Given are the bond enthalpies: $ϵ_{N \equiv N} = 945 {kJ mol}^{- 1}, ϵ_{H - H} = 436 {kJ mol}^{- 1}$ and $ϵ_{(N - H)} = 391 {kJ mol}^{- 1}$ . The enthalpy change of the reaction
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is

- 93 {kJ mol}^{- 1}

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- 89 {kJ mol}^{- 1}

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- 105 {kJ mol}^{- 1}

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105 {kJ mol}^{- 1}

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Solution

The correct option is A $- 93 {kJ mol}^{- 1}$
Given are the bond enthalpies: $ϵ_{N \equiv N} = 945 {kJ mol}^{- 1}, ϵ_{H - H} = 436 {kJ mol}^{- 1}$ and $ϵ_{(N - H)} = 391 {kJ mol}^{- 1}$ .
Hence, $Δ H = ϵ_{(N \equiv N)} + 3 ϵ_{(H - H)} - 6 ϵ_{(N - H)}$
$= (945 + 3 \times 436 - 6 \times 391) {kJ mol}^{- 1} = - 93 {kJ mol}^{- 1}$

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Q. Given are the bond enthalpies:

ϵ_{N \equiv N} = 945 {kJ mol}^{- 1}, ϵ_{H - H} = 436 {kJ mol}^{- 1}

and

ϵ_{(N - H)} = 391 {kJ mol}^{- 1}

. The enthalpy change of the reaction

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

The standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom is $- 436 k J m o l^{- 1}$ and that of $N_{2}$ is $- 712 k J m o l^{- 1}$ , the average bond enthalpy of $N - H$ bond in $N H_{3}$ is

Given the bond energies $N \equiv N$ , $H - H$ and $N - H$ bonds are $945, 436$ and $391 k J m o l e^{- 1}$ respectively, the enthalpy of the following reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is