Question

Calculate the standard enthalpy change for the reaction between methane and oxygen to form carbon dioxide and water vapor at 298 K and 1 atm of pressure, given the following reaction equation:CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)The standard enthalpies of formation for CH4(g), CO2(g), and H2O(g) are -74.8 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol, respectively.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the given reaction:CH4 g  + 2O2 g   CO2 g  + 2H2O g The standard enthalpies of formation are:Hf CH4  = -74.8 kJ/molHf CO2  = -393.5 kJ/molHf H2O  = -241.8 kJ/molSince O2 is an element in its standard state, its enthalpy of formation is 0 kJ/mol.Now, we can plug these values into the equation:H = [1   -393.5  + 2   -241.8 ] - [1   -74.8  + 2  0]H =  -393.5 - 483.6  -  -74.8 H = -877.1 + 74.8H = -802.3 kJ/molTherefore, the standard enthalpy change for the reaction between methane and oxygen to form carbon dioxide and water vapor at 298 K and 1 atm of pressure is -802.3 kJ/mol.