Question

Heat of Formation Calculations: 32) Use a standard enthalpies of formation (Ho) table to determine the change in enthalpy for each of these reactions Hrxn [n. Ho(products) - n. Ho(products)] CO (g): -110.5 kJ/mol; CO2 (g): -393.5 kJ/mol CH4 (g): -890.4 kJ/mol H2O (l): -285.8 kJ/mol; H2O (g): -241.8 kJ/mol H2S (g): -20.6 kJ/mol; NO: -90.2 kJ/mol NO2: +33.9 kJ/mol; HCl (g): -92.3 kJ/mol NaOH (s): -426.7 kJ/mol; SO2 (g): -296.8 kJ/mol

a) CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l)

b) 2 H2S(g) + 3 O2(g) ---> 2 H2O(l) + 2 SO2(g)

c) 2 NO(g) + O2(g) ---> 2 NO2(g)

Answer 1

CH4 (g): -890.4 kJ/mol , H2O (l): -285.8 KJ/mole , CO2 (g): -393.5 kJ/mol

a) CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l)

$\Delta$H⁰rxn     = $\Delta$ H⁰f products - $\Delta$ H⁰f reactants

                 = 2*-285.8-393.5-(-890.4 + 2*0)

                 = -74.7KJ/mole

b) 2 H2S(g) + 3 O2(g) ---> 2 H2O(l) + 2 SO2(g)

$\Delta$H⁰rxn     = $\Delta$ H⁰f products - $\Delta$ H⁰f reactants

                  = 2*-296.8+2*-285.8-(2*-20.6+3*0)

                    = -1124KJ/mole

c) 2 NO(g) + O2(g) ---> 2 NO2(g)

$\Delta$H⁰rxn     = $\Delta$ H⁰f products - $\Delta$ H⁰f reactants

                   = 2*33.9-(2*-90.2+0)

                   = 248.2KJ/mole