Question

For the process O2(g)→2O(g), ΔH° = 498 kJ·mol–1. What would be the predicted sign of ΔS°rxn and the conditions under which this reaction would be spontaneous? Question 13 options: 1) ΔS°rxn is positive, and the process is spontaneous at low temperatures only. 2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only. 3) ΔS°rxn is negative, and the process is spontaneous at high temperatures only. 4) ΔS°rxn is negative, and the process is spontaneous at low temperatures only.

Answer 1

Answer:

2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only.

Explanation:

ΔS = ΔH / T , ΔS is change in entropy , ΔH is change in enthalpy

Since ΔH is positive , ΔS is positive .

ΔG = ΔH - TΔS

For spontaneous reaction . ΔG should be negative .

As ΔS is positive , at high temperature the value of TΔS will be more and hence the value of TΔS will be higher than Δ H . Hence ΔG will be negative.

Hence at higher temperature , the reaction will be spontaneous.

Answer 2

The predicted sign of ΔS°rxn and the conditions under which this reaction would be spontaneous should be option  2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only.

Predicted sign of ΔS°rxn and its conditons:

We know that

ΔS = ΔH / T

Here ΔS means a change in entropy

And, ΔH means a change in enthalpy

Since ΔH is positive so  ΔS is positive.

Now

ΔG = ΔH - TΔS

However, For spontaneous reaction. ΔG should be negative .

Since ΔS is positive, at high temperature the value of TΔS should be more and due to this the value of TΔS will be more than Δ H . Thus,  ΔG will be negative.

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