Question

The heats of atomisation of PH_3(g) and P₂H_4(g) are 954 kJ mol⁻¹ and 1485 kJ mol respectively. The P-P bond energy in kJ mol⁻¹ is ____________.

Options

• 213

• 426

• 318

• 1272

Answer 1

The heats of atomisation of PH_3(g) and P₂H_4(g) are 954 kJ mol⁻¹ and 1485 kJ mol respectively. The P-P bond energy in kJ mol⁻¹ is 213.

Explanation:

\[\ce{PH3_{(g)} -> P_{(g)} + 3H_{(g)}}\] ∆_atomH⁰ = 954 kJ mol⁻¹

The bond energy is the average energy for the three P-H bonds.

∴ Average bond energy of P-H bond

= `(954  "kJ mol"^-1)/3`

= 318 kJ mol⁻¹

P₂H₄ has four P-H bonds and one P-P bond.

Bond Energy (P-P) + 4 Bond Energy (P-H) = 1485 kJ mol⁻¹

Bond Energy (P-P) = 1485 kJ ⁻¹ − 4 × Bond Energy (P-H)

= 1485 kJ mol⁻¹ − (4 × 318 kJ mol⁻¹)

= 213 kJ mol⁻¹