Chapter 6, Problem 6.86P

(a)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ of the following reaction is to be determined.

${\text{2PbSO}}_4\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to \text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}_2{\text{SO}}_4\left(l\right)$

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.

(b)

Interpretation Introduction

Interpretation:

$\Delta H_{\text{rxn}}^{°}$ of the following reaction from the given reactions is to be determined.

${\text{2PbSO}}_4\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\to \text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+2{\text{H}}_2{\text{SO}}_4\left(l\right)$

Concept introduction:

Hess’s law is used to calculate the enthalpy change of an overall reaction that can be derived as a sum of two or more reaction. According to Hess’s law $\Delta H$ of an overall reaction is equal to the sum of the enthalpy change for each individual reaction. $\Delta H_{\text{overall rxn}}=\Delta H_1+\Delta H_2+\mathrm{.......}+\Delta H_{\text{n}}$

Enthalpy is a state function so the value depends upon the initial state and final state not on the path so $\Delta H$ of an overall reaction can be calculated by the addition or subtraction of the individual steps whose $\Delta H$ is known.