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) $

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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 $ .



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