Chapter 17, Problem 41A

Interpretation Introduction

Interpretation The way in which the value of ${\text{K}}_{\text{eq}}$ is to be changed, if the reaction is rewritten in reverse direction.

Concept introduction

The reversible reaction is that reaction in which the forward reaction gets reversed. So, the equilibrium would be reversed

Answer to Problem 41A

The reversible reaction is that reaction in which the equilibrium constant values for both reactions are inversely proportional to each other.

Explanation of Solution

Consider an example of chemical reaction between hydrogen and iodine:

${\text{H}}_{\text{2}}\left(\text{g}\right)\text{+ I2}\left(\text{g}\right)\underset{}{\rightleftharpoons }\text{2HI}\left(\text{g}\right)$

$\text{K=}\frac{{\left[\text{HI}\right]}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]}\text{= 48}$

Then for the decomposition of hydrogen iodide is the inverse of the above equilibrium constant.

$\text{2HI }\left(\text{g}\right)\underset{}{\rightleftharpoons }{\text{H}}_{\text{2}}\left(\text{g}\right){\text{ + I}}_{\text{2}}\left(\text{g}\right)$

$\text{K'= }\frac{\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]}{{\left[\text{HI}\right]}^{\text{2}}}$

$\text{ = }\frac{\text{1}}{\text{K}}\text{=}\frac{\text{1}}{\text{48}}\text{=0}\text{.02}$

In general, $\text{ Kf = }\frac{\text{1}}{\text{Kb}}$

Hence, the equilibrium constant for forward and reverse equations are inversely proportional to each other.

Conclusion

In a reverse reaction the equilibrium constant value is decreases.