A) 20.8 kcal
B) 19.8 kcal
C) 18.8 kcal
D) 20.2 kcal
Correct Answer: D
Solution :
[d] Use \[\Delta H=\Delta E+\Delta n\,RT\] |
\[\Delta H=19+2\times 2\times {{10}^{-3}}\times 300\]\[=20.2\,kcal\]; \[\Delta n=2\]. |
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