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Question

Standard entropy of X2,Y2 and XY3 are 60,40 and 50 JK1 mol1, respectively. For the reaction,
12X2+32Y2XY3.ΔH=30kJ to be at equilirbium, the temperature will be:

A
1250K
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B
500K
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C
750K
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D
1000K
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Solution

The correct option is C 750K
Solution:- (C) 750K
12X2+32Y2XY3ΔH=30kJ
For the above reaction-
ΔS=ΔS(Product)ΔS(reactant)
ΔS=50(12×60+32×40)
ΔH=5090=40
As we know that,
ΔG=ΔHTΔS
But at equilibrium,
ΔG=0
ΔHTΔS=0
T=ΔHΔS
Given:- ΔH=30kJ=30000J
T=3000040=750K
Hence the temperature of the given reaction is 750K.

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