Question

Calculate the enthalpy change of the following reaction: 2C(s) + 2H2(g) + 5O2(g) -> 2CO2(g) + 2H2O(l) Given the enthalpies of formation of CO2(g), H2O(l), and H2(g) are -393.5 kJ/mol, -285.8 kJ/mol, and -241.8 kJ/mol, respectively.

Answer 1

To calculate the enthalpy change of the reaction, we will use the following formula:H_reaction =  H_f products  -  H_f reactants where H_f is the enthalpy of formation of each substance.For the products:2 moles of CO2 g : 2   -393.5 kJ/mol  = -787.0 kJ2 moles of H2O l : 2   -285.8 kJ/mol  = -571.6 kJFor the reactants:2 moles of C s : Since the enthalpy of formation of an element in its standard state is zero, the enthalpy of formation of C s  is 0 kJ/mol.2 moles of H2 g : 2   -241.8 kJ/mol  = -483.6 kJ5 moles of O2 g : Since the enthalpy of formation of an element in its standard state is zero, the enthalpy of formation of O2 g  is 0 kJ/mol.Now, we can calculate the enthalpy change of the reaction:H_reaction =  -787.0 kJ + -571.6 kJ  -  -483.6 kJ H_reaction = -1358.6 kJ + 483.6 kJH_reaction = -875.0 kJThe enthalpy change of the reaction is -875.0 kJ.