Question

A student needs to calculate the enthalpy change for a chemical reaction given the enthalpies of formation of the reactants and products. The specific problem is: Calculate the enthalpy change for the reaction:CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)Given the following enthalpies of formation:ΔHf(CH4) = -74.8 kJ/molΔHf(CO2) = -393.5 kJ/molΔHf(H2O) = -285.8 kJ/mol

Answer 1

To calculate the enthalpy change for the reaction, we can use the following formula:H reaction  =  Hf products  -  Hf reactants For this reaction, we have:H reaction  = [Hf CO2  + 2 * Hf H2O ] - [Hf CH4  + 2 * Hf O2 ]Since the enthalpy of formation for an element in its standard state is zero, Hf O2  = 0 kJ/mol. Now we can plug in the given values:H reaction  = [ -393.5 kJ/mol  + 2 *  -285.8 kJ/mol ] - [ -74.8 kJ/mol  + 2 *  0 kJ/mol ]H reaction  =  -393.5 kJ/mol - 571.6 kJ/mol  -  -74.8 kJ/mol H reaction  =  -965.1 kJ/mol  + 74.8 kJ/molH reaction  = -890.3 kJ/molSo, the enthalpy change for the reaction is -890.3 kJ/mol.