Question

Calculate the standard enthalpy change of formation for the compound methane (CH4) given the following information:- The standard enthalpy change of formation for carbon dioxide (CO2) is -393.5 kJ/mol.- The standard enthalpy change of formation for water (H2O) is -285.8 kJ/mol.- The enthalpy of combustion for methane is -890.4 kJ/mol.

Answer 1

To calculate the standard enthalpy change of formation for methane  CH4 , we can use the following combustion reaction:CH4 g  + 2O2 g   CO2 g  + 2H2O l The enthalpy change for this reaction is given as -890.4 kJ/mol. We can use Hess's Law to find the enthalpy change of formation for methane. According to Hess's Law, the enthalpy change of a reaction is 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  =  Hf products  -  Hf reactants We are given the standard enthalpy change of formation for CO2  -393.5 kJ/mol  and H2O  -285.8 kJ/mol . The enthalpy change of formation for O2 is 0 kJ/mol since it is an element in its standard state. We can now plug in the values into the equation:-890.4 kJ/mol = [ -393.5 kJ/mol  + 2 -285.8 kJ/mol ] - [Hf CH4  + 2 0 kJ/mol ]-890.4 kJ/mol = -393.5 kJ/mol - 571.6 kJ/mol - Hf CH4 Now, solve for Hf CH4 :Hf CH4  = -393.5 kJ/mol - 571.6 kJ/mol + 890.4 kJ/molHf CH4  = -74.7 kJ/molTherefore, the standard enthalpy change of formation for methane  CH4  is -74.7 kJ/mol.