Answer:
(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
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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.
