Chapter 9, Problem 9.84QP

Interpretation Introduction

Interpretation:

From the given information the standard enthalpy of formation of methane has to be calculated.

Concept Introduction:

$\Delta H°$ refers to change in enthalpy.  Change in enthalpy in a reaction and bond energy (BE) are related as,

$\text{ΔH°}\text{=}\text{ΣBE(reactants)-ΣBE(products)}$

Change in enthalpy in a reaction and enthalpy of formation are related by the formula,

$\Delta H_{rxn}^{°}=\Sigma \Delta H_f^{°}(product)-\Sigma \Delta H_f^{°}(reac\tan t)$

Where,

$\begin{array}{l}\Delta H_{rxn}^{°}=\text{ standard}\text{enthalpy change in reaction}\\ \Delta H_f^{°}(product)=\text{ standard}\text{enthalpy of formation of product}\\ \Delta H_f^{°}(reac\tan t)=\text{ standard}\text{enthalpy of formation of }reac\tan t\end{array}$

Hess’s law is applied to calculate the enthalpy changes in a reaction.  According to Hess’s law – “The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes involving in each and every individual steps in the reaction.”  Thus if a reaction involves ‘n’ steps then enthalpy change $\Delta H°$ of the reaction is,

$\Delta H°=\Delta {\text{H}}_1^{°}+\Delta {\text{H}}_2^{°}+\Delta {\text{H}}_3^{°}\mathrm{....}+\Delta {\text{H}}_n^{°}$