Chapter 17, Problem 62A

Interpretation Introduction

Interpretation:

The effect on the pressure of an equilibrium system when products are favored needs to be explained.

Concept introduction:

According to Le Chatelier’s principle, the effect of a pressure changes on solids and liquids can be ignored because the volume of a solution or liquid is approximately independent of pressure

Answer to Problem 62A

There is a decrease in pressure for a given reaction because on the reactants side there are no gaseous molecule and on product side there is one gaseous molecule $\left({\text{CO}}_{\text{2}}\right)$. So, the equilibrium is shifted towards the direction of larger number of gaseous molecule.

Explanation of Solution

The given reaction is

${\text{MgCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons \text{MgO}\left(\text{s}\right){\text{+CO}}_{\text{2}}\left(\text{g}\right)$

According to Le Chatelier’s principle, increase of pressure on a system at equilibrium will shift the equilibrium in the direction in which pressure is reduced. Now, as the pressure of the system is increased the volume occupied by the system decreases so that total number of moles per unit volume increases. By applying this principle, the effect of a pressure changes on solids and liquids can be ignored because the volume of a solution or liquid is approximately independent of pressure.

In the given reaction there is a decrease in pressure for a given reaction because on the reactants side there is no gaseous molecule and on product side there is one gaseous molecule $\left({\text{CO}}_{\text{2}}\right)$. So, the equilibrium is shifted towards the direction of larger number of gaseous molecule.

Conclusion

There is a decrease in pressure for a given reaction because on the reactants side there are no gaseous molecule and on product side there is one gaseous molecule $\left({\text{CO}}_{\text{2}}\right)$. So, the equilibrium is shifted towards the direction of larger number of gaseous molecule.