Question

Given that the equilibrium constant for the reaction, $2S{{O}_{2(g)}}\ +\ {{O}_{2(g)}}\ \to \ \ 2S{{O}_{3(g)}}$ has a value of 278 at a particular temperature.
What is the value of the equilibrium constant for the following reaction at the same temperature?
$S{{O}_{3(g)}}\ \to \ \ S{{O}_{2(g)}}\ +\ \dfrac{1}{2}{{O}_{2(g)}}$

Answer 1

Identify the change in the second reaction from the original reaction. Write the equilibrium constant for both the chemical reaction in terms of the products and reactants of the reaction. Now substitute the value obtained in first reaction in second reaction and simplify to arrive at the final answer.

Complete step by step solution:
The formula for equilibrium constant is given below.
Formula: ${{\text{K}}_{\text{c}}}\text{=}\dfrac{\text{Concentration}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{products}}{\text{Concentration}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{reactants}}$
The value of dissociation constant for the reaction, $2S{{O}_{2(g)}}\ +\ {{O}_{2(g)}}\ \to \ \ 2S{{O}_{3(g)}}$ is given to us as 278. We will now write the concentrations of reactants and products in the formula above and equate it to 278.
$\text{278 =}\dfrac{{{[S{{O}_{3}}]}^{2}}}{[{{O}_{2}}]{{[S{{O}_{2}}]}^{2}}}$
Now we will discuss about the second reaction i.e. $S{{O}_{3(g)}}\ \to \ \ S{{O}_{2(g)}}\ +\ \dfrac{1}{2}{{O}_{2(g)}}$.
In this reaction we see the following changes:
- The stoichiometric coefficient of reactants as well as the products is divided by 2.
- The products and reactants are interchanged i.e. products become reactant and vice versa.
The dissociation constant for the second reaction is:
${{\text{K}}_{\text{c}}}\text{=}\dfrac{[S{{O}_{2}}]{{[{{O}_{2}}]}^{\dfrac{1}{2}}}}{[S{{O}_{3}}]}$
We can see that there is similarity between the two reactions. We will substitute the value of dissociation constant for reaction one in reciprocal and root. The simplified form is given below to avoid confusion.
${{\text{K}}_{\text{c}}}\text{=}\dfrac{1}{{{(278)}^{\dfrac{1}{2}}}}$ $= 0.059976014$ $\approx$ $0.06$

Therefore, the correct answer is option (B).

Note: It is important to know that the concentration of reactant/product is raised to the power of its stoichiometric coefficient before substituting it into the equation for dissociation constant.