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For a given reaction, DeltaH = 35.5 kJ m...

For a given reaction, `DeltaH = 35.5 kJ mol^(-1)` and `Delta S = 83.6 JK^(-1) mol^(-1)`. The reaction is spontaneous at (Assume that `DeltaH and DeltaS` do not vary with temperature)

A

1118 K

B

1008 K

C

1200 K

D

845 K

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaG = DeltaH^(@) - TDeltaS^(@) =(179.1 - (298 xx 1602)/1000) kJ//mol = 179.1 - 47.74 = 131.36` kJ = +ve
`rArr` non-spontaneous at 298 K
Let at T, reaction is spontaneous
`DeltaG le 0 rArr DeltaH - TDeltaS le 0`
`T ge (DeltaH)/(DeltaS) = (17.9 xx 1000)/(160.2) K`
`T ge 1118 K`
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