Chemical equilibria: reversible reactions, dynamic equilibrium — practice questions

a) More nitrogen gas added:

• The concentration of a reactant, $N_2$, has been increased.
• According to Le Chatelier's principle, the system will act to decrease this concentration.
• The position of equilibrium will shift to the right, favouring the forward reaction to use up the extra $N_2$.

b) The pressure is decreased:

• The system will act to increase the pressure.
• The forward reaction produces 2 moles of gas from 4 moles of gas (1 mole of $N_2$ + 3 moles of $H_2$).
• The position of equilibrium will shift to the left, favouring the reverse reaction, as this produces more moles of gas (4 moles), thereby increasing the pressure.

c) A catalyst is added:

• A catalyst increases the rate of the forward and reverse reactions equally.
• It does not change the position of equilibrium.
• The only effect is that equilibrium is reached more quickly.

Temperature:

• The forward reaction is exothermic ($ΔH$ is negative).
• According to Le Chatelier's principle, a decrease in temperature favours the exothermic direction to release heat and counteract the change.
• Therefore, a low temperature would give the maximum yield of $SO_3$.

Pressure:

• There are 3 moles of gas on the reactant side ($2SO_2 + O_2$) and 2 moles of gas on the product side ($2SO_3$).
• According to Le Chatelier's principle, an increase in pressure favours the side with fewer moles of gas to reduce the overall pressure.
• Therefore, a high pressure would give the maximum yield of $SO_3$.