1. Understand the reaction

The reaction is:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

For ammonia synthesis, the reaction is exothermic:

$\Delta H < 0$

We are asked how the % yield of ammonia vs time changes when the reaction is conducted at the same pressure $P$ but at a higher temperature $T_2$, where $T_2 > T_1$.

2. Effect of temperature on equilibrium yield

Since the forward reaction is exothermic, increasing temperature shifts equilibrium in the reverse direction by Le Chatelier’s principle.

So, at higher temperature $T_2$:

• the equilibrium yield of $NH_3$ decreases
• hence the final plateau in the yield-vs-time graph must be lower than at $T_1$

Thus, if two curves are compared:

• curve at $T_1$ should end at a higher final yield
• curve at $T_2$ should end at a lower final yield

3. Effect of temperature on rate

When temperature increases, the rate constants increase according to Arrhenius equation:

$k = Ae^{-E_a/RT}$

So at higher temperature $T_2$:

• the reaction reaches equilibrium faster
• the initial rise of ammonia yield with time is steeper

Therefore, compared to $T_1$, the curve at $T_2$ should:

• rise more quickly initially
• but settle at a lower equilibrium yield

4. Required graph characteristics

So the correct graph for $(P, T_2)$ relative to $(P, T_1)$ must show:

1. steeper initial slope
2. lower final % yield

This is the standard behavior for an exothermic reversible reaction when temperature is increased.

5. Match with options

Among the given choices, the graph satisfying these two conditions is Option B.

6. Comparison with stored answer

Stored correct answer = B.

Our derived answer = B.

Hence, they agree.