Chemical Thermodynamics Question 100
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 million
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 $ = 7.0 kcal.
BETA
