Question

Use the data at the back of this book to verify the values of $∆H$and $∆G$ quoted above for the lead-acid reaction 5.13.

Short answer

The value of Gibbs free energy = -315.72 kJ.

Step-by-step solution

Given Information

T= 298 K and P=1 bar.

The information from the table

Explanation

Gibbs energy is given by

G= H-TS

where G= Gibbs energy, H= Enthalpy, T =temp and S =entropy.

Assume there is infinitesimal change in Gibbs energy , then

$∆G=∆H-T∆S.............\left(1\right)$

Now write equation for change in enthalpy for the given reaction

$\Delta H=2\Delta H_{{\mathrm{PbSO}}_4}+2\Delta H_{{\mathrm{H}}_2\mathrm{O}}-\Delta H_{\mathrm{Pb}}-\Delta H_{{\mathrm{PbO}}_2}+4\Delta H_{{\mathrm{H}}^{+}}-2\Delta H_{S{O}_4^{2-}}$

Now substitute the values from the given table, we get

$$\begin{gathered} \Delta H=\left(\begin{array}{c}2(-920.0\mathrm{kJ})+2(-285.83\mathrm{kJ})-0\\ -(-277.4\mathrm{kJ})-4\left(0\right)-2(-909.27\mathrm{kJ})\end{array}\right) \\ =-315.72\mathrm{kJ} \end{gathered}$$

Similarly write equation for change in Gibbs energy for the given reaction

$\Delta G=(2\Delta G_{{\mathrm{PbSO}}_4}+2\Delta G_{{\mathrm{H}}_2\mathrm{O}}-\Delta G_{\mathrm{Pb}}-\Delta G_{{\mathrm{PbO}}_2}+4\Delta G_{{\mathrm{H}}^{+}}-2\Delta G_{S{O}_4^{2-}})$

Put the values from the table, we get

$∆G=2(-813.0\mathrm{kJ})+2(-237.13\mathrm{kJ})-0-(-217.33\mathrm{kJ})-4\left(0\right)-2(-744.53\mathrm{kJ})$

= -315.72 kJ.