Question

Consider the reaction 2KHSO3 (s) ⇌ K2SO3 (s) + H2O(g) + SO2 (g) at equilibrium. What happens to the partial pressure of 2O(g) if the volume of the reaction flask is doubled? (increases, decreases, or nothing)?

Answer 1

First, let us state the Le Chatelier principle which deals with changes in an equilibrium:

The statement is as follows:

If any equilibrium is disturbed, that is, change in conditions such as P,T, concentration, partial pressure, etc.., the system will counterbalance such change in order to favour the system's equilbirium.

For the system shown previously,

there are 2 mol of solids forming 1 mol of solid + 2 mol of gases

if we double the volume, the Presure of the system will DECREASE

therefore, the gas formation can be allowed... that is, more PRODUCTS will form due to the decrease in pressure

answered by: anonymous