ThermochemistryHard

Question

For the given two processes

(i) ½P4(s) + 3Cl2(g) → 2PCl3(l): ΔH = −635 kJ (ii) PCl3(l) + Cl2(g) → PCl5(s): ΔH = −137 kJ

the value of ΔfH of PCl5(s) is

Options

A.454.5 kJ mol–1
B.−454.5 kJ mol−1
C.−772 kJ mol−1
D.−498 kJ mol−1

Solution

$\Delta_{f}H_{PCl_{5}(s)} = \frac{1}{2} \times (i) + (ii) = \frac{1}{2}( - 635) + ( - 137) = - 454.5\text{ kJ}$

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