Use the standard reaction enthalpies given below to determine δh°rxn for the following reaction:

p4(g) + 10 cl2(g) → 4pcl5(s) δh°rxn = ?
given:
pcl5(s) → pcl3(g) + cl2(g) δh°rxn = +157 kj
p4(g) + 6 cl2(g) → 4 pcl3(g) δh°rxn = -1207 kj

(a) -1050. kj
(b) -2100. kj
(c) -1364 kj
(d) -1786 kj
(e) -1835 kj

(E) -1835 kJ

Explanation:

Use Hess law to answer this question by manipulating algebraically the equations :

Eq .1   PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = +157

Eq.2    P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ

to arrive to the equation in our problem:

P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°rxn = ?

Notice PCl₅ is the product in our equation and in Eq. 1 is a reactant so we need to reverse it. We need also to multiply it by 4 since that is the number required in the question.

P₄ in equation 2 is a reactant and is also a reactant in our desired equation also it is 1 mol in both. That is telling us to leave it as it is and we can then proceed to add it  the the  -4 times Eq1 and check everything is right when we add them together.

4PCl3(g) + 4Cl2(g)  → 4PCl5(s)   ΔH°rxn = -4 x (157 kJ)  (change the sign)

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

P4(g) + 10 Cl2(g) → 4PCl5(s)        ΔH°rxn =   -4 x (157 kJ) +   -1207 kJ

                                                     ΔH°rxn =   -1835 kJ

See how the PCl3 cancel and the Cl add correctly.

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