Chapter 14, Problem 14.94QE

(a)

Interpretation Introduction

Interpretation:

Whether the system reaches equilibrium or not has to be determined.

Concept Introduction:

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

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 relation of equilibrium constant $K_{\text{p}}$ and $K_{\text{c}}$ is as follows:

  $K_{\text{p}}=K_{\text{c}}{\left(RT\right)}^{\Delta n}$        (1)

Here, $\Delta n$ is change in number of moles of gas.

The formula to calculate the value of $\Delta n$ is as follows:

  $\Delta n=\left(\begin{array}{l}\text{total number of moles of}\\ \text{gas on the product side}\end{array}\right)-\left(\begin{array}{l}\text{total number of moles of}\\ \text{gas on the reactant side}\end{array}\right)$        (2)

The value of $\Delta n$ can be zero, negative and positive.

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

  $Q=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here,

$Q$ is the reaction quotient.

$\left[\text{C}\right]$ is the concentration of $\text{C}$.

$\left[\text{D}\right]$ is the concentration of $\text{D}$.

$\left[\text{A}\right]$ is the concentration of $\text{A}$.

$\left[\text{B}\right]$ is the concentration 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 14.94QE

The system is not in equilibrium.

Explanation of Solution

The given reaction is as follows:

  ${\text{SO}}_{\text{2}}\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)$

Substitute 1 for the total number of moles of gas on product side and 2 for the total number of moles of gas on reactant side in equation (2).

  $\begin{array}{c}\Delta n=1-\text{2}\\ =-1\end{array}$

Rearrange equation (1) to calculate $K_{\text{c}}$.

  $K_{\text{c}}=\frac{K_{\text{p}}}{{\left(RT\right)}^{\Delta n}}$        (3)

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

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\text{ K}$        (4)

Substitute $\text{227 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (4) to calculate $T$.

  $\begin{array}{c}T\left(\text{K}\right)=\text{227 }°\text{C}+273.15\text{ K}\\ =500.15\text{ K}\end{array}$

Substitute $-1$ for $\Delta n$, $5.1\times {10}^{-2}$ for $K_{\text{p}}$, $500.15\text{ K}$ for $T$ and $\text{0}\text{.08206 L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (3).

  $\begin{array}{c}{K}_{\text{c}}=\frac{5.1\times {10}^{-2}}{{\left(\left(\text{0}\text{.08206 L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(500.15\text{ K}\right)\right)}^{-1}}\\ =2.093\end{array}$

The concentration of ${\text{SO}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{SO}}_2\right]=\left(\frac{\text{1 g}}{1\text{ L}}\right)\left(\frac{1\text{ mol}}{64.066\text{ g}}\right)\\ =0.0156M\end{array}$

The concentration of ${\text{Cl}}_2$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{Cl}}_2\right]=\left(\frac{\text{1 g}}{1\text{ L}}\right)\left(\frac{1\text{ mol}}{70.9\text{ g}}\right)\\ =0.0141M\end{array}$

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

  $\begin{array}{c}\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]=\left(\frac{\text{45}\text{.5 g}}{1\text{ mL}}\right)\left(\frac{{10}^{-6}\text{ g}}{1\text{ g}}\right)\left(\frac{1\text{ mL}}{{10}^{-3}\text{ L}}\right)\left(\frac{1\text{ mol}}{134.96\text{ g}}\right)\\ =3.37\times {10}^{-4}M\end{array}$

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

  $Q=\frac{\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}{\left[{\text{SO}}_2\right]\left[{\text{Cl}}_{\text{2}}\right]}$        (5)

Substitute $3.37\times {10}^{-4}M$ for $\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]$, $0.0156M$ for $\left[{\text{SO}}_2\right]$ and $0.0141M$ for $\left[{\text{Cl}}_{\text{2}}\right]$ in equation (5).

  $\begin{array}{c}Q=\frac{3.37\times {10}^{-4}}{\left(0.0156\right)\left(0.0141\right)}\\ =1.53\end{array}$

Since $Q$ is not equal to $K_{\text{c}}$, the system is not in equilibrium.

(b)

Interpretation Introduction

Interpretation:

The expected mass of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ at equilibrium has to be determined.

Concept Introduction:

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

  $K_{\text{c}}=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Here,

$K_{\text{c}}$ is the equilibrium constant.

$\left[\text{C}\right]$ is the concentration of $\text{C}$.

$\left[\text{D}\right]$ is the concentration of $\text{D}$.

$\left[\text{A}\right]$ is the concentration of $\text{A}$.

$\left[\text{B}\right]$ is the concentration of $\text{B}$.

Answer to Problem 14.94QE

The expected mass of ${\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}$ is $0.058\text{ g}$.

Explanation of Solution

The given reaction is as follows:

  ${\text{SO}}_{\text{2}}\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)$

The ICE table for the above reaction is as follows:

$\begin{array}{cccccc}\text{Equation}& {\text{SO}}_{\text{2}}& +& {\text{Cl}}_2& \rightleftharpoons & {\text{SO}}_2{\text{Cl}}_2\\ \text{Initial}\left(M\right)& 0.0156& & 0.0141& & 0\\ \text{Change}\left(M\right)& -x& & -x& & +x\\ \text{Equilibrium}\left(M\right)& 0.0156-x& & 0.0141-x& & x\end{array}$

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

  $K_{\text{c}}=\frac{\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]}{\left[{\text{SO}}_{\text{2}}\right]\left[{\text{Cl}}_2\right]}$        (6)

Substitute $x$ for $\left[{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\right]$, $0.0156-x$ for $\left[{\text{SO}}_{\text{2}}\right]$, $0.0141-x$ for $\left[{\text{Cl}}_{\text{2}}\right]$ and 2.093 for $K_{\text{c}}$ in equation (6).

  $\begin{array}{l}2.093=\frac{\left(x\right)}{\left(0.0156-x\right)\left(0.0141-x\right)}\\ x=\left(2.093\right)\left(0.0156-x\right)\left(0.0141-x\right)\end{array}$

Rearrange the above equation as follows:

  $x^2-0.507x+0.000219=0$

Solve the quadratic equation for $x$.

  $x=0.00043M$

Or,

  $x=0.507M$

But $x=0.507M$ is rejected as it will result in negative concentrations of both ${\text{SO}}_{\text{2}}$ and ${\text{Cl}}_{\text{2}}$.

Therefore the concentration of ${\text{SO}}_2{\text{Cl}}_2$ is $0.00043M$. This means 0.00043 moles of ${\text{SO}}_2{\text{Cl}}_2$ are present in one litre of solution.

The mass of ${\text{SO}}_2{\text{Cl}}_2$ can be calculated as follows:

  $\begin{array}{c}{\text{Mass of SO}}_{\text{2}}{\text{Cl}}_{\text{2}}=\left(0.00043\text{ mol}\right)\left(\frac{134.96\text{ g}}{1\text{ mol}}\right)\\ =0.058\text{ g}\end{array}$