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Q.

Given the bond energies $N \equiv N, H - H and N - H$ bonds are 945, 436 and 391 kJ ${mol}^{- 1}$ respectively, the enthalpy change of the reaction
$N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 {NH}_{3} (g) is$

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a

$- 93 kJ$

b

102 kJ

c

90 kJ

d

105 kJ

answer is A.

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Detailed Solution

$∆ H = (945 + 3 \times 436) - (2 \times 3 \times 391) = 2253 - 2346 = - 93 kJ$

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