White phosphorus is a tetra-atomic solid $P_{4} (s)$ at room temperature.

If $Δ H_{s u b l ima t i o n}$ of P₄(s) = 61 kJ/mol and $Δ H_{a t o mi s a t i o n}$ of P₄(s) = 1321 kJ/mol, the bond enthalpy of (P–P) bond (in kJ/mol) is.

Question

White phosphorus is a tetra-atomic solid $P_{4} (s)$ at room temperature.

If $Δ H_{s u b l ima t i o n}$ of P₄(s) = 61 kJ/mol and $Δ H_{a t o mi s a t i o n}$ of P₄(s) = 1321 kJ/mol, the bond enthalpy of (P–P) bond (in kJ/mol) is.

TestHub · Accepted Answer

Concepts: Sublimation, atomisation, bond enthalpy. P₄ has 6 P-P bonds.

Solution: Bond enthalpy = $(Δ H_{a t o mi s a t i o n} - Δ H_{s u b l ima t i o n}) /6 = (1321 - 61) /6 = 210$ kJ/mol.

Chemistry - Thermochemistry Question with Solution | TestHub

Doubts & Discussion

JEE Main 2027 (Class 12+)