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For the reaction
$A (g) + 2 B (g) \overset{}{\to} 2 C (g) + 3 D (g),$
The value of $∆ H at 27^{0} C is 19.0 kcal .$ The value of $∆ E$ for the reaction would be
$(R = 2.0 cal K^{- 1} {mol}^{- 1})$

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20.8 kcal

19.8 kcal

18.8 kcal

17.8 kcal

answer is D.

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Detailed Solution

$∆ H = ∆ E + ∆ n_{g} RT 19 = ∆ E + (\frac{2 \times 2 \times 300}{1000}) kcal 19 = ∆ E + 1.2 ∆ E = 19 - 1.2 kcal = 17.8 kcal$

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