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

Question

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

Tardigrade Admin · Accepted Answer

Given (1/2) N 2+(3/2) H 2 leftharpoons NH 3 Δ Hf=-46.0 kJ / mol H+H leftharpoons H2 ; Δ Hf=-436 kJ / mol N+N leftharpoons N2 ; Δ Hf=-712 kJ / mol Δ Hf(N H3)=(1/2) Δ HN-N+(3/2) Δ HH-H-Δ HN-H -46=(1/2)(-712)+(3/2)(-436)-Δ HN-H On calculation Δ HN-H=-964 k J / m o l

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

Given21​N2​+23​H2​​NH3​; ΔHf​=−46.0kJ/mol H+H​H2​;ΔHf​=−436kJ/mol N+N​N2​;ΔHf​=−712kJ/mol ΔHf​(NH3​)=21​ΔHN−N​+23​ΔHH−H​−ΔHN−H​ −46=21​(−712)+23​(−436)−ΔHN−H​ On calculation ΔHN−H​=−964kJ/mol

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