91Ó°ÊÓ

To form the balanced chemical equations, we first need to write the reactants and products for the three reactions. In all reactions, \(\mathrm{CH}_4\) (methane) and \(\mathrm{O}_2\)(oxygen) will be the reactants. a) Methane reacts with oxygen to form soot (graphite) and water. Reaction: \(\mathrm{CH}_4(g) + \mathrm{O}_2(g) \rightarrow \mathrm{C(s)} + \mathrm{H_2O(l)}\) Now, to balance the equation, we need to add coefficients to the reactants and products such that the number of atoms of each element is equal on each side of the equation. Balanced equation: \(\mathrm{CH}_4(g) + \frac{3}{2}\mathrm{O}_2(g) \rightarrow \mathrm{C(s)} + 2\mathrm{H_2O(l)}\) b) Methane reacts with oxygen to form CO and water. Reaction: \(\mathrm{CH}_4(g) + \mathrm{O}_2(g) \rightarrow \mathrm{CO(g)} + \mathrm{H_2O(l)}\) Balanced equation: \(\mathrm{CH}_4(g) + \frac{3}{2}\mathrm{O}_2(g) \rightarrow \mathrm{CO(g)} + 2\mathrm{H_2O(l)}\) c) Methane reacts with oxygen to form CO2 and water. Reaction: \(\mathrm{CH}_4(g) + \mathrm{O}_2(g) \rightarrow \mathrm{CO_2(g)} + \mathrm{H_2O(l)}\) Balanced equation: \(\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \rightarrow \mathrm{CO_2(g)} + 2\mathrm{H_2O(l)}\)

To calculate the standard enthalpy change for each reaction, we can use the equation: \(\Delta H^\circ_\text{reaction} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})\) Here, \(\Delta H_f^\circ\) refers to the standard enthalpy of formation of the species. We need to look up their values in a standard table or textbook. In this case, the standard enthalpies of formation are given as: \(\Delta H_f^\circ(\mathrm{CH}_4(g)) = -74.8\,\text{kJ/mol}\) \(\Delta H_f^\circ(\mathrm{H_2O(l)}) = -285.8\,\text{kJ/mol}\) \(\Delta H_f^\circ(\mathrm{CO(g)}) = -110.5\,\text{kJ/mol}\) \(\Delta H_f^\circ(\mathrm{CO_2(g)}) = -393.5\,\text{kJ/mol}\) Note: Since graphite (soot) is the most stable form of carbon, its standard enthalpy of formation is zero. The same applies to oxygen gas. a) For reaction 1: \(\Delta H^\circ_\text{reaction} = [\Delta H_f^\circ(\mathrm{C(s)}) + 2\Delta H_f^\circ(\mathrm{H_2O(l)})] - [\Delta H_f^\circ(\mathrm{CH}_4(g)) + \frac{3}{2}\Delta H_f^\circ(\mathrm{O_2(g)})]\) Since the standard enthalpies of formation of C(s) and O2(g) are zero: \(\Delta H^\circ_\text{reaction} = [-285.8\times2 - (-74.8)]\, \text{kJ/mol}\) \(\Delta H^\circ_\text{reaction} = -496.8\,\text{kJ/mol}\) b) For reaction 2: \(\Delta H^\circ_\text{reaction} = [\Delta H_f^\circ(\mathrm{CO(g)}) + 2\Delta H_f^\circ(\mathrm{H_2O(l)})] - [\Delta H_f^\circ(\mathrm{CH}_4(g)) + \frac{3}{2}\Delta H_f^\circ(\mathrm{O_2(g)})]\) \(\Delta H^\circ_\text{reaction} = [-110.5 + 2\times(-285.8) - (-74.8)]\, \text{kJ/mol}\) \(\Delta H^\circ_\text{reaction} = -802.3\,\text{kJ/mol}\) c) For reaction 3: \(\Delta H^\circ_\text{reaction} = [\Delta H_f^\circ(\mathrm{CO_2(g)}) + 2\Delta H_f^\circ(\mathrm{H_2O(l)})] - [\Delta H_f^\circ(\mathrm{CH}_4(g)) + 2\Delta H_f^\circ(\mathrm{O_2(g)})]\) \(\Delta H^\circ_\text{reaction} = [-393.5 + 2\times(-285.8) - (-74.8 + 0)]\, \text{kJ/mol}\) \(\Delta H^\circ_\text{reaction} = -802.3\,\text{kJ/mol}\)

The formation of CO2 results in the release of more energy than the formation of CO or soot. This can be seen from the magnitude of the standard enthalpies calculated in step 2, which indicates that the combustion reaction producing CO2 (reaction 3) has a more negative standard enthalpy change than the reactions producing CO or soot. Thus, the reaction is more exothermic and thermodynamically favorable when CO2 is generated as the carbon-containing product. When the oxygen supply is adequate, the combustion reaction will follow the pathway that releases more energy, leading to the predominant formation of CO2.