Q.

A process has ∆H=200 J/mole and ∆S=40 J/mole. The minimum temperature above which the process will be spontaneous is

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a

20K

b

4K

c

5K

d

12K

answer is C.

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Detailed Solution

∆G=∆H-T∆SThe process will be spontaneous,when ∆G = -ve ;|T∆S| >∆HT > |∆H∆S|T =20040=5K
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