Burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), CO(g), and \(\mathrm{CO}_{2}(g) .\) (a) Write three balanced equations for the reaction of methane gas with oxygen to produce these three products. In each case assume that \(\mathrm{H}_{2} \mathrm{O}(l)\) is the only other product. (b) Determine the standard enthalpies for the reactions in part (a).(c) Why, when the oxygen supply is adequate, is \(\mathrm{CO}_{2}(g)\) the predominant carbon-containing product of the combustion of methane?

We need to write the balanced equations for the reaction of methane (CH4) with oxygen (O2) to produce: 1. Soot (graphite particles) 2. CO(g) 3. CO2(g) (a) Reaction 1: Methane + Oxygen → Soot (graphite) + water \( CH_{4}(g) + O_{2}(g) \rightarrow C(graphite) + 2H_{2}O(l) \) To balance this equation, we insert a coefficient in front of the oxygen molecule: \( CH_{4}(g) + 2O_{2}(g) \rightarrow C(graphite) + 2H_{2}O(l) \) (b) Reaction 2: Methane + Oxygen → CO(g) + water \( CH_{4}(g) + O_{2}(g) \rightarrow CO(g) + 2H_{2}O(l) \) This equation is already balanced. (c) Reaction 3: Methane + Oxygen → CO2(g) + water \( CH_{4}(g) + O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l) \) To balance this equation, we insert a coefficient in front of the oxygen molecule: \( CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l) \)

The standard enthalpies for each reaction can be calculated using the standard enthalpies of formation, which can be looked up in a reference table. We will denote the standard enthalpy of formation as ΔHf°. For methane (CH4(g)): ΔHf° = -74.8 kJ/mol For oxygen (O2(g)): ΔHf° = 0 kJ/mol (it's an element in its standard state) For graphite (C): ΔHf° = 0 kJ/mol (it's an element in its standard state) For CO(g): ΔHf° = -110.5 kJ/mol For CO2(g): ΔHf° = -393.5 kJ/mol For water (H2O(l)): ΔHf° = -285.8 kJ/mol Using Hess's Law, the standard enthalpy of a reaction (ΔH°) can be calculated as the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants: (a) Reaction 1: ΔH° = [ΔHf°(C) + 2ΔHf°(H2O(l))] - [ΔHf°(CH4(g)) + 2ΔHf°(O2(g))] ΔH° = [0 + 2(-285.8)] - [(-74.8) + 0] ΔH° = -497.6 kJ/mol (b) Reaction 2: ΔH° = [ΔHf°(CO) + 2ΔHf°(H2O(l))] - [ΔHf°(CH4(g)) + ΔHf°(O2(g))] ΔH° = [(-110.5) + 2(-285.8)] - [(-74.8) + 0] ΔH° = -607.3 kJ/mol (c) Reaction 3: ΔH° = [ΔHf°(CO2) + 2ΔHf°(H2O(l))] - [ΔHf°(CH4(g)) + 2ΔHf°(O2(g))] ΔH° = [(-393.5) + 2(-285.8)] - [(-74.8) + 0] ΔH° = -890.3 kJ/mol

When the oxygen supply is adequate, CO2(g) is the predominant product of methane combustion because the formation of CO2(g) results in the greatest release of energy (has the lowest standard enthalpy of the three reactions). In other words, the reaction that produces CO2(g) is the most exothermic and thermodynamically favorable among the three reactions under analysis. Thus, when there is enough oxygen, the reaction will predominantly proceed to form CO2(g) to minimize the overall energy of the system and maximize stability.