac{\text{Weight}\text{in}\text{grams}}{\text{Gram}\text{molecular}\text{weight}}\\ =\frac{35.8}{208.22}\\ =0.172\end{array}$

The concentration of ${\text{PCl}}_{\text{5}}$ is calculated by plugging in the values of and in the given equation.

$\begin{array}{l}{\text{Concentration of PCl}}_5\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}}\\ =\frac{0.172}{5.0L}\\ =0.0344M\end{array}$

To find the equilibrium composition.

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

$\begin{array}{l}\text{Amount}\text{(M)}{\text{PCl}}_{\text{3(g)}}\text{+}{\text{Cl}}_{\text{2(g)}}\rightleftharpoons {\text{PCl}}_{\text{(g)}}\\ \end{array}$

$\begin{array}{l}\text{Initial}00\rightleftharpoons \text{0}\text{.0344}\text{}\\ \text{Change}+x+x\rightleftharpoons -x\\ \text{Equillibrium}xx\rightleftharpoons 0.0344-x\end{array}$

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

${\text{K}}_c\text{=4}\text{.1=}\frac{\left[{\text{PCl}}_{\text{3}}\right]}{\left[{\text{PCl}}_{\text{3}}\right]\left[{\text{Cl}}_{\text{2}}\right]}$

$4.1=\frac{0.0344-x}{x^2}$

On rearranging we get a quadratic equation.

$4.1x^2+x-0.0344=0$

On solving the quadratic equation the value of x obtained.

$x=\frac{-1\pm \sqrt{{(1)}^2-4(4.1)(-0.0344)}}{2(4.1)}$

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

$\text{x}=0.0306$

Hence,

$\text{The equilibrium concentration of}{\text{PCl}}_{\text{3}}{\text{=Cl}}_2=x=0.031M$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{PCl}}_{\text{5}}\text{= 0}\text{.0344-0}\text{.0306}\\ \text{=0}\text{.004M}\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium composition of the given reaction mixture has to be found and the fraction of decomposition of ${\text{PCl}}_5$ also has to be found. The effect of addition of more amount of ${\text{PCl}}_5$ to the vessel with less volume has to be explained.

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.

Answer to Problem 14.103QP

The fraction of ${\text{PCl}}_5$ decomposed is found to be $0.89$

Explanation of Solution

To find the fraction of ${\text{PCl}}_{\text{5}}$decomposed.

$\begin{array}{l}\text{The fraction of decomposed=}\frac{\text{Initial}\text{Amount}}{\text{Final}\text{Amount}}\\ =\frac{0.0306}{0.0344}\\ =0.89\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium composition of the given reaction mixture has to be found and the fraction of decomposition of ${\text{PCl}}_5$ also has to be found. The effect of addition of more amount of ${\text{PCl}}_5$ to the vessel with less volume has to be explained.

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.

Answer to Problem 14.103QP

Increase in pressure will shift equilibrium towards left and the rate of decomposition of ${\text{PCl}}_5$ decreases.

Explanation of Solution

Initially $35.8g$ of ${\text{PCl}}_{\text{5}}$ is added to $\text{5}\text{.0L}$ vessel.  In the second case the volume of the vessel is reduced to 1 L.  When we are adding the same initial amount of ${\text{PCl}}_{\text{5}}$, the pressure in the system increase. Increase in pressure will shift equilibrium towards left and the rate of decomposition decreases.