IMAT 2018 Q51 [Haber Process]

When nitrogen and hydrogen are mixed under suitable conditions of temperature and pressure, and in the presence of a catalyst, an equilibrium state is attained:

N_2(g) + 3H_2(g) ⇌ 2NH_3(g) ΔH is negative

Which of the following statements explain(s) why, when the pressure on a mixture of nitrogen and hydrogen is increased while the temperature remains constant, the time taken to attain equilibrium is decreased?

  1. The molecules have a higher collision frequency.
  2. The yield of ammonia (NH_3(g)) decreases.
  3. A greater proportion of molecules in the mixture have energy greater than the activation energy.

A. 2 only
B. 3 only
C. 1 and 3 only
D. 1, 2 and 3
E. 1 only


Important considerations:

N_2(g) + 3H_2(g) : 4 moles total

2NH_3(g) : 2$ moles total

Negative ΔH

(1) Molecules have a higher collision frequency: more pressure = more collisions

With more collisions = more chance of successful reaction = faster reaction

Not all collisions result in the formation of products, there needs to be enough energy and the molecules need to be in the correct orientation. If there is more pressure, it increases the number of collisions and this gives us more attempts at successful reactions. This explains why Le Chatelier’s principle favours the side with fewer moles when there is an increase in pressure. The correction is a result of more collisions/breakdowns between the many moles on one side.

(2) To continue on the idea of Le Chatelier’s principle applied to an increase in pressure, we can analyse the moles of products and reactants to determine where the shift is going to be. Here we have 4 moles on the reactant side and 2 moles on the product side, so the reaction is going to shift to the side with 2 moles (ammonia side). This shift causes an increase in ammonia, which contradicts what is said in the question, invalidating that statement.

(3) The proportion of the reactant molecules with energy greater than activation energy will stay the same, why?

Because KE is only dependent on temperature. We are told in the body of the question that temperature remains constant, therefore the kinetic energy will stay the same. Can we explain this further? PV = nRT. We are told that n, R, and T are constant, therefore if P is increasing, it will only affect the volume, so the change will be accounted for.

The rate of reaction depends on the number of collisions (frequency) and the number of particles with KE > Ea (activation energy). In this question, the number of collisions is increased in 1) and no other change/reason given to us will result in a change in the time needed to reach the rate of reaction faster.

\fcolorbox{red}{grey!30}{Therefore the answer is E, $1$ only.}