Question

Calculate the standard N-H bond enthalpy from following data:
N2(g)+3H2(g) gives 2NH3(g)
∆H° = -83 kJ
∆H°(N triple bond N) = 946kJ/mol, ∆H°(H-H) = 435 kJ/mol
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Answer 1

Answer:

The given reaction is shown below. Two moles of ammonia are formed during the reaction. Thus, the reaction for formation of one mole of ammonia is shown below. The standard enthalpy of formation of ammonia is calculated as, Therefore, the standard enthalpy of formation of ammonia gas is -46.2 kJ mol-1.Read more on Sarthaks.com - https://www.sarthaks.com/380058/given-n2-g-3h2-g-2nh3-g-rh-0-92-4-kj-mol-1-what-is-the-standard-enthalpy-of-formation-of-nh3-g?show=449825#a449825