For A Certain Process, Δ H=280 KJ And Δ S = 140 JK - Tardigrade App

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4. For a certain process, Δ H=280 kJ and Δ S = 140 JK -1 mol -1. What is the minimum temperature at which the process will be spontaneous?

Q. For a certain process, ΔH=280kJ and ΔS=140JK−1mol−1. What is the minimum temperature at which the process will be spontaneous? 3582 230 Thermodynamics Report Error A

2000K

84% B

200K

6% C

140K

6% D

420K

4%

Solution:

Temperature has minimum value at equilibrium i.e., at ΔG=0 Since ΔG=ΔH−TΔS,∴T=ΔSΔH​ T=140280×103​=2000K