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Question

Consider a reaction:
N2 (g)+3H2 (g)2NH3 (g)
Given that ,
Keq=7×105
ΔSNH3=23.66 cal mol1K1
Both the values are at 300K
Find the value of ΔHNH3 at 300 K
Take ln(7)=1.95



A
15.16 kcal mol1
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B
35.46 kcal mol1
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C
85.2 kcal mol1
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D
65.4 kcal mol1
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Solution

The correct option is A 15.16 kcal mol1

At temperature T, equilibrium constant is Keq ,we know:

ln(Keq)=ΔHoRT+ΔSoR
R=2 cal mol1 K1
Putting the given values
ln(7×105)=ΔH2×300+(23.66)2(1.95+5×2.3)=ΔH60011.83(13.45+11.83)×600=ΔHΔH=15168 cal mol1=15.16 kcal mol1

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