Question

If bond energies of $ N\equiv N, $ $ H-H $ and $ N-H $ bonds are 945, 437 and 389 kJ respectively, AH for the following gaseous reaction is $ {{N}_{2}}+3{{H}_{2}}\xrightarrow{{}}2N{{H}_{3}} $

• A. $ -78\,kJ $
• B. $ -156\,kJ $
• C. $ -1478\,kJ $
• D. $ +1478\,kJ $

Answer 1

Answer: (A)

$ {{N}_{2}}+3{{H}_{2}}\xrightarrow{{}}2N{{H}_{3}} $ For this reaction, Heat evolved $ =6\times $ ( $ N-H $ bond energy) $ - $ [3 ( $ H-H $ bond energy) + ( $ N\equiv N $ ) bond energy] = 6 (389) $ - $ [3 (437) + (945)] = 2334 $ - $ (1311 + 945) = 2334 $ - $ 2256 = 78 kJ $ \therefore $ $ \Delta H=-78\,kJ $