Question

determine the enthalpy of the reaction of
\ce{p4o10(s) + 6 pcl5(g) -> 10 pocl3(g)}
using the equations:
\ce{p4(s) + 6 cl2(g) -> 4 pcl3(g)} \quad $\delta h^\circ = -1225.6\\ \mathrm{kj/mol}$ (note: original had \kj/mx\ likely typo)
\ce{p4(s) + 5 o2(g) -> p4o10(s)} \quad $\delta h^\circ = -2967.3\\ \mathrm{kj/mol}$
\ce{pcl3(g) + cl2(g) -> pcl5(g)} \quad $\delta h^\circ = -84.2\\ \mathrm{kj/mol}$
\ce{pcl3(g) + \frac{1}{2} o2(g) -> pocl3(g)} \quad $\delta h^\circ = -285.72\\ \mathrm{kj/mol}$

Explanation:

Step1: Identify target reaction and given reactions

Target: $\ce{P4O10(s) + 6 PCl5(g) -> 10 POCl3(g)}$
Given reactions:

1. $\ce{P4(s) + 6 Cl2(g) -> 4 PCl3(g)}$ $\Delta H^\circ_1 = -1225.6\ \mathrm{kJ/mol}$
2. $\ce{P4(s) + 5 O2(g) -> P4O10(s)}$ $\Delta H^\circ_2 = -2967.3\ \mathrm{kJ/mol}$ (reverse: $\ce{P4O10(s) -> P4(s) + 5 O2(g)}$ $\Delta H^\circ_{2r} = +2967.3\ \mathrm{kJ/mol}$)
3. $\ce{PCl3(g) + Cl2(g) -> PCl5(g)}$ $\Delta H^\circ_3 = -84.2\ \mathrm{kJ/mol}$ (reverse and multiply by 6: $\ce{6 PCl5(g) -> 6 PCl3(g) + 6 Cl2(g)}$ $\Delta H^\circ_{3r} = 6 \times 84.2 = 505.2\ \mathrm{kJ/mol}$)
4. $\ce{PCl3(g) + \frac{1}{2} O2(g) -> POCl3(g)}$ $\Delta H^\circ_4 = -285.72\ \mathrm{kJ/mol}$ (multiply by 10: $\ce{10 PCl3(g) + 5 O2(g) -> 10 POCl3(g)}$ $\Delta H^\circ_{4m} = 10 \times (-285.72) = -2857.2\ \mathrm{kJ/mol}$)

Step2: Combine reactions

Add reversed reaction 2, reversed & multiplied reaction 3, multiplied reaction 4, and reaction 1:

• Reversed 2: $\ce{P4O10(s) -> P4(s) + 5 O2(g)}$ $\Delta H = +2967.3$
• Reversed & ×6 3: $\ce{6 PCl5(g) -> 6 PCl3(g) + 6 Cl2(g)}$ $\Delta H = +505.2$
• ×10 4: $\ce{10 PCl3(g) + 5 O2(g) -> 10 POCl3(g)}$ $\Delta H = -2857.2$
• Reaction 1: $\ce{P4(s) + 6 Cl2(g) -> 4 PCl3(g)}$ $\Delta H = -1225.6$

Sum the equations:
$\ce{P4O10(s) + 6 PCl5(g) + 10 PCl3(g) + 5 O2(g) + P4(s) + 6 Cl2(g) -> P4(s) + 5 O2(g) + 6 PCl3(g) + 6 Cl2(g) + 10 POCl3(g) + 4 PCl3(g)}$

Simplify (cancel common terms):
$\ce{P4O10(s) + 6 PCl5(g) -> 10 POCl3(g)}$ (matches target)

Step3: Calculate total $\Delta H$

Sum the $\Delta H$ values:
$\Delta H^\circ = 2967.3 + 505.2 - 2857.2 - 1225.6$

Calculate step-by-step:
$2967.3 + 505.2 = 3472.5$
$3472.5 - 2857.2 = 615.3$
$615.3 - 1225.6 = -610.3$

Answer:

The enthalpy of the reaction is $\boxed{-610.3\ \mathrm{kJ/mol}}$