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.

Question

Chemistry

Cynthia Day

19 September, 01:26

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.

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Chaz Montes · Answer 1

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.