Chapter 6.6, Problem 6.11BFP

Interpretation Introduction

Interpretation:

$\Delta H_{\text{f}}^{°}$ of ${\text{CH}}_3\text{OH}\left(l\right)$ in the following reaction when $\Delta H_{\text{rxn}}^{°}$ is $-638.6\text{kJ}$ is to be calculated.

${\text{CH}}_3\text{OH}\left(l\right)+\frac{3}{2}{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of product at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum m\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.