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Calculate the standard enthalpy of formation (ΔH°f) of methane (CH4) given the following data: 2H2(g) + O2(g) → 2H2O(l)         ΔH°rxn = -572 kJ/mol C(graphite) + O2(g) → CO2(g)   ΔH°rxn = -394 kJ/mol Note: The enthalpy of formation of an element in its standard state is zero (0).

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To calculate the standard enthalpy of formation  Hf  of methane  CH4 , we need to use Hess's Law, which states that the enthalpy change of a reaction is the same whether it occurs in one step or several steps. We can manipulate the given reactions to create a reaction that forms methane  CH4 .First, we need to find the reaction that forms CH4. The balanced equation for the formation of methane is:C graphite  + 2H2 g   CH4 g Now, we need to manipulate the given reactions to match this equation. For the first reaction, we need to reverse it and divide by 2 to get:H2O l   H2 g  + 1/2O2 g          Hrxn = +286 kJ/molFor the second reaction, we need to reverse it and add 2 moles of hydrogen to both sides:CO2 g  + 2H2 g   C graphite  + 2H2O g    Hrxn = +394 kJ/molNow, we can add these two manipulated reactions together:H2O l   H2 g  + 1/2O2 g          Hrxn = +286 kJ/molCO2 g  + 2H2 g   C graphite  + 2H2O g    Hrxn = +394 kJ/mol--------------------------------------------------------------CO2 g  + H2O l   CH4 g  + 1/2O2 g          Hrxn = +680 kJ/molFinally, we can calculate the standard enthalpy of formation  Hf  of methane  CH4  by subtracting the enthalpy of formation of CO2 and H2O from the enthalpy of the reaction:Hf CH4  = Hrxn - Hf CO2  - Hf H2O Hf CH4  = +680 kJ/mol -  -394 kJ/mol  - 0Hf CH4  = +680 kJ/mol + 394 kJ/molHf CH4  = +1074 kJ/mol

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