Animated Solution for Chemistry - Chemical Equilibrium: The % yield of ammonia as a function of time in the reaction $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g),\Delta H<0$ at $(P,T_1)$ is given below - If this reaction is conducted at $(P,T_2)$, with $T_2>T_1$, the % yield of ammonia as a function of time is represented by -

Title: The Kinetics-Thermodynamics Tug-of-War: Analyzing the Haber Process Graph

Introduction: The synthesis of ammonia via the Haber process is one of the most important chemical reactions in the world. It's also a classic textbook example of the delicate balance between chemical kinetics (how fast a reaction goes) and chemical thermodynamics (how far a reaction goes). In this problem, we are asked to predict how the percentage yield versus time graph changes when we increase the temperature. Let's break down the physics and chemistry behind this curve.

This means the forward reaction is exothermic—it releases heat into the surroundings.

The provided graph shows the percentage yield of ammonia over time at a constant temperature $T_1$. The curve starts at zero, rises as the reaction proceeds, and eventually flattens out into a horizontal plateau. This plateau represents the equilibrium state, where the forward and backward reaction rates are equal, and the concentration of ammonia no longer changes.

Now, we conduct the exact same experiment at a higher temperature $T_2$ ($T_2>T_1$). To predict the new curve, we must analyze the initial slope and the final plateau separately.

The Need for Speed Let's look at the very beginning of the reaction ($t=0$). At this point, we only have reactants ($N_2$ and $H_2$), so the net rate of reaction is simply the rate of the forward reaction:

What happens to the rate constant $k_f$ when we increase the temperature? According to the Arrhenius equation, an increase in temperature provides more molecules with sufficient kinetic energy to overcome the activation energy barrier. Therefore, the rate constant always increases with temperature, regardless of whether the reaction is exothermic or endothermic.

Because $k_f$ is larger at $T_2$, the initial rate of ammonia production is much faster. Visually, this means the curve for $T_2$ must start with a steeper slope than the curve for $T_1$. It will shoot up more rapidly in the initial stages.

The Equilibrium Compromise Now, let's look at the final plateau, which is governed by thermodynamics. This is where the exothermic nature of the reaction ($\Delta H<0$) becomes critical.

According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in temperature, the system will shift to counteract that change. Since our forward reaction releases heat, increasing the temperature to $T_2$ acts as a stress. The system responds by favoring the backward, endothermic reaction to absorb the excess heat.

Mathematically, the equilibrium constant $K_c$ decreases as temperature increases for an exothermic reaction. A smaller $K_c$ means a lower concentration of products at equilibrium. Therefore, the maximum percentage yield of ammonia at $T_2$ will be strictly lower than at $T_1$. The horizontal plateau of our new curve must sit below the original plateau.

We now have our two defining characteristics for the $T_2$ curve: 1. Steeper initial slope (due to faster kinetics). 2. Lower final plateau (due to unfavorable thermodynamics).

If we examine the given options, only Option (B) exhibits this exact behavior. The solid curve for $T_2$ starts off rising faster than the dashed $T_1$ curve, meaning it initially sits above it. However, because its maximum yield is lower, it eventually crosses the $T_1$ curve and flattens out at a lower horizontal asymptote.

This problem beautifully illustrates the classic industrial compromise: we often run exothermic reactions at higher temperatures to get the product faster, even though we get less of it in the end!