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P(g)+2Q(g)toPQ(2)(g),DeltaH=18kJmol^(-1)...

`P(g)+2Q(g)toPQ_(2)(g),DeltaH=18kJmol^(-1)`
The entropy change of the above reaction `(DeltaS_("system")` is 60 `JK^(-1)mol^(-1))`. At what temperature the reaction becomes spontaneous?

A

Below 200 K

B

Above 300 K

C

Below 300 K

D

Above 200 K and below 300 K.

Text Solution

Verified by Experts

The correct Answer is:
B
For spontaneous process `DeltaGlt0`
`thereforeDeltaH-TDeltaSlt0,DeltaHltTDeltaS`
`thereforeTgt(DeltaH)/(DeltaS)=(18000)/(60)=300K`
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