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For a given reaction, DeltaH=35.5kJ*mol^...

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

A

`T gt 425K`

B

all temperature

C

`T gt 298K`

D

`T lt 425K`.

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaG=DeltaH-TDeltaS=35.5xx10^(3)-Txx83.6`
Reaction is to be spontaneous if `DeltaG lt0`.
thus, `35.5 xx10^(3)-Txx83.6lt0`
Therefore, `Txx83.6gt35.5xx10^(3)`
or, `T gt 424.64K`.
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