Yield of Ammonia as Function of Time at Different Temperatures

To determine the correct curve representing the percent yield of ammonia over time at temperature $T_2$ ($T_2 > T_1$), we need to analyze both the kinetic and thermodynamic aspects of the reaction:

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), \quad \Delta H < 0$

1. Kinetic Aspect (Effect on Reaction Rate):

• According to the Arrhenius equation, the rate constant $k$ is given by: $k = A e^{-\frac{E_a}{RT}}$
• As temperature increases from $T_1$ to $T_2$ ($T_2 > T_1$), the kinetic energy of the molecules increases, increasing the frequency of effective collisions.
• Consequently, the initial rate of the reaction is faster at higher temperature $T_2$. Thus, the initial slope of the percent yield vs. time curve for $T_2$ will be steeper than that for $T_1$, meaning the curve for $T_2$ initially lies above the curve for $T_1$.

2. Thermodynamic Aspect (Effect on Equilibrium Yield):

• The formation of ammonia is an exothermic reaction ($\Delta H < 0$).
• According to the van 't Hoff equation: $\frac{d \ln K_{eq}}{dT} = \frac{\Delta H^\circ}{R T^2}$
• Since $\Delta H^\circ < 0$, an increase in temperature causes the equilibrium constant $K_{eq}$ to decrease ($K_{eq}(T_2) < K_{eq}(T_1)$).
• According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the backward (endothermic) direction to absorb heat.
• Therefore, the equilibrium yield (maximum percentage yield at $t \to \infty$) of $\text{NH}_3$ at $T_2$ will be lower than at $T_1$.

Conclusion:

The curve for $T_2$ must:

1. Rise faster initially (higher rate of reaction, steeper slope at small $t$).
2. Level off at a lower final plateau than $T_1$ (lower equilibrium yield).

This behavior is correctly depicted in Option (B).

Correct Answer: (B)