Question

How does a decrease in the container volume affect the amount of C present at equilibrium for the following reaction at equilibrium? 2 A(g) + B(g) ↔ C(g) + 3 D(g) ΔH° = -50 kJ How does a decrease in the container volume affect the amount of C present at equilibrium for the following reaction at equilibrium? 2 A(g) + B(g) ↔ C(g) + 3 D(g) ΔH° = -50 kJ Cannot be predicted. The amount of C will not change. The amount of C will increase. The amount of C will decrease.

Answer 1

we know that

for a equilibrium

according to Le chatlier principle

if any change is made to the equilibrium

the equilibrium will shift in a direction to counter the change

now

consider the given equilibrium

2A + B --> C + 3D

given

volume of the container is decreased

as a result

the pressure is increased

So

according to Le chatlier principle

the equilibrium will shift in a direction to reduce the pressure

So

the equilibrium will shift in a direction with less number of gaseous moles

we can see that

in this reaction

reactants have lesser gaseous moles than products

so

the equilibrium will shift towards the reactants

As a result

the amount of C will decrease