215k views
5 votes
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

User Asura
by
5.5k points

1 Answer

3 votes
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 PCl
₃ 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
User Jay Kreps
by
5.3k points