# Chemistry

You wish to know the enthalpy change for the formation of liquid PCl3 from the elements shown below.

P4(s) + 6 Cl2(g)--> 4 PCl3(l) ΔH°f = ?
The enthalpy change for the formation of PCl5 from the elements can be determined experimentally, as can the enthalpy change for the reaction of PCl3(l) with more chlorine to give PCl5(s).

P4(s) + 10 Cl2(g)--> 4 PCl5(s) ΔH°rxn= -1774.0 kJ

PCl3(l) + Cl2(g)--> PCl5(s) ΔH°rxn = -123.8 kJ
Use these data to calculate the enthalpy change for the formation of 1.50 mol of PCl3(l) from phosphorus and chlorine.

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asked by Kat
1. P4(s) + 10 Cl2(g)--> 4 PCl5(s) ΔH°rxn= -1774.0 kJ

PCl3(l) + Cl2(g)--> PCl5(s) ΔH°rxn = -123.8 kJ

Use eqn 1 as is. Multiply eqn 2 by 4 (multiply delta H rxn by 4 also), reverse it (change the sign of delta H x 4), and add to eqn 1. That will give you the reaction you want. Add the delta Hs to arrive at delta H for the desired reaction. That will be delta H for 4 moles of PCl3 and you change that to 1.50 moles PCl3.

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posted by DrBob222

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