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From the following enthalpy changes,

2PCl3(l) → 2P(s) + 3Cl2(g) ∆H = -640 kJ
2P(s) + 5Cl2(g) → 2PCl5(s) ∆H = -886 kJ

calculate the value of ∆H for the reaction
PCl3(l) + Cl2(g) → PCl5(s) ∆H = ??

Sagot :

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Answer:

∆H = -763kJ

Explanation:

Using Hess's law we can determine the ΔH of a reaction from the sum of similar reactions. Using the reactions:

(1) 2PCl3(l) → 2P(s) + 3Cl2(g) ∆H = -640 kJ

(2) 2P(s) + 5Cl2(g) → 2PCl5(s) ∆H = -886 kJ

The sum of (1)/2 + (2)/2 gives:

(1) / 2 = PCl3(l) → P(s) + 3/2Cl2(g) ∆H = -640 kJ/2 = -320kJ

(2) / 2 = P(s) + 5/2Cl2(g) → PCl5(s) ∆H = -886 kJ/2 = -443kJ

PCl3(l) + Cl2(g) → PCl5(s) ∆H = -320kJ - 443kJ =

∆H = -763kJ

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