Calculate the change in enthalpy (ΔH) for the combustion of methane gas (CH4) at 25°C and constant pressure of 1 atm. The equation for the combustion of methane is:CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) Given the standard enthalpies of formation at 25°C are:ΔHf[CH4 (g)] = -74.87 kJ/mol ΔHf[CO2 (g)] = -393.5 kJ/mol ΔHf[H2O (l)] = -285.8 kJ/mol Assume that water is in its liquid phase at the end of the reaction.

Question

Calculate the change in enthalpy (ΔH) for the combustion of methane gas (CH4) at 25°C and constant pressure of 1 atm. The equation for the combustion of methane is:CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) Given the standard enthalpies of formation at 25°C are:ΔHf[CH4 (g)] = -74.87 kJ/mol ΔHf[CO2 (g)] = -393.5 kJ/mol ΔHf[H2O (l)] = -285.8

Answer 1

To calculate the change in enthalpy  H  for the combustion of methane, we can use the following equation:H =  Hf products  -  Hf reactants For the given reaction, the products are CO2  g  and 2H2O  l , and the reactants are CH4  g  and 2O2  g . The standard enthalpies of formation for O2  g  are zero because it is in its standard state.H = [1  Hf CO2  + 2  Hf H2O ] - [1  Hf CH4  + 2  Hf O2 ]Plugging in the given values:H = [1   -393.5 kJ/mol  + 2   -285.8 kJ/mol ] - [1   -74.87 kJ/mol  + 2   0 kJ/mol ]H =  -393.5 kJ/mol - 571.6 kJ/mol  -  -74.87 kJ/mol H = -965.1 kJ/mol + 74.87 kJ/molH = -890.23 kJ/molThe change in enthalpy  H  for the combustion of methane gas  CH4  at 25C and constant pressure of 1 atm is approximately -890.23 kJ/mol.