For the reaction, A(g)+2B(g)→2C(g)+3D(g)
The value of ΔH at 27∘C is 19.0kcal. The value of ΔE for the reaction would be
A
20.8kcal
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B
19.8kcal
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C
18.8kcal
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D
17.8kcal
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Solution
The correct option is D17.8kcal Given reaction is, A(g)+2B(g)→2C(g)+3D(g) ∴Δn(g)=5−3=2
We know, ΔH=ΔE+ΔngRT⇒ΔE=ΔH−ΔngRT=19−(2×2×10−3×300)=19−1.2kcal=17.8kcal.