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Calculate the standard enthalpy change for the formation of one mole of methane (CH4) gas given the following information: - Heat of combustion of methane gas is -890.36 kJ/mol - Heat of formation of water (H2O) liquid is -285.83 kJ/mol - Heat of formation of carbon dioxide (CO2) gas is -393.52 kJ/mol

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To calculate the standard enthalpy change for the formation of one mole of methane  CH4  gas, we can use the following balanced chemical equation for the combustion of methane:CH4 g  + 2 O2 g   CO2 g  + 2 H2O l We are given the heat of combustion of methane gas  -890.36 kJ/mol , the heat of formation of water liquid  -285.83 kJ/mol , and the heat of formation of carbon dioxide gas  -393.52 kJ/mol .We can use Hess's law to find the heat of formation of methane gas. Hess's law states that the enthalpy change of a reaction is the same, whether it occurs in one step or several steps. Therefore, we can write the following equation:H formation of CH4  = H combustion of CH4  + H formation of CO2  + 2  H formation of H2O Now, we can plug in the given values:H formation of CH4  = -890.36 kJ/mol +  -393.52 kJ/mol  + 2   -285.83 kJ/mol H formation of CH4  = -890.36 kJ/mol + 393.52 kJ/mol +  -571.66 kJ/mol H formation of CH4  = -1068.50 kJ/molTherefore, the standard enthalpy change for the formation of one mole of methane  CH4  gas is -1068.50 kJ/mol.

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