For the reaction at 298 k, 2no2(g) n2o4(g) the values of δh° and δs° are -58.03 kj and -176.6 j/k, respectively. calculate the value of δg° at 298 k. δg° = kj assuming that δh° and δs° do not depend on temperature, at what temperature is δg° = 0? t = k

ΔG° = -5.4032 kJ

T = 328.6 K

Explanation:

Data

ΔH°: -58.03 kJ

ΔS: -176.6 J/K = -0.1766 kJ/K

The change in Gibbs free energy is defined as:

ΔG° = ΔH° − T*ΔS°

When T = 298 K:

ΔG° = -58.03 − 298*(-0.1766) = -5.4032 kJ

if ΔG° = 0 kJ, then:

0 = -58.03 − T*(-0.1766)

58.03 = T*0.1766

T = 58.03/0.1766

T = 328.6 K

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