Chapter 14, Problem 14.107QP

Sulfuryl chloride is used in organic chemistry as a chlorinating agent. At moderately high temperatures it decomposes as follows:

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

with K_c = 0.045 at 650 K.

1. a A sample of 8.25 g of SO₂Cl₂ is placed in a 1.00-L reaction vessel and heated to 650 K. What are the equilibrium concentrations of all of the species?
2. b What fraction of SO₂Cl₂ has decomposed?
3. c If 5 g of chlorine is inserted into the reaction vessel, what qualitative effect would this have on the fraction of SO₂Cl₂ that has decomposed?

Interpretation Introduction

Interpretation:

The equilibrium composition of the given reaction mixture has to be found and the fraction of decomposition of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ also has to be found. The effect of addition of more amount of chlorine to the vessel 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 14.107QP

The equilibrium mixture contains $\text{0}\text{.03M}$ of ${\text{SO}}_{\text{2}}$, $\text{0}\text{.03M}$ of ${\text{Cl}}_2$ and $\text{0}\text{.026M}$ $\text{0}\text{.004M}$ of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$

Explanation of Solution

Given,

The weight of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}=8.25g$

Molecular weight of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}=134.96{\text{gmol}}^{\text{-1}}$

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

Equilibrium constant ${\text{K}}_{\text{c}}\text{=0}\text{.045}$

To find the initial concentration of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles=}\frac{\text{Weight}\text{in}\text{grams}}{\text{Gram}\text{molecular}\text{weight}}\\ =\frac{8.25}{134.96}\\ =\text{0}\text{.0612}\text{mol}\end{array}$

The concentration of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is calculated by plugging in the values of and in the given equation.

$\begin{array}{l}{\text{Concentration of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}}\\ =\frac{0.0612}{1.0L}\\ =0.06113M\end{array}$

To find the equilibrium composition.

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

$\begin{array}{l}\text{Amount}\text{(M)}{\text{SO}}_2{\text{Cl}}_{\text{2(g)}}\rightleftharpoons {\text{SO}}_{\text{2}}{}_{\text{(g)}}\text{+}{\text{Cl}}_{\text{2(g)}}\\ \end{array}$

$\begin{array}{l}\text{Initial}0.06113\rightleftharpoons \text{0}\text{0}\text{}\\ \text{Change}-x\rightleftharpoons +x+x\\ \text{Equillibrium}0.06113-x\rightleftharpoons xx\end{array}$

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

${\text{K}}_c\text{=}\frac{\left[{\text{SO}}_{\text{2}}\right]\left[{\text{Cl}}_{\text{2}}\right]}{\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}$

$0.045=\frac{x^2}{(0.06113-x)}$

On rearranging we get a quadratic equation.

$x^2+0.045x-0.00275=0$

On solving the quadratic equation the value of x obtained.

$x=\frac{-(0.045)\pm \sqrt{{(0.045)}^2-4(1)(-0.002759)}}{2(1)}$

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

$\text{x}=0.03457$

Hence,

$\text{The equilibrium concentration of}{\text{SO}}_{\text{2}}{\text{=Cl}}_{\text{2}}\text{=x=0}\text{.03M}$

$\begin{array}{l}\text{The equilibrium concentration of}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{= 0}\text{.06113-0}\text{.03457}\\ \text{=0}\text{.026M}\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium composition of the given reaction mixture has to be found and the fraction of decomposition of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ also has to be found. The effect of addition of more amount of chlorine to the vessel 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 14.107QP

The fraction of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ decomposed is found to be $0.6$

Explanation of Solution

To find the fraction of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$decomposed.

$\begin{array}{l}\text{The fraction of decomposed=}\frac{\text{Initial}\text{Amount}}{\text{Final}\text{Amount}}\\ =\frac{0.03457}{0.06113}\\ =0.6\end{array}$

Interpretation Introduction

Interpretation:

The equilibrium composition of the given reaction mixture has to be found and the fraction of decomposition of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ also has to be found. The effect of addition of more amount of chlorine to the vessel 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 14.107QP

Addition of more amount of chlorine will shift equilibrium towards left and the rate of decomposition of  ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ decreases.

Explanation of Solution

According to Le Chatelier’s principle a system at equilibrium is disturbed, it will try to get the equilibrium state back by nullifying the disturbance.  Addition of more product to the reaction mixture will shift the equilibrium towards left and the rate of decomposition decreases.