Chapter 14, Problem 14.87QE

(a)

Interpretation Introduction

Interpretation:

The concentration of all species after the attainment of equilibrium in the following reaction has to be determined.

  ${\text{2SO}}_3\left(\text{g}\right)\rightleftharpoons 2{\text{SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

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 balances 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.

Answer to Problem 14.87QE

The concentration of ${\text{O}}_{\text{2}}$ is $0.0020M$, that of ${\text{SO}}_{\text{2}}$ is $0.0040M$ and that of ${\text{SO}}_{\text{3}}$ is $0.001M$.

Explanation of Solution

The given reaction occurs as follows:

  ${\text{2SO}}_3\left(\text{g}\right)\rightleftharpoons 2{\text{SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

The concentration of ${\text{SO}}_3$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{SO}}_3\right]=\frac{\text{0}\text{.010 mol}}{2\text{ L}}\\ =0.005M\end{array}$

The ICE table for the above reaction is as follows:

$\begin{array}{cccccc}\text{Equation}& {\text{2SO}}_3& \rightleftharpoons & {\text{2SO}}_2& +& {\text{O}}_{\text{2}}\\ \text{Initial}\left(M\right)& 0.005& & 0& & 0\\ \text{Change}\left(M\right)& -x& & +2x& & +x\\ \text{Equilibrium}\left(M\right)& 0.005-x& & 2x& & x\end{array}$

The concentration of ${\text{O}}_{\text{2}}$ at equilibrium can be calculated as follows:

  $\begin{array}{c}\left[{\text{O}}_{\text{2}}\right]=\frac{\text{0}\text{.0040 mol}}{2\text{ L}}\\ =0.0020M\end{array}$

The concentration of ${\text{O}}_{\text{2}}$ at equilibrium is $0.0020M$. Therefore the value of $x$ is $0.0020M$.

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

  $\begin{array}{c}\left[{\text{SO}}_{\text{2}}\right]=2\left(0.0020M\right)\\ =0.0040M\end{array}$

Therefore the equilibrium concentration of ${\text{SO}}_{\text{2}}$ is $0.0040M$.

The equilibrium concentration of ${\text{SO}}_{\text{3}}$ can be calculated as follows:

  $\begin{array}{c}\left[{\text{SO}}_{\text{2}}\right]=\left(0.005M-2\left(0.0020M\right)\right)\\ =0.001M\end{array}$

Therefore the equilibrium concentration of ${\text{SO}}_{\text{3}}$ is $0.001M$.

(b)

Interpretation Introduction

Interpretation:

The value of $K_{\text{c}}$ for the following reaction has to be determined.

  ${\text{2SO}}_3\left(\text{g}\right)\rightleftharpoons 2{\text{SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

Concept Introduction:

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:

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

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 14.87QE

The value of $K_{\text{c}}$ for the given reaction is 0.032.

Explanation of Solution

The given reaction occurs as follows:

  ${\text{2SO}}_3\left(\text{g}\right)\rightleftharpoons 2{\text{SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

The expression of $K_{\text{c}}$ for the above reaction is as follows:

  $K_{\text{c}}=\frac{{\left[{\text{SO}}_2\right]}^2\left[{\text{O}}_2\right]}{{\left[{\text{SO}}_3\right]}^2}$        (1)

Substitute $0.0040M$ for $\left[{\text{SO}}_{\text{2}}\right]$, $0.0020M$ for $\left[{\text{O}}_{\text{2}}\right]$ and $0.001M$ for $\left[{\text{SO}}_{\text{3}}\right]$ in equation (1).

  $\begin{array}{c}{K}_{\text{c}}=\frac{{\left(0.0040\right)}^2\left(0.0020\right)}{{\left(0.001\right)}^2}\\ =0.032\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $K_{\text{p}}$ for the following reaction has to be determined.

  ${\text{2SO}}_3\left(\text{g}\right)\rightleftharpoons 2{\text{SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

Concept Introduction:

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

  $K_{\text{p}}=K_{\text{c}}{\left(RT\right)}^{\Delta n}$        (2)

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

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

  $\Delta n=\left(\begin{array}{l}\text{total number of moles of}\\ \text{gas on the product side}\end{array}\right)-\left(\begin{array}{l}\text{total number of moles of}\\ \text{gas on the reactant side}\end{array}\right)$        (3)

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

Answer to Problem 14.87QE

The value of $K_{\text{p}}$ for the given reaction is 2.9.

Explanation of Solution

The given reaction occurs as follows:

  ${\text{2SO}}_3\left(\text{g}\right)\rightleftharpoons 2{\text{SO}}_2\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

Substitute 3 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 (3).

  $\begin{array}{c}\Delta n=3-\text{2}\\ \text{=1}\end{array}$

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 $\text{832 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (4) to calculate $T$.

  $\begin{array}{c}T\left(\text{K}\right)=\text{832 }°\text{C}+273.15\text{ K}\\ =1105.15\text{ K}\end{array}$

Substitute 1 for $\Delta n$, 0.032 for $K_{\text{c}}$, $1105.15\text{ K}$ for $T$ and $\text{0}\text{.08206 L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (2).

  $\begin{array}{c}{K}_{\text{p}}=\left(0.032\right){\left(\left(\text{0}\text{.08206}\right)\left(1105.15\right)\right)}^1\\ =2.9\end{array}$