Question

Calculate the enthalpy of oxidation of methane gas ($CH_4$) when it is burned completely in the presence of excess oxygen gas ($O_2$), given that the standard enthalpy of formation of $CH_4$ is -74.8 kJ/mol and the standard enthalpy of formation of $H_2O$ is -285.8 kJ/mol.

Answer 1

To calculate the enthalpy of oxidation of methane gas, we first need to write the balanced chemical equation for the complete combustion of methane:$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$Next, we will use Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of formation of the products minus the sum of the enthalpy changes of formation of the reactants.H_reaction =  H_f products  -  H_f reactants We are given the standard enthalpy of formation of $CH_4$  -74.8 kJ/mol  and $H_2O$  -285.8 kJ/mol . The standard enthalpy of formation of $O_2$ is 0 kJ/mol, as it is in its elemental form. The standard enthalpy of formation of $CO_2$ is -393.5 kJ/mol.Now, we can plug in the values into the equation:H_reaction = [1   -393.5 kJ/mol  + 2   -285.8 kJ/mol ] - [1   -74.8 kJ/mol  + 2  0 kJ/mol]H_reaction =  -393.5 - 571.6  -  -74.8 H_reaction = -965.1 + 74.8H_reaction = -890.3 kJ/molTherefore, the enthalpy of oxidation of methane gas when it is burned completely in the presence of excess oxygen gas is -890.3 kJ/mol.