Chapter 9, Problem 70E

Interpretation Introduction

Interpretation: The $\Delta \text{H}$ value for the following reaction needs to be determined:

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s)+6PCl}}_5(g)\to 10C{l}_{3}PO(g)$

Concept Introduction: For the overall processes, the enthalpy change can be determined by adding the enthalpy change of all the steps involved in the process is known as Hess’s Law. The equation to show Hess’s law is:

  ${\text{ΔH}}_{\text{total}}{\text{= ΔH}}_{\text{1}}{\text{+ΔH}}_{\text{2}}{\text{+ΔH}}_{\text{3}}\text{+}\mathrm{...}{\text{+ΔH}}_{\text{n}}$

Answer to Problem 70E

  $\Delta \text{H = }-610.1\text{ kJ}$

Explanation of Solution

Given:

  $\begin{array}{l}{\text{P}}_4(s)+6C{l}_2(g)\to 4PC{l}_3(g)\Delta H=-1225.6\text{ kJ -(1)}\\ {\text{P}}_4(s)+5O_2(g)\to P_4{O}_{10}(s)\Delta H=-2967.3\text{ kJ -(2)}\\ {\text{PCl}}_3(g)+C{l}_2(g)\to PC{l}_5(g)\Delta H=-84.2\text{ kJ -(3)}\\ {\text{PCl}}_3(g)+\frac{1}{2}{O}_2(g)\to C{l}_{3}PO(g)\Delta H=-285.7\text{ kJ -(4)}\end{array}$

Rules related to enthalpies of the reaction are:

• When a reaction is inverted, then the sign of enthalpy is also inverted.
• When a reaction is multiplied by ‘n’ coefficient, then the value of enthalpy is multiplied by ‘n’ value.

The required equation is:

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s)+6PCl}}_5(g)\to 10C{l}_{3}PO(g)$

In order to obtain the required equation, reaction (3) is multiplied by 6 and added to the reaction (2):

  $\begin{array}{l}{\text{P}}_4(s)+5O_2(g)\to P_4{O}_{10}(g)\Delta H=-2967.3\text{ kJ }\\ \underset{\_}{{\text{6PCl}}_3(g)+6C{l}_2(g)\to 6PC{l}_5(g)\Delta H=6(-84.2\text{ kJ) = -505}\text{.2 kJ }}\\ {\text{P}}_4(s)+5O_2(g)+{\text{6PCl}}_3(g)+6C{l}_2(g)\to P_4{O}_{10}(g)+6PC{l}_5(g)\Delta H=-3472.5\text{ kJ -(5)}\end{array}$

Reversing reaction (5):

  $P_4{O}_{10}(g)+6PC{l}_5(g)\to {\text{P}}_4(s)+5O_2(g)+{\text{6PCl}}_3(g)+6C{l}_2(g)\Delta H=+3472.5\text{ kJ -(6)}$

Multiply reaction (4) with 10 and add reaction (6):

  $\begin{array}{l}{\text{10PCl}}_3(g)+5O_2(g)\to 10C{l}_{3}PO(g)\Delta H=-2857\text{ kJ }\\ \underset{\_}{P_4{O}_{10}(s)+6PC{l}_5(g)\to {\text{P}}_4(s)+5O_2(g)+{\text{6PCl}}_3(g)+6C{l}_2(g)\Delta H=+3472.5\text{ kJ }}\\ P_4{O}_{10}(s)+6PC{l}_5(g)+10{\text{PCl}}_3(g)\to 10C{l}_{3}PO(g)+{\text{P}}_4(s)+{\text{6PCl}}_3(g)+6C{l}_2(g)\\ \Delta H=+3472.5\text{ kJ-}2857\text{ kJ =615}\text{.5 kJ}\end{array}$

Now, adding equation (1) with the observed equation:

  $\begin{array}{l}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s)+6PCl}}_{\text{5}}\text{(g)}+10{\text{PCl}}_3(g)\to {\text{10Cl}}_{\text{3}}{\text{PO(g) +P}}_{\text{4}}{\text{(s)+6PCl}}_{\text{3}}{\text{(g)+6Cl}}_{\text{2}}\text{(g)}\\ \Delta H\text{ = 615}\text{.5 kJ}\\ \underset{\_}{{\text{P}}_{\text{4}}{\text{(s)+6Cl}}_{\text{2}}\text{(g)}\to {\text{4PCl}}_{\text{3}}\text{(g) }\Delta H=-1225.6\text{ kJ }}\\ {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}{\text{(s)+6PCl}}_{\text{5}}\text{(g)}\to {\text{10Cl}}_{\text{3}}\text{PO(g) }\Delta H=-1225.6\text{ kJ+615}\text{.5 kJ}=-610.1\text{ kJ }\\ \end{array}$