Decoding the Concentration-Time Graphs
Imagine you are watching a chemical reaction unfold in real-time. The graphs provided in the question are exactly that—a window into the dynamic world of molecules. We are observing the reaction A⇌P at two different temperatures, T1 and T2.
The first thing to notice is what happens as time stretches towards infinity. The curves flatten out. This horizontal plateau is the hallmark of chemical equilibrium—the state where the forward and backward reactions are happening at the exact same rate.
If we look closely at the y-axis, the dashed line represents a concentration of 5 mol L−1. At equilibrium, for both temperatures, the concentration of our product [P] is soaring above 5, while our reactant [A] has dipped below 5.
What does this tell us? The equilibrium constant, K, is defined as the ratio of products to reactants:
Since the numerator is larger than the denominator, K is strictly greater than 1. And from the master equation of thermodynamics, ΔG∘=−RTlnK, a K>1 guarantees that ΔG∘<0. The reaction is spontaneous in the forward direction under standard conditions.
The Temperature Dependence of Equilibrium
Now, let's compare the two temperatures. The problem explicitly states that T2>T1.
Look at the plateau levels again. At the lower temperature T1, the product concentration is higher, and the reactant concentration is lower compared to T2. This means the equilibrium constant at T1 is larger than at T2 (K1>K2).
Think about what this implies physically. We heated the system up (from T1 to T2), and the system responded by producing less product. The equilibrium shifted backward. According to Le Chatelier's Principle, a system shifts to absorb added heat. If heating it shifts it backward, the forward reaction must be releasing heat. It is an exothermic process! Therefore, the standard enthalpy change is negative: ΔH∘<0.
Unlocking the Entropy Mystery
We have conquered ΔG∘ and ΔH∘. Now for the final boss: Entropy (ΔS∘). The question hands us a peculiar mathematical key:
Since we already established that both K1 and K2 are greater than 1, their natural logarithms are positive. This allows us to safely cross-multiply without flipping the inequality sign:
Let's multiply both sides by the universal gas constant, R:
Does RTlnK look familiar? It is exactly −ΔG∘. Substituting this in, we get:
Now, we unleash the fundamental definition of Gibbs free energy, ΔG∘=ΔH∘−TΔS∘. Substituting this into our inequality:
−(ΔH∘−T1ΔS∘)>−(ΔH∘−T2ΔS∘)
Expanding the negative sign:
The −ΔH∘ terms beautifully cancel out from both sides, leaving us with a pure entropy relationship:
Bringing everything to one side and factoring out ΔS∘:
The Final Verdict
Here is where the logic snaps into place. We know that T2>T1, which means the term (T1−T2) is strictly negative.
We have a product of two numbers that is greater than zero (positive). If one of those numbers is negative, the other must also be negative to make the overall product positive.
Therefore, the standard entropy change must be negative: ΔS∘<0.
Piecing it all together, we have ΔG∘<0, ΔH∘<0, and ΔS∘<0. This perfectly matches options (A) and (C). A brilliant problem that seamlessly weaves graphical analysis with rigorous thermodynamic inequalities!
