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Given: PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = + 157 kJ

Given: PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = + 157 kJ 

P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = - 1207 kJ 

What is the standard-state enthalpy change for the following reaction? 

P4(g) + 10 Cl2(g) → 4PCl5(s)

a -2100 kJ

b-1835 kJ

c-1364 kJ

D -1786 kJ

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Answer #1

Reverse the reaction-1 and multiply with 4

4 PCl3(g) + 4 Cl2(g) ------> 4 PCl5(s) delta H = -4*157 = -628 kJ

P4(g) + 6 Cl2(g) ? 4 PCl3(g) ?H°rxn = -1207 kJ

_------------------------------------------------------_

P4(g) + 10 Cl2(g) ? 4PCl5(s)

Delta H = -1207 kJ - 628 kJ = -1835 kJ

Thus, correct option is (b)

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