Determine the enthalpy change for the reaction represented by the equation below:

Question

Determine the enthalpy change for the reaction represented by the equation below: 

P4O10(s) +6PCl5(g) → 10POCl3(g)

Use Hess's Law along with the following equations:

P4(s) + 6Cl2(g) →4PCl3(g)                      ∆H=-1226 kJ

P4(s) + 5O2(g) → P4O10(g)                      ∆H=-2967 kJ

PCl3(g) + Cl2(g) → PCl5(g)                      ∆H=-84 kJ

PCl3(g) + 12O2(g) → POCl3(g)                 ∆H=-286 kJ

 Give your answer rounded to the nearest kJ. Don't forget the + or - sign.

Answer 1

So, this is a little different from the others you've posted. For this, I'll show you how to do it, but you should learn it so you can apply it to other similar problems.

For Hess' Law, we use the given reactions and ∆H values to find the ∆H for the "target" equation. Here, the

TARGET EQUATION is P₄O₁₀ + 6PCl₅ ==> 10POCl₃

Given:

Eq. 1: P4(s) + 6Cl2(g) →4PCl3(g)                      ∆H=-1226 kJ

Eq. 2: P4(s) + 5O2(g) → P4O10(g)                      ∆H=-2967 kJ

Eq. 3: PCl3(g) + Cl2(g) → PCl5(g)                      ∆H=-84 kJ

Eq. 4: PCl3(g) + 1/2 O2(g) → POCl3(g)                 ∆H=-286 kJ (NOTE: It should be 1/2 O₂ not 12 O₂. Big error)

Hess' Law:

copy Eq. 1: P4 + 6Cl2 ==> 4PCl3 ∆H = -1226 kJ

reverse Eq. 2: P4O10 ==> P4 + 5O2 ∆H = +2967 kJ (change sign)

reverse Eq. 3 and x 6: 6PCl5 ==> 6PCl3 + 6Cl2 ∆H = +504 kJ (change sign and multiply by 6)

copy Eq. 4 x10: 10PCl3 + 5 O2 ==> 10POCl3 ∆H = -2860 kJ (multiply by 10)

Add up all the equations and cancel and combine like items. You end up with...

P₄O₁₀ + 6PCl₅ ==> 10POCl₃ -> TARGET EQUATION

To get the ∆Hreaction, add up the individual ∆H values;

-1226 + 2967 + 504 + (-2860) = -615 kJ