The standard enthalpy of formation of NH3 is -46.0 kJ mol-1. If the enthalpy of formation of H2 from its atoms is -436 kJ mol-1 and that of N2 is -712 kJ mol-1, the average bond enthalpy of N-H bond in NH3 is

Question

The standard enthalpy of formation of NH3 is -46.0 kJ mol-1. If the enthalpy of formation of H2 from its atoms is -436 kJ mol-1 and that of N2 is -712 kJ mol-1, the average bond enthalpy of N-H bond in NH3 is (A) -964 kJ mol-1 (B) +352 kJ mol-1 (C) + 1056 kJ mol-1 (D) - 1102 kJ mol-1

Tardigrade Admin · Accepted Answer

Enthalpy of formation of NH3 = -46 kJ/mole ∴ N2 + 3H2 → 2NH3 Δ Hf = - 2 x 46 kJ mol Bond breaking is endothermic and Bond formation is exothermic Assuming ‘x’ is the bond energy of N-H bond (kJ mol-1) ∴ 712 + (3 x 436)- 6x = -46 x 2 ∴ x = 352 kJ/mol

Q. 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 atoms 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

$- 964 k J m o l^{- 1}$

$+ 352 k J m o l^{- 1}$

$+ 1056 k J m o l^{- 1}$

$- 1102 k J m o l^{- 1}$

Q. The standard enthalpy of formation of NH3​ is −46.0kJmol−1. If the enthalpy of formation of H2​ from its atoms is −436kJmol−1 and that of N2​ is −712kJmol−1, the average bond enthalpy of N−H bond in NH3​ is

−964kJmol−1

+352kJmol−1

+1056kJmol−1

−1102kJmol−1

Enthalpy of formation of NH3​=−46kJ/mole ∴N2​+3H2​→2NH3​ΔHf​=−2x46kJmol Bond breaking is endothermic and Bond formation is exothermic Assuming ‘x’ is the bond energy of N−H bond (kJmol−1) ∴712+(3x436)−6x=−46x2 ∴x=352kJ/mol

Q. 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 atoms 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

$- 964 k J m o l^{- 1}$

$+ 352 k J m o l^{- 1}$

$+ 1056 k J m o l^{- 1}$

$- 1102 k J m o l^{- 1}$