QuestionDetails: standard molar enthalpies of formation Using the componds given

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QuestionDetails: standard molar enthalpies of formation
Using the componds given below, calculate the standard enthalpy(kJ/mole) of a reaction that occurs at 298.15 K and produces 11.6gof H2O(l).
reactants: H2S(g) O2(g)

products: H2O(l) SO2(g) QuestionDetails: standard molar enthalpies of formation
Using the componds given below, calculate the standard enthalpy(kJ/mole) of a reaction that occurs at 298.15 K and produces 11.6gof H2O(l).
reactants: H2S(g) O2(g)

products: H2O(l) SO2(g) QuestionDetails: standard molar enthalpies of formation
Using the componds given below, calculate the standard enthalpy(kJ/mole) of a reaction that occurs at 298.15 K and produces 11.6gof H2O(l).
reactants: H2S(g) O2(g)

products: H2O(l) SO2(g)

Explanation / Answer

Chemical equation :                              2 H2S(g)+ 3 O2(g) ------------ > 2H2O(l) + 2 SO2(g) H0   =H0   products - H0   reactants           = 3mols * -258..83 kJ / mol + 2 mol * -296.9 kJ / mol - ( 2 mol *-20.17 kJ / mol + 3 mol * 0 kJ / mol )           = -776.49 kJ - 593.8 kJ - ( -40.34 kJ )           =-1329.95 kJ For 1 mol H0 = -1329.95 kJ No.of mols of H2O(l). = 11.6g / 18.015 g /mol                                  = 0.64 mol 0.64 mols of H2O(l)H0 = - 1329.95 kJ * 0.64mol / 1 mol                                            = -851.168 kJ

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