Question

The maximum work (\(|W_{max}|\), in kJ) that can be obtained by complete combustion of \(1.0\) mol of \(\mathrm{CH_4}\) at constant pressure and \(25^{\circ}\mathrm{C}\) is (rounded off to one decimal place).

(Given: \(\Delta S = -241.60\ \mathrm{J\,K^{-1}\,mol^{-1}}\); \(\Delta H = -890.01\ \mathrm{kJ\,mol^{-1}}\); \(R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\))

Hint:

Maximum total work from a reaction at constant T is \(|\Delta A| = |\Delta U - T\Delta S|\), not \(|\Delta G|\); get \(\Delta U\) from \(\Delta H\) using the change in gas moles.

Answer 1

Correct Answer: 813

Here is a quicker route to the same result: start from the constant-pressure free energy change $\Delta G$, then correct it for the fact that the gas volume shrinks during the reaction.

Since $G=H-TS$ and $A=U-TS$, the two differ by $G-A=PV$, and for an ideal-gas change at constant $T$, $\Delta(PV)=\Delta n_{gas}RT$. That gives a direct shortcut:

\[ \Delta A = \Delta G - \Delta n_{gas}RT \]

First compute $\Delta G$ the ordinary way:

\[ \Delta G = \Delta H - T\Delta S = -890.01 - (298.15)(-241.60\times10^{-3}) = -890.01+72.03 = -817.98\ \mathrm{kJ} \]

Now find $\Delta n_{gas}$ for $\mathrm{CH_4(g)}+2\mathrm{O_2(g)}\to\mathrm{CO_2(g)}+2\mathrm{H_2O(l)}$: 3 mol of gas before, only 1 mol of gas after (water condenses), so $\Delta n_{gas}=1-3=-2$.

Apply the correction:

\[ \Delta n_{gas}RT = (-2)(8.314\times10^{-3})(298.15) = -4.958\ \mathrm{kJ} \] \[ \Delta A = -817.98-(-4.958) = -813.02\ \mathrm{kJ} \]

This lands on the same number as computing $\Delta U$ first, since both routes are the same rearrangement done in a different order. The maximum total work obtainable from burning 1 mol of methane at constant pressure and $25^{\circ}\mathrm{C}$ is about $813.0\ \mathrm{kJ}$, and the gap between this and the naive $|\Delta G|\approx818.0\ \mathrm{kJ}$ is exactly the $PV$ work tied up in the 2 mol drop in gas amount.