Question

thermodynamics of ammonia
formation of ammonia
the formation of ammonia from hydrogen and nitrogen gases becomes less and less spontaneous as the temperature is increased, eventually becoming non - spontaneous at sufficiently high temperatures.
\ce{n_{2}(g) + 3h_{2}(g) -> 2nh_{3}(g)}
which of the following statements must be true?

• the reaction is exothermic.
• the change in entropy is positive.
• there is no change in entropy.
• the reaction is endothermic.

Explanation:

To determine the correct statement, we use the Gibbs free energy equation for spontaneity: $\Delta G = \Delta H - T\Delta S$. A reaction is spontaneous when $\Delta G < 0$. The problem states the reaction becomes less spontaneous (approaches $\Delta G = 0$ or positive) as temperature ($T$) increases.

1. Analyze entropy change ($\Delta S$):

For the reaction $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$, the number of moles of gas decreases (4 moles reactants → 2 moles products). Since entropy is related to disorder, a decrease in moles of gas means $\Delta S < 0$ (negative entropy change). Thus, options about positive or no entropy change are false.

1. Analyze enthalpy change ($\Delta H$):

The spontaneity decreases with increasing $T$. From $\Delta G = \Delta H - T\Delta S$, if $\Delta S < 0$ (as established), the term $-T\Delta S$ becomes positive (because $\Delta S$ is negative, $-T\Delta S = -T(-|\Delta S|) = +T|\Delta S|$). For $\Delta G$ to become less negative (or non - spontaneous) as $T$ increases, the $\Delta H$ term must be negative (exothermic). If $\Delta H$ were positive (endothermic), increasing $T$ would make $\Delta G$ more negative (more spontaneous), which contradicts the problem statement. Thus, the reaction must be exothermic ($\Delta H < 0$).

Answer:

The reaction is exothermic.