Chemical Thermodynamics Question 269

Question: The heat change for the following reaction at $ 298^{o}K $ and at constant pressure is $ +7.3,kcal $

$ A_2B(s),\to ,2A(s)+1/2,B_2(g) $ , $ \Delta H=+7.3,kcal $ The heat change at constant volume would be [DCE 2000]

Options:

A) 7.3 kcal

B) More than 7.3

C) Zero

D) None of these

Show Answer

Answer:

Correct Answer: D

Solution:

$ \Delta H=\Delta E+\Delta 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.



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