A) 7.3 kcal
B) More than 7.3
C) Zero
D) None of these
Correct Answer: D
Solution :
\[\Delta H=\Delta \text{E}+\Delta \text{nRT}\] or\[\Delta E=\Delta H-\Delta nRT\] \[\therefore \Delta E=+7.3-\frac{1}{2}\times 0.002\times 298\]\[=7.3-0.298\]= 7 kcal.You need to login to perform this action.
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