Consider the reaction 2KHSO3(s) ⇌ K2SO3(s) + H2O(g) + SO2(g) at equilibrium. What happens to the...

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

Expert Solution

Le Chatelier’s Principle:

According to this principle, whenever an equilibrium system is subjected to change, certain processes takes place which have a tendency to partially neutralize the initial change, which result in bringing the system to a new position of equilibrium. This principle mainly explains three different equilibrium changes:

Equilibrium change due to change in the concentration

Equilibrium change due to change in the volume or pressure

Equilibrium change due to change in the temperature

Equilibrium change due to change in the concentration:

Changing the concentration of a species will shift the equilibrium that side which would reduce that change in concentration.

Equilibrium change to change in the volume or pressure:

According to this law a change in the volume of an equilibrium system that one or more gases changes the concentration of those gases.

If equilibrium of gaseous of molecules is condensed than the pressure of equilibrium is increased. This increased pressure will be partially relieved by a shift of equilibrium in the less gaseous molecules.

If equilibrium of gaseous of molecules is expanded than the pressure of equilibrium is decreased. This decreased pressure will be partially restored by a shift of equilibrium in the more gaseous molecules.

The reaction:

2KHSO3(s) ⇌ K2SO3(s) + H2O(g) + SO2(g)

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

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

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

queen_honey_blossom answered 2 years ago

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