I have a helpful trick for such problems.
Step 1:
Write the equation asked with exact moles given in question,
PCl₃ + Cl₂ → PCl₅ --------(1)
Step 2:
Write equations given in data along with energies,
P₄ + 6Cl₂ → 4PCl₃ ΔH = −1280 kj -----(2)
P₄ + 10Cl₂ → 4PCl₅ ΔH = −1774 kj -----(3)
Step 3:
Compare the equation 1 with eq 2 and 3 and modify eq 2 and 3 according to eq 1 both in terms of reactant and product and number of moles. For example in eq 1 PCl₃ is on left hand side and its number of moles is one, so, modify eq, 2 according to eq 1 both in terms of moles and reactant product sides. i.e.
By inter converting eq 2, the sign of energy will change from negative to positive, and divide eq 2 with 4, to have its ratio's equal to eq 1
PCl₃ → 1/4 P₄ + 1.5 Cl₂ ΔH = +320 KJ
Same steps are done for eq 3, but in this case signs are not changed only moles are changed
1/4P₄ + 2.5Cl₂ → 1/4PCl₅ ΔH = −443.5 kj
Now,
Compare both last equations,
PCl₃ → 1/4 P₄ + 1.5 Cl₂ ΔH = +320 KJ
1/4P₄ + 2.5Cl₂ → 4/4PCl₅ ΔH = −443.5 kj
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PCl₃ + Cl₂ → PCl₅ ΔH = -123.5 KJ
