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Enthalpy of formation of 2 mol of ${NH}_{3} (g) is - 90 kJ, and ∆ H_{H - H} and ∆ H_{N - H}$ are respectively 435 kJ ${mol}^{- 1}$ and 390 kJ ${mol}^{- 1}$ . The value of $∆ H_{N \equiv N} is$

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-472.75 KJ

-945 KJ

472.5 kJ

945 kJ ${mol}^{- 1}$

answer is D.

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Detailed Solution

$N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 {NH}_{3} (g); ∆ H = - 90 kJ Now - 90 = ∆ H_{N \equiv N} + 3 ∆ H_{H - H} - 6 ∆ H_{N - H} ∆ H_{N \equiv N} = 2340 - 1305 - 90 = 945$

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