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

Step 1: Write the balanced combustion reaction and get the change in gas moles.
Complete combustion of methane at constant pressure gives liquid water:
\[ \mathrm{CH_4(g)} + 2\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\mathrm{H_2O(l)} \]
Moles of gas: reactants have \(1+2=3\) mol of gas, products have only \(1\) mol of gas (water is liquid). So
\[ \Delta n_{gas} = 1 - 3 = -2 \]

Step 2: Recall which thermodynamic function gives the maximum work.
For a process run reversibly at constant temperature, the total maximum work obtainable (which includes any pressure-volume work, not only the useful non-expansion part that \(\Delta G\) tracks) equals the change in Helmholtz free energy:
\[ |W_{max}| = |\Delta A| = |\Delta U - T\Delta S| \]
So the first job is to get \(\Delta U\), not \(\Delta H\), since the given \(\Delta H\) is a constant-pressure quantity.

Step 3: Convert \(\Delta H\) to \(\Delta U\).
At constant \(T\), \(\Delta H = \Delta U + \Delta n_{gas}RT\), so
\[ \Delta U = \Delta H - \Delta n_{gas}RT \]
With \(T = 298.15\ \mathrm{K}\) (\(25^{\circ}\mathrm{C}\)) and \(R = 8.314\times10^{-3}\ \mathrm{kJ\,mol^{-1}\,K^{-1}}\):
\[ \Delta n_{gas}RT = (-2)(8.314\times10^{-3})(298.15) = -4.958\ \mathrm{kJ} \]
\[ \Delta U = -890.01 - (-4.958) = -885.05\ \mathrm{kJ} \]

Step 4: Find \(\Delta A\).
\[ T\Delta S = (298.15\ \mathrm{K})(-241.60\times10^{-3}\ \mathrm{kJ\,K^{-1}}) = -72.03\ \mathrm{kJ} \]
\[ \Delta A = \Delta U - T\Delta S = -885.05 - (-72.03) = -813.02\ \mathrm{kJ} \]

Final Answer:
\[ \boxed{|W_{max}| \approx 813.0\ \mathrm{kJ}} \]
Using \(\Delta G = \Delta H - T\Delta S\) directly (about \(818\) kJ) would be wrong here, since that only accounts for the non-expansion work; the actual maximum total work needs \(\Delta A\), built from \(\Delta U\).