## Enter your answer in the provided box. phosphorus pentachloride is used in the industrial preparation of organic phosphorus compounds. equat

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

in progress 0

1. I have a helpful trick for such problems.

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

PClu2083+ Clu2082u2192 PClu2085 ——–(1)
Step 2:
Write equations given in data along with energies,

Pu2084+ 6Clu2082 u2192 4PClu2083 u0394H = u22121280 kj —–(2)

Pu2084+ 10Clu2082 u2192 4PClu2085 u0394H= u22121774 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 PClu2083 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

PClu2083 u2192 1/4 Pu2084 + 1.5 Clu2082 u0394H = +320 KJ

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

1/4Pu2084+ 2.5Clu2082u2192 1/4PClu2085 u0394H= u2212443.5 kj
Now,
Compare both last equations,

PClu2083 u2192 1/4 Pu2084 + 1.5 Clu2082 u0394H = +320 KJ

1/4Pu2084+ 2.5Clu2082u2192 4/4PClu2085 u0394H= u2212443.5 kj
_____________________________________________

PClu2083 + Clu2082 u2192 PClu2085 u0394H = -123.5 KJ