Chapter 15, Problem 56GQ

Sulfuryl chloride, SO₂Cl₂ is used as a reagent in the synthesis of organic compounds. When heated to a sufficiently high temperature, it decomposes to SO₂ and Cl₂.

SO₂Cl₂(g) ⇄ SO₂(g) + Cl₂(g)    Kc = 0.045 at 375 °C

1. (a) A 10.0-L flask containing 6.70 g of SO₂Cl₂ is heated to 375 °C. What is the concentration of each of the compounds in the system when equilibrium is achieved? What fraction of SO₂Cl₂ has dissociated?
2. (b) What are the concentrations of SO₂Cl₂, SO₂, and Cl₂ at equilibrium in the 10.0-L flask at 375 °C if you begin with a mixture of SO₂Cl₂ (6.70 g) and Cl₂ (0.10 atm)? What fraction of SO₂Cl₂ has dissociated?
3. (c) Compare the fractions of SO₂Cl₂ in parts (a) and (b). Do they agree with your expectations based on Le Chatelier’s principle?

Interpretation Introduction

Interpretation:

The concentration of the compounds in the decomposition of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at equilibrium and the fraction of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ that has dissociated has to be identified.

Concept Introduction:

Equilibrium constant in terms of concentration${\text{K}}_{\text{C}}$:

$\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{C}}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}\text{×}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}\text{×}{\left[\text{B}\right]}^{\text{b}}}\end{array}$

$\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{(L)}}$

$\text{Ideal}\text{gas}\text{equation}\text{:}\text{PV}\text{=}\text{nRT}$

Le Chatelier’s principle: If equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Factor’s that effect chemical equilibria:

Concentration – Equilibrium will be affected by changing the concentration of reactant or product. If we increase the concentration of reactant system will try to reverse the change by favouring forward reaction and thus increase the concentration of products. Like wise adding products increase yield of reactants.

Temperature – When the temperature increases equilibrium will shift in the endothermic direction, in the direction that absorbs heat. When the temperature decreases equilibrium will shift in the exothermic direction, in the direction that releases heat.

Explanation of Solution

Given:

$\begin{array}{l}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftarrows {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\\ {\text{K}}_{\text{C}}\text{=}\text{0}\text{.045}\\ \text{Mass}\text{of}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{=}\text{6}\text{.70}\text{g}\\ \text{T}\text{=}\text{375°C}\text{=}\text{648}\text{K}\\ \text{V}\text{=}\text{10}\text{L}\end{array}$

$\begin{array}{l}{\text{n}}_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\text{=}\frac{\text{mass}\text{of}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{6}\text{.7}\text{g}}{\text{135}\text{g}/\text{mol}}\\ \text{=}\text{0}\text{.0496}\text{mol}\end{array}$

$\begin{array}{l}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftarrows {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\\ \text{Initial}\text{0}\text{.04963}\text{0}\text{0}\\ \text{Change}\text{x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{0}\text{.04963}\text{-x}\text{x}\text{x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{C}}\text{=}\frac{\left[{\text{SO}}_{\text{2}}\right]\left[{\text{Cl}}_{\text{2}}\right]}{\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}\\ \text{0}\text{.045}\text{=}\frac{{\text{x}}^{\text{2}}}{\left(\text{0}\text{.04963}\text{-x}\right)}\end{array}$

Solving this equation we get

$\text{x}\text{=}\text{0}\text{.02984}$

$\begin{array}{l}\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]\text{=}\text{0}\text{.04963}\text{-x}\\ \text{=}\text{0}\text{.04963}\text{-}\text{0}\text{.02984}\\ \text{=}\text{0}\text{.01979}\text{M}\end{array}$

$\left[{\text{SO}}_{\text{2}}\right]\text{=}\left[{\text{Cl}}_{\text{2}}\right]\text{=}\text{x}\text{=}\text{0}\text{.02984}\text{M}$

$\begin{array}{l}\text{Fraction}\text{of}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{dissociation}\text{=}\frac{\text{x}}{\text{0}\text{.04963}}\\ \text{=}\frac{\text{0}\text{.02984}}{\text{0}\text{.04963}}\\ \text{=}\text{0}\text{.601}\end{array}$

Interpretation Introduction

Interpretation:

The concentration of the compounds in the decomposition of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at equilibrium and the fraction of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ that has dissociated has to be identified.

Concept Introduction:

Equilibrium constant in terms of concentration${\text{K}}_{\text{C}}$:

$\begin{array}{l}\text{aA}{}_{\text{(g)}}\text{+}{\text{bB}}_{\text{(g)}}\rightleftarrows \text{cC}{}_{\text{(g)}}\text{+}\text{dD}{}_{\text{(g)}}\\ {\text{K}}_{\text{C}}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}\text{×}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}\text{×}{\left[\text{B}\right]}^{\text{b}}}\end{array}$

$\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{(L)}}$

$\text{Ideal}\text{gas}\text{equation}\text{:}\text{PV}\text{=}\text{nRT}$

Le Chatelier’s principle: If equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Factor’s that effect chemical equilibria:

Concentration – Equilibrium will be affected by changing the concentration of reactant or product. If we increase the concentration of reactant system will try to reverse the change by favouring forward reaction and thus increase the concentration of products. Likewise adding products increase yield of reactants.

Temperature – When the temperature increases equilibrium will shift in the endothermic direction, in the direction that absorbs heat. When the temperature decreases equilibrium will shift in the exothermic direction, in the direction that releases heat.

Answer to Problem 56GQ

The fraction of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ dissociated when reaction start with a mixture of $\text{6}\text{.70}\text{g}$ of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{and}{\text{Cl}}_{\text{2}}\text{with}\text{a}\text{pressure}\text{of}\text{0}\text{.10}\text{atm}$ is

Explanation of Solution

Given:

$\begin{array}{l}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftarrows {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\\ {\text{K}}_{\text{C}}\text{=}\text{0}\text{.045}\\ \text{Mass}\text{of}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{=}\text{6}\text{.70}\text{g}\\ \text{T}\text{=}\text{375°C}\text{=}\text{648}\text{K}\\ \text{V}\text{=}\text{10}\text{L}\\ {\text{P}}_{{\text{Cl}}_{\text{2}}}\text{=}\text{0}\text{.10}\text{atm}\end{array}$

$\begin{array}{l}{\text{n}}_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}\text{=}\frac{\text{mass}\text{of}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}{\text{Molar}\text{mass}}\\ \text{=}\frac{\text{6}\text{.7}\text{g}}{\text{135}\text{g}/\text{mol}}\\ \text{=}\text{0}\text{.0496}\text{mol}\end{array}$

We can calculate the concentration of chlorine using ideal gas equation

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

$\begin{array}{l}\left[{\text{Cl}}_{\text{2}}\right]\text{=}\frac{\text{P}}{\text{RT}}\\ \text{=}\frac{\text{1}}{\text{0}\text{.082×648}}\\ \text{=}\text{0}\text{.0188}\text{M}\end{array}$

$\begin{array}{l}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\rightleftarrows {\text{SO}}_{\text{2}}\text{(g)}\text{+}{\text{Cl}}_{\text{2}}\text{(g)}\\ \text{Initial}\text{0}\text{.04963}\text{0}\text{0}\text{.0188}\\ \text{Change}\text{x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{0}\text{.04963}\text{-x}\text{x}\text{0}\text{.0188}\text{+x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{C}}\text{=}\frac{\left[{\text{SO}}_{\text{2}}\right]\left[{\text{Cl}}_{\text{2}}\right]}{\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}\\ \text{0}\text{.045}\text{=}\frac{\text{x}\text{×}\left(\text{0}\text{.0188+x}\right)}{\left(\text{0}\text{.04963}\text{-x}\right)}\end{array}$

Solving this equation we get

$\text{x}\text{=}\text{0}\text{.02511}$

$\begin{array}{l}\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]\text{=}\text{0}\text{.04963}\text{-x}\\ \text{=}\text{0}\text{.04963}\text{-}\text{0}\text{.02511}\\ \text{=}\text{0}\text{.02452}\text{M}\end{array}$

$\begin{array}{l}\left[{\text{SO}}_{\text{2}}\right]\text{=}\text{x}\text{=}\text{0}\text{.02511}\text{M}\\ \left[{\text{Cl}}_{\text{2}\text{.}}\right]\text{=}\text{0}\text{.0188+x}\\ \text{=}\text{0}\text{.0188}\text{+}\text{0}\text{.02511}\\ \text{=}\text{0}\text{.04391}\text{M}\end{array}$

$\begin{array}{l}\text{Fraction}\text{of}{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{dissociation}\text{=}\frac{\text{x}}{\text{0}\text{.04963}}\\ \text{=}\frac{\text{0}\text{.02511}}{\text{0}\text{.04963}}\\ \text{=}\text{0}\text{.506}\end{array}$

Interpretation Introduction

Interpretation:

The fraction of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ that has dissociated has to be compared.

Concept Introduction:

Le Chatelier’s principle: If equilibrium is disturbed by changing conditions, the system will moves the equilibrium to reverse the change.

Factor’s that effect chemical equilibria:

Concentration – Equilibrium will be affected by changing the concentration of reactant or product. If we increase the concentration of reactant system will try to reverse the change by favouring forward reaction and thus increase the concentration of products. Like wise adding products increase yield of reactants.

Temperature – When the temperature increases equilibrium will shift in the endothermic direction, in the direction that absorbs heat. When the temperature decreases equilibrium will shift in the exothermic direction, in the direction that releases heat.

Answer to Problem 56GQ

Yes, the fractions found are agree with principle.

Explanation of Solution

According to Le Chatelier’s principle: If an equilibrium is disturbed by changing conditions, the system will move the equilibrium to reverse the change.

If we increase the concentration of reactant, system will try to reverse the change by favouring forward reaction and thus increase the concentration of products. Like wise adding products increase yield of reactants.

Initially only ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is present in the system , so inorder to achieve equilibrium ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ has to decompose more. But in the part (b) chlorine gas is present initially. So ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ decompose less to reach equilibrium.