Chapter 6, Problem 6.58P

The common lead-acid car battery produces a large burst of current, even at low temperatures, and is rechargeable. The reaction that occurs while recharging a “dead” battery is $2\text{P}b\text{S}{\text{O}}_4(s)+2{\text{H}}_2\text{O}(\text{l})\to \text{P}b(s)+\text{P}b{\text{O}}_2(s)+2{\text{H}}_2\text{S}{\text{O}}_4(\text{l})$

1. Use $\text{Δ}{\text{H}}_f^o$ values from Apendix B to calculate $\text{Δ}{\text{H}}_{rxn}^o.$ Use the following equations to check your answer to part (a):

(1) $\text{P}b(s)+\text{P}b{\text{O}}_2(s)+2\text{S}{\text{O}}_3(g)\to 2\text{P}b\text{S}{\text{O}}_4(s)$ $\text{Δ}{\text{H}}_{rxn}^o=-768\text{ kJ}$ $\text{Δ}{\text{H}}_{rxn}^o=-132\text{ kJ}$ (2) $\text{S}{\text{O}}_3(g)+{\text{H}}_2\text{O}(\text{l})\to {\text{H}}_2\text{S}{\text{O}}_4(\text{l})$

Interpretation Introduction

Interpretation:The value of $\Delta H^0{}_{\text{rxn}}$ for the following reaction needs to be determined using $\Delta H^0{}_{\text{f}}$ values.

  $2{\text{PbSO}}_{\text{4}}(s)+2{\text{H}}_{\text{2}}\text{O(}l\text{)}\to \text{Pb(}s{\text{)+PbO}}_2(s)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(l)$

Concept Introduction : The amount of energy released on combustion of an organic compound is known as enthalpy of combustion. The formula of determination of $\Delta H_{\text{rxn}}$ is as follows:

  $\Delta H^0{}_{\text{rxn}}={\displaystyle \sum _{}^{}m\Delta H^0{}_{\text{f}}(\text{products})-}{\displaystyle \sum n\Delta H^0{}_{\text{f}}}(\text{reactants})$

Here, ${\displaystyle \sum }$ represents the sum of terms, m and n are number of moles of products and reactants respectively. The term $H^0{}_{\text{f}}$ is standard heat of formation.

Answer to Problem 6.58P

  $\Delta H^0{}_{\text{rxn}}$ of the reaction is $\text{504}\text{kJ}$ .

Explanation of Solution

  $\begin{array}{c}\Delta H^0{}_{\text{rxn}}=\left[1\text{mol}\Delta H^0{}_{\text{f}}(\text{Pb}(s))+1\text{mol }\Delta H^0{}_{\text{f}}({\text{PbO}}_{\text{2}}(s))+2\text{mol }\Delta H^0{}_{\text{f}}({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(l))\right]-\\ \left[2\text{mol}\Delta H^0{}_{\text{f}}({\text{PbSO}}_{\text{4}}\text{(}s\text{)})+2\text{mol}\Delta H^0{}_{\text{f}}({\text{H}}_2\text{O}(l))\right]\\ =\left[1\text{mol }\times \left(0\right)+1\text{mol}\times \left(-276.6\frac{\text{kJ}}{\text{mol}}\right)+2\text{mol}\left(-813.9\frac{\text{kJ}}{\text{mol}}\right)\right]-\left[\begin{array}{l}2\text{mol}\times \left(-918.4\frac{\text{kJ}}{\text{mol}}\right)+2\text{mol}\times \\ \left(-285.8\frac{\text{kJ}}{\text{mol}}\right)\end{array}\right]\\ =\left[-276.6\text{kJ}-1627.8\text{kJ}\right]-\left[-1836.8\text{kJ}-571.6\text{kJ}\right]\\ =-1904.4\text{kJ+2408}\text{.4}\text{kJ}\\ \text{=504}\text{kJ}\end{array}$

Interpretation Introduction

Interpretation:Using the following reaction, the result of part (a.) needs to be verified.

  $\begin{array}{l}\text{Pb}(s)+{\text{PbO}}_{\text{2}}(s)+2{\text{SO}}_{\text{3}}(g)\to 2{\text{PbSO}}_{\text{4}}(s)\Delta H_{\text{rxn}}=-768\text{kJ}\\ {\text{SO}}_{\text{3}}(g)+{\text{H}}_{\text{2}}\text{O(}l\text{)}\to {\text{H}}_2{\text{SO}}_{\text{4}}(l)\Delta H_{\text{rxn}}=-132\text{kJ}\end{array}$

Concept Introduction: According to Hess’s law, the change of enthalpy in a process is the summation of enthalpy changes of individual steps of the reaction.

  $\Delta H_{\text{overall}}=\Delta H_1+\Delta H_2+\mathrm{.......}+\Delta H_n$

Answer to Problem 6.58P

Part (a) answer is verified.

Explanation of Solution

The thermochemical equations given as follows:

  $\text{Pb}(s)+{\text{PbO}}_{\text{2}}(s)+2{\text{SO}}_{\text{3}}(g)\to 2{\text{PbSO}}_{\text{4}}(s)\Delta H_{\text{rxn}}=-768\text{kJ}$..... (1)

  ${\text{SO}}_{\text{3}}(g)+{\text{H}}_{\text{2}}\text{O(}l\text{)}\to {\text{H}}_2{\text{SO}}_{\text{4}}(l)\Delta H_{\text{rxn}}=-132\text{kJ}$..... (2)

The target equation is,

  $2{\text{PbSO}}_{\text{4}}(s)+2{\text{H}}_{\text{2}}\text{O(}l\text{)}\to \text{Pb(}s{\text{)+PbO}}_2(s)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(l)$..... (3)

To obtain the equation (3), first reverse the equation (1).

  $2{\text{PbSO}}_{\text{4}}(s)\to \text{Pb}(s)+{\text{PbO}}_{\text{2}}(s)+2{\text{SO}}_{\text{3}}(g)\Delta H_{\text{rxn}}=768\text{kJ}$..... (4)

Now multiply the equation (2) by 2.

  ${\text{2SO}}_{\text{3}}(g)+2{\text{H}}_{\text{2}}\text{O(}l\text{)}\to 2{\text{H}}_2{\text{SO}}_{\text{4}}(l)\Delta H_{\text{rxn}}=2\times -132\text{kJ=}-\text{264}\text{kJ}$ …… (5)

Now add equation (4) and (5) to obtain the equation (3).

  $\begin{array}{c}2{\text{PbSO}}_{\text{4}}(s)+\cancel{{\text{2SO}}_{\text{3}}(g)}+2{\text{H}}_{\text{2}}\text{O(}l\text{)}\to \text{Pb}(s)+{\text{PbO}}_{\text{2}}(s)+\cancel{2{\text{SO}}_{\text{3}}(g)}+2{\text{H}}_2{\text{SO}}_{\text{4}}(l)\Delta H_{\text{rxn}}=768\text{kJ}-\text{264}\text{kJ}\\ 2{\text{PbSO}}_{\text{4}}(s)+2{\text{H}}_{\text{2}}\text{O(}l\text{)}\to \text{Pb}(s)+{\text{PbO}}_{\text{2}}(s)+2{\text{H}}_2{\text{SO}}_{\text{4}}(l)\Delta H_{\text{rxn}}=504\text{kJ}\end{array}$

Hence, answer of part (a) is correct.