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

Answer 1

Answer: (B)

• 352 kJ mol–1

Answer 2

N2(g) + $\frac{3}{2}$ H2(g) $\longrightarrow$ NH3 (g) ;

$\Delta$Hfŗ = – 46.0 kJ mol–1

2H(g) $\longrightarrow$ H2(g) ;$\Delta$Hfŗ = – 436 kJ mol–1

2N(g) $\longrightarrow$ N2(g) ;$\Delta$Hfŗ = – 712 kJ mol–1

NH3(g) $\longrightarrow$N2(g) + H2 (g);$\Delta$H = + 46

$\frac{3}{2}$H2 $\longrightarrow$3 H ;$\Delta$H = + 436 × $\frac{3}{2}$

$\frac{1}{2}$ N2 $\longrightarrow$ N;$\Delta$H = + 712 × $\frac{1}{2}$

NH3(g) $\longrightarrow$ N (g) + 3H (g);$\Delta$H = + 1056 kJ mol–

Average bond enthalpy of N–H bond = $\frac{1056}{3}$ = + 352 kJ mol–1