Question

Enter your answer in the provided box. phosphorus pentachloride is used in the industrial preparation of organic phosphorus compounds. equation 1 shows its preparation from pcl3(l) and cl2(g): (1) pcl3(l) + cl2(g) → pcl5(g) use equations (2) and (3) to calculate δhrxn of equation (1): (2) p4(s) + 6cl2(g) → 4pcl3(l) δh = −1280 kj (3) p4(s) + 10cl2(g) → 4pcl5(g) δh = −1774 kj

Answer 1

I have a helpful trick for such problems. 

Step 1:
Write the equation asked with exact moles given in question,

                                       PCl₃ + Cl₂   →   PCl₅            --------(1)
Step 2:
Write equations given in data along with energies,

                               P₄ + 6Cl₂     →   4PCl₃       ΔH = −1280 kj  -----(2)

                               P₄ + 10Cl₂   →   4PCl₅       ΔH = −1774 kj  -----(3)

Step 3:
Compare the equation 1 with eq 2 and 3 and modify eq 2 and 3 according to eq 1 both in terms of reactant and product and number of moles. For example in eq 1 PCl₃ is on left hand side and its number of moles is one, so, modify eq, 2 according to eq 1 both in terms of moles and reactant product sides. i.e.
By inter converting eq 2, the sign of energy will change from negative to positive, and divide eq 2 with 4, to have its ratio's equal to eq 1

                             PCl₃   →   1/4 P₄  +  1.5 Cl₂            ΔH  =  +320 KJ

Same steps are done for eq 3, but in this case signs are not changed only moles are changed

                            1/4P₄ + 2.5Cl₂   →   1/4PCl₅            ΔH = −443.5 kj
Now, 
Compare both last equations,

                             PCl₃   →   1/4 P₄  +  1.5 Cl₂            ΔH  =  +320 KJ


                            1/4P₄ + 2.5Cl₂   →   4/4PCl₅            ΔH = −443.5 kj
                        _____________________________________________

                          PCl₃  +  Cl₂     →     PCl₅                   ΔH = -123.5 KJ