Question

Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. Predict the effect (shift right, shift left, or no effect) of the indicated volume change.

CO(g) + H2O(g) <=> CO2(g) + H2(g)
(volume is decreased)


PCl3(g) + Cl2(g) <=> PCl5(g)
(volume is increased)


CaCO3(s)<=> CaO(s) + CO2(g)
(volume is increased)

Answer 1

CO(g) + H₂O(g) <=> CO₂(g) + H₂(g), (volume is decreased) .. No effect.

PCl₃(g) + Cl₂(g) <=> PCl₅(g) , (volume is increased) .. Shift left.

CaCO₃(s) <=> CaO(s) + CO₂(g) , (volume is increased) .. Shift right.

Step-by-step explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

CO(g) + H₂O(g) <=> CO₂(g) + H₂(g) (volume is decreased)

• When volume is decreased, the pressure will increase:

• When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

• The reactants side (left) has 2.0 moles of gases and the products side (right) has 2.0 moles of gases.

So, decreasing the volume will have no effect on the reaction.

PCl₃(g) + Cl₂(g) <=> PCl₅(g) , (volume is increased)

• When volume is increased, the pressure will decrease:

• When there is an decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

• The reactants side (left) has 2.0 moles of gases and the products side (right) has 1.0 mole of gases.

• So, decreasing the pressure will shift the reaction to the side with more moles of gas (left side).

so, increasing the volume will shift the reaction left.

CaCO₃(s) <=> CaO(s) + CO₂(g) , (volume is increased)

• When volume is increased, the pressure will decrease:

• When there is an decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

• The reactants side (left) has 0 moles of gases and the products side (right) has 1.0 mole of gases.

• So, decreasing the pressure will shift the reaction to the side with more moles of gas (right side).

so, increasing the volume will shift the reaction right.