AIEEE 2008 Solved Paper

For a reaction to be at equilibrium ∆G=0.
Since ∆G=∆H−T∆S
so at equilibrium ∆H−T∆S=0
or ∆H=T∆S
‌

1

2

X2+‌

3

2

Y2→XY3
∆H=−30kJ (given )
Calculating ∆S for the above reaction, we get
∆S=50−[‌

1

2

×60+‌

3

2

×40]JK−1
=50−(30+60)JK−1
=−40JK−1
At equilibrium, T∆S=∆H
[ as ∆G=0 ]
∴‌‌T×(−40)=−30×1000
[ as 1kJ=1000J ]
or T=‌

−30×1000

−40

or ‌‌750K