The fuel in high-efficiency natural gas vehicles consists primarily of methane \(\left(\mathrm{CH}_{4}\right) .\) (a) How much heat is produced in burning 1 mol of \(\mathrm{CH}_{4}(g)\) under standard conditions if reactants and products are brought to \(298 \mathrm{~K}\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed? (b) What is the maximum amount of useful work that can be accomplished under standard conditions by this system?

Using the formula for the combustion of a hydrocarbon, we can find the balanced chemical equation for the combustion of CH4: CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)

We can calculate the standard enthalpy change of the reaction (∆H°) as follows: ∆H° = [n * ∆H°f(products)] - [n * ∆H°f(reactants)] Standard enthalpy of formation (∆H°f) values for the compounds are as follows: ∆H°f(CH4 (g)) = -74.8 kJ/mol ∆H°f(O2 (g)) = 0 kJ/mol (since it's an element in its standard state) ∆H°f(CO2 (g)) = -393.5 kJ/mol ∆H°f(H2O (l)) = -285.8 kJ/mol Now, we can calculate ∆H° for the reaction: ∆H° = [1(-393.5) + 2(-285.8)] - [1(-74.8) + 2(0)] ∆H° = (-393.5 - 2 * 285.8) - (-74.8) ∆H° = -965.1 kJ/mol The heat produced in burning 1 mol of CH4(g) is 965.1 kJ/mol.

Now we need to find the standard Gibbs free energy change of the reaction (∆G°) to determine the maximum amount of useful work that can be accomplished under standard conditions. We can do this by using the following equation: ∆G° = ∆H° - T∆S° First, we need to find the standard entropy change (∆S°) for the reaction: ∆S° = [n * S°(products)] - [n * S°(reactants)] Standard molar entropy (S°) values for the compounds are as follows: S°(CH4 (g)) = 186.3 J/mol·K S°(O2 (g)) = 205.2 J/mol·K S°(CO2 (g)) = 213.8 J/mol·K S°(H2O (l)) = 69.9 J/mol·K Now, we can calculate ∆S° for the reaction: ∆S° = [1(213.8) + 2(69.9)] - [1(186.3) + 2(205.2)] ∆S° = (213.8 + 2 * 69.9) - (186.3 + 410.4) ∆S° = -270.8 J/mol·K Now we can calculate the ∆G° for the reaction at 298 K: ∆G° = ∆H° - T∆S° ∆G° = -965.1 kJ/mol - (298 K)(-270.8 J/mol·K) / 1000 (to convert J to kJ) ∆G° = -965.1 kJ/mol + 80.8 kJ/mol ∆G° = -884.3 kJ/mol The maximum amount of useful work that can be accomplished under standard conditions by this system is 884.3 kJ/mol.