Chapter 15, Problem 15.1P

The oxidation of sulfur dioxide to give sulfurtrioxide is an important step in the industrial process for thesynthesis of sulfuric acid.
What is the equilibrium constant expression, $k_c$ ,for the reaction below?
   $2{\text{SO}}_2(g)+{\text{O}}_2\rightleftharpoons 2{\text{SO}}_{\text{3}}(g)$
What is the equilibrium constant expression, $k_c$ (reverse), forthe reaction below?
   $2{\text{SO}}_{\text{3}}(g)+\rightleftharpoons 2{\text{SO}}_{\text{2}}(g)+{\text{O}}_{\text{2}}(g)$
How is the equilibrium constant $k_c$ (reverse) related to $k_c$ ?

Interpretation Introduction

Interpretation:

The necessary expressions need to be given for the equilibrium constant & reverse equilibrium constant for the given reaction.

Concept introduction:

The equilibrium constant of a chemical reaction is the value of the reaction quotient of the specific reaction at chemical equilibrium.

For a general reaction as follows:

$aA+bB\rightleftharpoons cC+dD$

The expression for the equilibrium constant is as follows:

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

Answer to Problem 15.1P

$K_c=\frac{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{] }}{\text{mol}}^{\text{-1}}{\text{ dm}}^{\text{3}}$

Also,

${\text{K}}_{\text{c(reverse)}}\text{=}\frac{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}{\text{] mol dm}}^{\text{-3}}}{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}$

Relation:

$K_c=\frac{1}{{\text{K}}_{\text{c(reverse)}}}$

Explanation of Solution

At equilibrium, there will be no change in the concentration of products and reactants takes place.

Consider the following reaction.

The forward half arrow indicates the forward reaction where A & B reacts to form C & D. The half-back arrow indicates the backward reaction where C & D reacts to form A & B. at the chemical equilibrium these forward & backward reactions occur at the same rate. Therefore, no change in chemical composition happens. In the above reaction a, b, c & d indicates the stoichiometric coefficients.

The equilibrium constant is the ratio between the concentration of the products with the power of the stoichiometric coefficient&the same of the reactants.

So, the equilibrium constant (K_c) of the forward reaction can be expressed as,

$K_c=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

Now consider the given reaction.

To derive the expression for the equilibrium constant, consider the forward reaction. The forward reaction is the reaction between SO₂& O₂ to produce SO₃.

So, according to the above description, the equilibrium constant can be expressed as follows.

$K_c=\frac{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}{\text{mol}}^{\text{2}}{\text{ dm}}^{\text{-6}}}{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{mol}}^{\text{2}}{\text{ dm}}^{\text{-6}}{\text{[O}}_{\text{2}}{\text{] mol dm}}^{\text{-3}}}$

$K_c=\frac{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{] }}{\text{mol}}^{\text{-1}}{\text{ dm}}^{\text{3}}$

The reverse equilibrium, K_c (reverse) is related to the backward reaction. The backward reaction is the reaction of SO₃to form SO₂& O₂.

So, the equilibrium constant of the backward reaction (K_c, reverse) can be expressed as

$K_{c(reverse)}=\frac{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{mol}}^{\text{2}}{\text{ dm}}^{\text{-6}}{\text{[O}}_{\text{2}}{\text{] mol dm}}^{\text{-3}}}{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}{\text{mol}}^{\text{2}}{\text{ dm}}^{\text{-6}}}$

$K_{c(reverse)}=\frac{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}{\text{] mol dm}}^{\text{-3}}}{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}$

Consider the derived expression for K_c which is,

$K_c=\frac{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{] }}{\text{mol}}^{\text{-1}}{\text{ dm}}^{\text{3}}$

$\frac{1}{K_c}=\frac{1}{\frac{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{] }}}$

$\frac{1}{K_c}=\frac{{{\text{[SO}}_{\text{2}}\text{]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{] }}{{{\text{[SO}}_{\text{3}}\text{]}}^{\text{2}}}$

This is equal to $K_{c(reverse)}$.

So, it is clear that $\frac{1}{K_c}=K_{c(reverse)}$.