1 / 6

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

keagan
Download Presentation

Hess’s Law

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related