Question

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

Answer 1

The enthalpy change for the reaction P4(g) + 10 Cl2(g) → 4PCl5(s) is -2257 kJ.

Step-by-step explanation:

To determine the enthalpy change (ΔH°rxn) for the reaction P4(g) + 10 Cl2(g) → 4PCl5(s), we can use the given standard reaction enthalpies:

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

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

We can combine these two reactions to cancel out the formation of PCl3(g) and find the enthalpy change for the target reaction. By multiplying the second reaction by 2 and reversing the direction of the first reaction, we get:

2PCl3(g) + 10 Cl2(g) → P4(g) + 14 Cl2(g)

Now, we can subtract the enthalpies of the two reactions:

ΔH°rxn = 2(-1207 kJ) - (+157 kJ) = -2414 kJ + 157 kJ = -2257 kJ

Therefore, the enthalpy change for the reaction P4(g) + 10 Cl2(g) → 4PCl5(s) is -2257 kJ.

Answer 2

(E) -1835 kJ

Step-by-step 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.