Question

Use the data at the back of this book to verify the values of ΔH and ΔGquoted above for the lead-acid reaction 5.13.

At temp 298K and pressure 1 bar.

Short answer

The value of Gibbs free energy = -315.72 kJ.

Step-by-step solution

Given information

The table is given as below

Temp = 298k and pressure =1 bar

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

ΔG = ΔH - TΔS ..........................................(1)

Now write equation for change in enthalpy for the given reaction

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

Substitute the values from 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
Put the values from the table, we get

$$\begin{gathered} ∆G=2(-813.0\mathrm{kJ})+2(-237.13\mathrm{kJ})-0-(-217.33\mathrm{kJ})-4\left(0\right)-2(-744.53\mathrm{kJ}) \\ ∆G=-315.72kJ \end{gathered}$$