Chapter 14, Problem 14.45QP

Obtain the value of K_c for the following reaction at 900 K:

${\text{2SO}}_2(g)+{\text{O}}_2(g)\stackrel{}{\rightleftharpoons }2{\text{SO}}_3(g)$

Use the data given in Problem 14.33.

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept introduction:

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 14.45QP

The equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ for the given reaction was found to be $3.1\times {10}^3$

Explanation of Solution

Given,

The initial amount of ${\text{SO}}_2\text{=}0.0400\text{mol}$

The initial amount of ${\text{O}}_{\text{2}}$ $\text{= 0}\text{.0200 mol}$

The number of moles of ${\text{SO}}_3$ at equilibrium $\text{= 0}\text{.0296 mol}$

The volume of the vessel $\text{=2}\text{.00}0\text{L}$

To find the equilibrium composition

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

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

$\begin{array}{l}\text{Initial}\text{0}\text{.0400}\text{ 0}\text{.0200}\rightleftharpoons \text{0}\text{}\\ \text{Change}\text{-x}-\frac{x}{2}\rightleftharpoons +x\\ \text{Equillibrium}\text{0}\text{.0400}-x\text{0}\text{.0200}\text{-}\frac{\text{x}}{2}\rightleftharpoons x\end{array}$

The amount ${\text{SO}}_{\text{3}}$ of at equilibrium is $\text{0}\text{.0296}\text{mol}$ which is equal to the change in the amount $\left(x\right)$.

From the equilibrium reaction we can see that the coefficient of ${\text{SO}}_2$ and ${\text{SO}}_{\text{3}}$ are same.  But the coefficient of ${\text{O}}_{\text{2}}$ is half of that of the product.  Hence, we can conclude that an amount of $\frac{\text{x}}{2}$ is lost from the ${\text{O}}_{\text{2}}$ to form the given amount of product.

The equilibrium compositions of each of the reactant and product are found as given below.

Equilibrium amount of ${\text{SO}}_2\text{(g)}\text{=}\text{0}\text{.0400- x}\text{=}0.0400\text{-0}\text{.0296 =}0.0104\text{mol}$

Equilibrium amount of ${\text{O}}_{\text{2}}(g)\text{=0}\text{.0200 - }\frac{x}{2}\text{=0}\text{.0200- }\left(\frac{0.0296}{2}\right)\text{= 0}\text{.0052mol}$

Equilibrium amount of ${\text{SO}}_3(g)\text{=x}\text{=}\text{0}\text{.0296mol}$

To find the concentrations of the reactants and products

The molar concentration of each can be obtained as given below.

$\begin{array}{l}{\text{The equilibrium concentration of SO}}_2\text{= }\frac{\text{Num of moles}}{\text{Volume}}\\ \text{=}\frac{\text{0}\text{.0104mol}}{\text{2}\text{.00L}}\\ =\text{0}\text{.0052M}\end{array}$

$\begin{array}{l}{\text{The concentration of O}}_{\text{2}}\text{at equilibrium}\text{= }\frac{\text{Num of moles}}{\text{Volume}}\\ \text{=}\frac{\text{0}\text{.0052mol}}{2.000L}\\ =\text{0}\text{.0026M}\end{array}$

$\begin{array}{l}{\text{The concentration of SO}}_3\text{at equilibrium}\text{= }\frac{\text{Num of moles}}{\text{Volume}}\\ \text{=}\frac{\text{0}\text{.0296mol}}{\text{2}\text{.000L}}\\ =\text{0}\text{.0148M}\end{array}$

To find the expression for the equilibrium constant

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

Equilibrium constant 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 of each terms in the chemical reaction.  The equilibrium constant for the given reaction is expressed as given below.

The equilibrium constant ${\text{K}}_{\text{c}}\text{=}\frac{{\left[{\text{SO}}_3\right]}^2}{{\left[{\text{SO}}_{\text{2}}\right]}^2\left[{\text{O}}_{\text{2}}\right]}$

To find the value for the equilibrium constant

The value of the equilibrium constant can be obtained by substituting the equilibrium concentrations of the reactant and product in the expression for the equilibrium constant.

$\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\frac{{\left[{\text{SO}}_3\right]}^2}{{\left[{\text{SO}}_{\text{2}}\right]}^2\left[{\text{O}}_{\text{2}}\right]}\\ =\frac{{\left[0.0148\right]}^2}{{\left[0.0052\right]}^2\left[0.0026\right]}\\ =3.1\times {10}^3\end{array}$

Conclusion

The equilibrium composition of the given reaction mixture was found from the initial amount of reactants and products.  The value of the equilibrium constant was then obtained by substituting the equilibrium concentrations of the products and the reactants in the expression for the equilibrium constant.