Chapter 15, Problem 92A

Interpretation Introduction

Interpretation:

The value of ${\text{ΔΗ}}_{\text{rxn}}^{\text{°}}\text{=}\text{-1343}\text{.9kJ}$ needs to be calculated using standard enthalpies of formation for the reaction

${\text{P}}_{\text{4}}^{}{\text{O}}_{\text{6}}^{}\text{(s)}\text{+}{\text{2O}}_{\text{2}}^{}\text{(g)}\to {\text{P}}_{\text{4}}^{}{\text{O}}_{\text{10}}^{}\text{(s)}$

Concept introduction:

Standard enthalpy of formation: It is the energy consumed or released during the formation of 1 mole of a substance under standard condition from it’s pure elements. Standard enthalpy of formation is can be written as ${\text{ΔΗ}}_{\text{f}}^{}$

Answer to Problem 92A

${\text{ΔΗ}}_{\text{rxn}}^{\text{°}}\text{=}\text{-1343}\text{.9kJ}$

Explanation of Solution

$\Delta {\text{H}}_{\text{rxn}}^{\text{o}}=\sum {\text{ΔH}}_{\text{f}}^{\text{o}}\left\{\text{products}\right\}-\Sigma {\text{ΔH}}_{\text{f}}^{\text{o}}\left\{\text{reactants}\right\}$

For the given reaction,

${\text{P}}_{\text{4}}^{}{\text{O}}_{\text{6}}^{}\text{(s)}\text{+}{\text{2O}}_{\text{2}}^{}\text{(g)}\to {\text{P}}_{\text{4}}^{}{\text{O}}_{\text{10}}^{}\text{(s)}$

Given values of ${\text{ΔH}}_{\text{f}}^{\text{o}}$ of the reactants and products are as follows,

${\text{ΔH}}_{\text{f}}^{\text{o}}{\text{{P}}_{\text{4}}^{}{\text{O}}_{\text{6}}^{}\text{}=}\text{-16}\text{.401kJ}$

${\text{ΔH}}_{\text{f}}^{\text{o}}{\text{{O}}_2^{}\text{}=}\text{-0kJ}$

${\text{ΔH}}_{\text{f}}^{\text{o}}{\text{{P}}_4^{}{\text{O}}_{10}^{}\text{}=}-2984.0\text{kJ}$

Let us calculate the standard enthalpies of the products and reactants

$\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{o}}{\text{{reactant}=ΔH}}_{\text{f}}^{\text{o}}{\text{{P}}_{\text{4}}^{}{\text{O}}_{\text{6}}^{}{\text{(s)}+ΔH}}_{\text{f}}^{\text{o}}{\text{{O}}_{\text{2}}^{}\text{(g)}}\\ \text{=(1)(-1640}\text{.1)+(2)(0)}\\ \text{=}\text{-1640}\text{.1kJ}\end{array}$

$\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{o}}{\text{{product}=ΔH}}_{\text{f}}^{\text{o}}{\text{{P}}_{\text{4}}^{}{\text{O}}_{10}^{}\text{(s)}}\\ \text{=(1)(-2984}\text{.0)}\\ \text{=}\text{-2984}\text{.0kJ}\end{array}$

$\begin{array}{c}\Delta {\text{H}}_{\text{rxn}}^{\text{o}}=\sum {\text{ΔH}}_{\text{f}}^{\text{o}}\left\{\text{products}\right\}-\Sigma {\text{ΔH}}_{\text{f}}^{\text{o}}\left\{\text{reactants}\right\}\\ \text{=}-2984.0-(-1640.1)\\ =-2984.0+1640.1\\ =-1343.9\text{kJ}\end{array}$

Conclusion

Hess’s law : According to this law, total enthalpy change of a reaction is the sum of all the reactions regardless of the steps or stages of reaction.