In the following hypothetical reaction A + B → C + D, the equilibrium constant, Keq is less than 1.0 at
25°C and decreases by 35% on changing the temperature to 45°C. What must be true according to this information?
A. The ΔH° for the reaction is negative.
B. The ΔS° for the reaction is positive.
C. As temperature increases, the ΔG° for the reaction does not change.
D. The ΔG° for the reaction at 25°C is negative.
E. The ΔG° for the reaction at 45°C is zero.
I'm sticking my neck out but go through this reasoning and see if you agree.
reactants --> products
k = P/R
For k to be < 1, reactants must be favored. Just try A; if dH is - the reaction would be R ==> P + heat
If we increase T the reaction will be shifted to the left, R increases and P decreases so k decreases and that agrees with the problem.
What about B? If dS is +.
dG = dH - TdS
If dS is + then increases T will make the -TdS term more negative and that added to a constant H (whatever it is) will make dG more negative and that means products are favored so I think no for B.
C? C can't possibly be right because if T changes dG must change since -TdS changes.
D? I don' think so. If dG were - then k would be greater than 1.
E. If dG = 0 it means k = 1; however, if that is true then increase in T would make products favored but the problem says k decreases, not increases from < 1 to 1
so A?
The correct answer is A, which I got it right, but I was torn between A & B. could you please explain why B is incorrect. Thank you.
Yes, A! Thank you for making it easy to understand.
To determine what must be true according to the provided information, let's analyze each answer choice:
A. The ΔH° for the reaction is negative.
To determine the sign of the enthalpy change (ΔH°), we need to evaluate the effect of temperature on the equilibrium constant. According to Le Chatelier's principle, if the equilibrium constant decreases when the temperature is increased, it implies that the reaction is exothermic (releases heat) and the ΔH° is negative. Therefore, choice A is correct.
B. The ΔS° for the reaction is positive.
The information provided does not give any direct indication about the entropy change (ΔS°) for the reaction. Therefore, we cannot determine whether the ΔS° is positive or not based on the given information. Thus, choice B cannot be conclusively determined.
C. As temperature increases, the ΔG° for the reaction does not change.
The change in temperature does affect the standard free energy change (ΔG°) for a reaction. In this case, since the equilibrium constant decreases as the temperature increases, it implies that the ΔG° for the reaction is becoming more negative. Therefore, choice C is incorrect.
D. The ΔG° for the reaction at 25°C is negative.
Since the equilibrium constant (Keq) is less than 1.0 at 25°C, it implies that the reaction predominantly favors the reactants. In this scenario, the ΔG° must be negative for the reaction at 25°C because a negative ΔG° indicates that the reaction is spontaneous in the forward direction (from reactants to products). Therefore, choice D is correct.
E. The ΔG° for the reaction at 45°C is zero.
The information provided does not give any direct indication about the standard free energy change (ΔG°) at 45°C. Therefore, we cannot determine whether the ΔG° is zero or not based on the given information. Thus, choice E cannot be conclusively determined.
Based on the analysis, the correct answers are A (The ΔH° for the reaction is negative) and D (The ΔG° for the reaction at 25°C is negative).