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Calculate the standard enthalpy of formation of methane gas (CH4) given the following data:- ∆H°combustion of methane = -890.36 kJ/mol - ∆H°formation of H2O (l) = -285.8 kJ/mol - ∆H°formation of CO2 (g) = -393.5 kJ/mol

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To calculate the standard enthalpy of formation of methane gas  CH4 , we can use the following combustion reaction:CH4  g  + 2 O2  g   CO2  g  + 2 H2O  l The standard enthalpy change for this reaction, Hcombustion, is given as -890.36 kJ/mol. We also have the standard enthalpies of formation for H2O  l  and CO2  g . We can use Hess's Law to determine the standard enthalpy of formation for CH4  g .Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction, regardless of the order of the steps. In this case, we can write the following equations:1. CH4  g   C  graphite  + 2 H2  g  ; Hf = x kJ/mol  unknown 2. C  graphite  + O2  g   CO2  g  ; Hf = -393.5 kJ/mol3. 2 H2  g  + O2  g   2 H2O  l  ; Hf = 2 *  -285.8  kJ/mol = -571.6 kJ/molNow, we can add equations 2 and 3 to get the combustion reaction:C  graphite  + 2 H2  g  + 2 O2  g   CO2  g  + 2 H2O  l The enthalpy change for this reaction is the sum of the enthalpy changes for equations 2 and 3:Hcombustion = -393.5 kJ/mol +  -571.6 kJ/mol  = -965.1 kJ/molHowever, we are given that the actual Hcombustion of methane is -890.36 kJ/mol. Therefore, the difference between these two values is the standard enthalpy of formation of CH4  g :x = -890.36 kJ/mol -  -965.1 kJ/mol  = 74.74 kJ/molSo, the standard enthalpy of formation of methane gas  CH4  is 74.74 kJ/mol.

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