Question

Ammonia is prepared industrially by the reaction:
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ for the reaction: ΔH° = –92.2 kJ and K (at 25°C) = 4.0 × 10⁸. When the temperature of the reaction is increased to 500°C, which of the following is true?
a. At equilibrium, more NH₃ is present at 500°C than at 25°C.
b. The reaction of N₂ with H₂ to form ammonia is endothermic.
c. K for the reaction will be larger at 500°C than at 25°C.
d. Product formation (at equilibrium) is not favored as the temperature is raised.
e. None of these choices is true.

Answer 1

Answer:

The correct option is d.

Explanation:

The reaction in the question is that for the formation of ammonia. The arrow in the reaction should be a reversible arrow. Because of the negative ΔH value, the forward reaction is exothermic while the reverse reaction is endothermic.

An increase in temperature of a system at equilibrium favors the endothermic reaction, hence the reverse reaction (the reactants) is favored. And when the reverse reaction is favored, more reactants are produced. Hence, product formation (at equilibrium) is not favored as the temperature is raised.