Question

If bond dissociation energies of N ŗN, H–H and N-H are x₁, x₂ and x₃ respectively, hence enthalpy of formation of NH₃ (g) is =

• A. x₁ + 3x₂ – 6x₃
• B. 3x₃ – $\frac{1}{2}$x₁ –$\frac{3}{2}$ x₂
• C. $\frac{x_{1}}{2}$ + $\frac{3}{2}$ x₂ – 3x₃₃.
• D. 6x₃ – x₁ –3x₂

Answer 1

Answer: (C)

$\frac{x_{1}}{2}$ + $\frac{3}{2}$ x₂ – 3x₃₃.

Answer 2

$\frac{1}{2}$ N₂ (g) + $\frac{3}{2}$ H₂ (g)) ¾® NH₃ (g)

D_rH = D_form H(NH₃,g)

D_­form H (NH_3, g) = $\frac{1}{2}$ D_{N ŗ N} + $\frac{3}{2}$ D_{H– H} – 3 D_{N – H}

= $\frac{1}{2}$ × x₁ + $\frac{3}{2}$ x₂– 3x₃