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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Hard
1 of 1
Application +4 | -1
Given the bond energies of N ≡ N, H─H and N─H bonds as 945, 436 and 391 kJ  mol−1kJ\,\,mo{l^{ - 1}}respectively, the enthalpy of the reaction, 2N2(g)+3H2(g)→2NH3(g)2{\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right) is :
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Correct option: A
-93 kJ
Option A (-93 kJ) is correct because it correctly calculates the enthalpy of the reaction using the given bond energies and the formula ΔH=∑BEreactants−∑BEproducts\Delta H = \sum{BE_{reactants}} - \sum{BE_{products}}.

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