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Question: The enthalpy of formation of ammonia when calculated from the following bond energy data is (B.E. of...

The enthalpy of formation of ammonia when calculated from the following bond energy data is (B.E. of N – H, H – H, N ≡\equivN is 389 kJ ,

389kJmol−1,435kJmol−1,945.36kJmol−1389kJmol^{- 1},435kJmol^{- 1},945.36kJmol^{- 1}respectively)

A

−41.82kJmol−1- 41.82kJmol^{- 1}

B

+83.64kJmol−1+ 83.64kJmol^{- 1}

C

−945.36kJmol−1- 945.36kJmol^{- 1}

D

−833kJmol−1- 833kJmol^{- 1}

Answer

−41.82kJmol−1- 41.82kJmol^{- 1}

Explanation

Solution

: N2+3H2→2NH3N_{2} + 3H_{2} \rightarrow 2NH_{3}

ΔH=ΔH(N≡N)+3×ΔH(H−H)−2×3ΔH(N−H)=945.36+3×435.0−6×389.0=−83.64kJ\Delta H = \Delta H(N \equiv N) + 3 \times \Delta H(H - H) - 2 \times 3\Delta H(N - H) = 945.36 + 3 \times 435.0 - 6 \times 389.0 = - 83.64kJ

Heat of formation of NH3=−83.642=−41.82NH_{3} = \frac{- 83.64}{2} = - 41.82kJ/mol