Question

Calculate the enthalpy change for the reaction: 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) given the enthalpies of formation of H2O(g), SO2(g), and H2S(g) are -241.8 kJ/mol, -296.8 kJ/mol, and -20.2 kJ/mol, respectively.

Answer 1

To calculate the enthalpy change for the reaction, we can use the formula:H_reaction =  H_f products  -  H_f reactants where H_f is the enthalpy of formation.For the products:2 moles of H2O g : 2   -241.8 kJ/mol  = -483.6 kJ2 moles of SO2 g : 2   -296.8 kJ/mol  = -593.6 kJFor the reactants:2 moles of H2S g : 2   -20.2 kJ/mol  = -40.4 kJ3 moles of O2 g : 3  0 kJ/mol  since O2 is in its standard state  = 0 kJNow, we can plug these values into the formula:H_reaction =  -483.6 kJ +  -593.6 kJ   -  -40.4 kJ + 0 kJ H_reaction = -1077.2 kJ + 40.4 kJH_reaction = -1036.8 kJThe enthalpy change for the reaction is -1036.8 kJ.