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Question

For a gaseous reaction,
A(g)+3B(g)3C(g)+3D(g). ΔE is 17 Kcal at 270C assuming R=2calK1mol. The value of ΔH for the above reaction is :

A
15.8 Kcal
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B
18.2 Kcal
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C
20.0 Kcal
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D
16.4 Kcal
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Solution

The correct option is B 18.2 Kcal
Solution:- (B) 18.2 kcal
Given that:-
ΔE=17 kcal
T=27=(273+27)=300K
R=2cal/molK=2×103kcal/molK
A(g)+3B(g)3C(g)+3D(g)
Δng=nPnR=(3+3)(1+3)=2
As we know that,
ΔH=ΔE+ΔngRT
ΔH=17+(2×300×2×103)=18.2 kcal
Hence the value of ΔH for the given reaction will be 18.2 kcal.

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