Question

Determine the standard enthalpy of formation of methane (CH4) given the following information: The standard enthalpy of formation of CO2 and H2O are -393.5 kJ/mol and -285.8 kJ/mol, respectively. The combustion of 0.50 g of methane in a bomb calorimeter resulted in a 1.20°C temperature increase of the surrounding water. The heat capacity of the bomb calorimeter is 5.85 kJ/°C.

Answer 1

First, let's find the heat released during the combustion of 0.50 g of methane using the heat capacity of the bomb calorimeter and the temperature increase:q = C * Tq = 5.85 kJ/C * 1.20Cq = 7.02 kJNow, let's convert the mass of methane to moles using its molar mass  12.01 g/mol for carbon and 1.01 g/mol for hydrogen :0.50 g CH4 *  1 mol CH4 /  12.01 g/mol C + 4 * 1.01 g/mol H   = 0.0312 mol CH4Next, we'll find the heat released per mole of methane:q/mol = 7.02 kJ / 0.0312 mol = 225 kJ/molThe balanced combustion reaction of methane is:CH4 + 2 O2 -> CO2 + 2 H2OUsing the standard enthalpy of formation values for CO2 and H2O, we can write the equation for the standard enthalpy change of the reaction:H_rxn = H_f[CO2] + 2 * H_f[H2O] - H_f[CH4]We know the standard enthalpy change of the reaction is -225 kJ/mol  since the heat released is positive, the enthalpy change is negative , and we have the values for CO2 and H2O:-225 kJ/mol =  -393.5 kJ/mol  + 2 *  -285.8 kJ/mol  - H_f[CH4]Now, we can solve for the standard enthalpy of formation of methane:H_f[CH4] = -393.5 kJ/mol + 2 * 285.8 kJ/mol + 225 kJ/molH_f[CH4] = -74.9 kJ/molTherefore, the standard enthalpy of formation of methane  CH4  is approximately -74.9 kJ/mol.