Chemical Thermodynamics Question 120
Question: The value of $ \Delta H-\Delta E $ for the following reaction at $ 27^{o}C $ will be $ 2NH_3(g),\to ,N_2(g)+3H_2(g) $ [Kerala (Med.) 2002]
Options:
A) $ 8.314\times 273\times (-2) $
B) $ 8.314\times 300\times (-2) $
C) $ 8.314\times 27\times (-2) $
D) $ 8.314\times 300\times (2) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \Delta H-\Delta E=\Delta nRT $ also, $ 2N{H_{3(g)}}\to {N_{2(g)}}+3{H_{2(g)}} $
$ \Delta n=2 $ .
