Chapter 15.4, Problem 36PP

Interpretation Introduction

Interpretation:

The $\Delta {{\rm H}}_{rxn}^0$ for the given reaction is to be calculated.

Concept introduction:

$\Delta {{\rm H}}_{rxn}^0$ is the standard enthalpy of reaction.

Standard enthalpy of a reaction is calculated by subtracting sum of heat of formation of reactants from sum of heat of formation of products.

$\Delta {{\rm H}}_{rxn}^0={\displaystyle \sum \Delta {{\rm H}}_f^0(products)-}{\displaystyle \sum \Delta {{\rm H}}_f^0(reac\tan ts)}$

Answer to Problem 36PP

$\Delta {{\rm H}}_{rxn}^0=-2183.2kJ/mol$

Explanation of Solution

The given reaction is:

$C_3{H}_{7}COOH(l)+5O_2(g)\to 4C{O}_2(g)+4H_{2}O(l)$

$\Delta {{\rm H}}_{rxn}^0=\left\{4{{\rm H}}_f^0(C{O}_2)+4{{\rm H}}_f^0(H_{2}O)\right\}-\left\{{{\rm H}}_f^0(C_3{H}_{7}COOH)+5{{\rm H}}_f^0(O_2)\right\}$

From table:

$\begin{array}{l}{{\rm H}}_f^0(C{O}_2)=-393.5kJ/mol\\ {{\rm H}}_f^0(H_{2}O)=-285.8kJ/mol\\ {{\rm H}}_f^0(O_2)=0\end{array}$

From given data:

${{\rm H}}_f^0(C_3{H}_{7}COOH)=-534kJ/mol$

Therefore,

$\begin{array}{c}\Delta {{\rm H}}_{rxn}^0=\left\{4{{\rm H}}_f^0(C{O}_2)+4{{\rm H}}_f^0(H_{2}O)\right\}-\left\{{{\rm H}}_f^0(C_3{H}_{7}COOH)+5{{\rm H}}_f^0(O_2)\right\}\\ =\left\{4\times -393.5+4\times -285.8\right\}-\left\{-534+7\times 0\right\}\\ =-1574-1143.2+534\\ =-2183.2kJ/mol\end{array}$

Conclusion

$\Delta {{\rm H}}_{rxn}^0=-2183.2kJ/mol$