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Hess’s Law

Hess’s Law. Hess’s Law. Hess’s Law states that the enthalpy change for a reaction is independent of the reaction pathway and depends only on the concentration of reactants . Examples. Combustion of Glucose The enthalpy change for the direct combustion of 1 mol glucose in one step

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Hess’s Law

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  1. Hess’s Law

  2. Hess’s Law • Hess’s Law states that the enthalpy change for a reaction is independent of the reaction pathway and depends only on the concentration of reactants

  3. Examples • Combustion of Glucose • The enthalpy change for the direct combustion of 1 mol glucose in one step • C6H12O6 + O2 H2O + CO2 DH -ve • Is equal to the enthalpy change when 1 mol of glucose is combusted in the body – approximately 50 chemical reactions • C6H12O6 + O2 H2O + CO2 DH -ve

  4. Questions – equation writing Write thermochemical equations to represent the following: • DfH°(CO2, g, 298K) = -393 kJmol-1 • DcH°(C, s, 298K) = -393 kJmol-1 • DfH°(CH4, g, 298K) = -75 kJmol-1 • DfH°(H2SO4, g, 298K) = -814 kJmol-1 • DcH°(C8H18, g, 298K) = -5464 kJmol-1

  5. Questions –When data given is enthalpies of formation only • Calculate the DcH° for CH4 given: DfH°(CH4(g)) = -75 kJmol-1, DfH°(CO2(g)) = -393 kJmol-1, DfH°(H2O(l)) = -285 kJmol-1 • Calculate the DrH° for CO(g) + 2H2(g) CH3OH(l) given: DfH°(CH3OH(g)) = -201 kJmol-1, DfH°(CO(g)) = -111 kJmol-1 • Calculate the DrH° for • 2 CH3OH(l) C2H4(g) + 2H2O(l) and • 4C2H4(g) + H2(g) C8H18(l) given: DfH°(CH3OH(l)) = -201 kJmol-1, DfH°(C2H4(g)) = +52 kJmol-1, DfH°(H2O(l)) = -285 kJmol-1,DfH°(C8H18(l)) = -208 kJmol-1 C) Calculate the enthalpy change for the reaction which produces 1 mol of octane from 8 moles of methanol. Is it exo or endothermic?

  6. Answers: • -888 kJmol-1 • -90 kJmol-1 • A) -116 kJmol-1 B)-416 kJmol-1 C) -880 kJmol-1, exothermic

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