Calculate the enthalpy change for the combustion of methane gas if 10 grams of methane is burned completely in the presence of excess oxygen gas. Given the following balanced chemical equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)The standard enthalpy of formation for methane gas is -74.8 kJ/mol, carbon dioxide gas is -393.5 kJ/mol, and water vapor is -241.8 kJ/mol.

Question

Calculate the enthalpy change for the combustion of methane gas if 10 grams of methane is burned completely in the presence of excess oxygen gas. Given the following balanced chemical equation: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)The standard enthalpy of formation for methane gas is -74.8 kJ/mol, carbon dioxide gas is

Answer 1

To calculate the enthalpy change for the combustion of methane gas, we can use the following formula:H_combustion =  H_f products  -  H_f reactants First, we need to determine the number of moles of methane  CH4  in 10 grams. The molar mass of CH4 is 12.01 g/mol  C  + 4 * 1.01 g/mol  H  = 16.05 g/mol.moles of CH4 = mass / molar mass = 10 g / 16.05 g/mol  0.623 molesNow, we can calculate the enthalpy change for the combustion of methane:H_combustion = [1 * H_f CO2  + 2 * H_f H2O ] - [H_f CH4  + 2 * H_f O2 ]Since the standard enthalpy of formation for oxygen gas  O2  is zero  it is in its standard state , the formula becomes:H_combustion = [1 *  -393.5 kJ/mol  + 2 *  -241.8 kJ/mol ] - [ -74.8 kJ/mol ]H_combustion =  -393.5 - 2 * 241.8 + 74.8  kJ/mol  -802.9 kJ/molNow, we can calculate the enthalpy change for the combustion of 0.623 moles of methane:H = H_combustion * moles of CH4 = -802.9 kJ/mol * 0.623 mol  -500.2 kJThe enthalpy change for the combustion of 10 grams of methane gas is approximately -500.2 kJ.