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If δh = -70.0 kj and δs = -0.400 kj/k , the reaction is spontaneous below a certain temperature. calculate that temperature.

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According to this formula, when:

ΔG = ΔH - T*ΔS

when the reaction is thermodynamically spontaneous ΔG < 0

∴ ΔH - T* ΔS = 0

∴T*ΔS = ΔH

∴ T = ΔH / ΔS

when we have:

ΔH = -70KJ

and ΔS = -0.4 KJ/K

So by substitution:

T = -70KJ /- 0.4

= 175 K

∴the certain temperature below which the reaction will be thermodynamically spontaneous is 175 K
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