ThermochemistryHard

Question

The enthalpy of atomization of PH3(g) is +954 kJ/ mol and that of P2H4 is +1.488 MJ/mol. The bond energy of the P–P bond is

Options

A.318 kJ/mol
B.372 kJ/mol
C.216 kJ/mol
D.534 kJ/mol

Solution

$PH_{3}(g) \rightarrow P(g) + 3H(g);\Delta H = 954\text{ kJ}$

$\therefore 3 \times B.E._{P - H} = 954 \Rightarrow B.E._{P - H} = 318\text{ kJ/mol} $$${P_{2}H_{4}(g) \rightarrow 2P(g) + 4H(g);\Delta H = 1488\text{ kJ} }{\therefore B.E._{P - P} + 4 \times B.E._{P - H} = 1488 \Rightarrow B.E._{P - P} = 216\text{ kJ/mol}}$$

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