Topic 5.3 and 5.4 Hess’s Law and Bond Enthalpies

Topic 5.3 and 5.4Hess’s Law and Bond Enthalpies

Hess’s LawTopic 5.3 • CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890 KJ

shows three different pathways: A  B A  C  B A  D  E  B • enthalpy change from reactants to products for all of these is the same

if a series of reactions are added together, the enthalpy change for the net reaction (Hfinal) will be the sum of the enthalpy change for the individual reactions (Hind + Hind + Hind ….) • the change in enthalpy is the same whether the reaction takes place in one step, or in a series of steps • H is independent of the reaction pathway • depends only on the difference between the enthalpy of the products and the reactants • H = Hproducts−Hreactants • provides a way to calculate enthalpy changes even when the reaction cannot be performed directly 8

energy in products energy in reactants

Problem-Solving Strategy • work backwards from the final reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal • if a reaction is reversed the sign on ΔH is reversed • N2 (g) + 2O2 (g) → 2NO2 (g) ΔH = 68kJ • 2NO2 (g) → N2 (g) + 2O2 (g) ΔH = - 68kJ • multiply reactions to give the correct numbers of reactants and products in order to get the final reaction. • the value of Δ H is also multiplied by the same integer • identical substances found on both sides of the summed equation cancel each other out

Example 1 • Given: N2 (g) + O2 (g)  2 NO (g) DH1 = +181 kJ 2 NO(g) + O2 (g)  2 NO2 (g) DH2 = -113 kJ • Find the enthalpy change for: N2 (g) + 2 O2 (g)  2 NO2 (g) • DH = DH1 + DH2 = +181 kJ + (-113 kJ) = + 68 kJ

Example 2 ΔH = (- 184 kJ) + (+ 1452 kJ) = + 1268 kJ

Example 3 • Given: C (s) + O2 (g)  CO2 (g) DH1 = - 393 kJ mol-1 H2 (g) + ½O2 (g)  H2O (g) DH2 = - 286 kJ mol-1 CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g) DH3 = - 890 kJ mol-1 • Find the enthalpy change for: C (s) + 2H2 (g)  CH4(g) This equation needs to be “flipped”. The CH4 is on the wrong side of the equation 13

Example 3 • Given: C (s) + O2 (g)  CO2 (g) DH1 = - 393 kJ mol-1 H2 (g) + ½O2 (g)  H2O (g) DH2 = - 286 kJ mol-1 CO2 (g) + 2H2O (g)  CH4 (g) + 2O2 (g) DH3 = + 890 kJ mol-1 • Find the enthalpy change for: C (s) + 2H2 (g)  CH4(g) • (- 393 kJ mol-1) + (- 572 kJ mol-1 ) + (+ 890 kJ mol-1 ) = - 75 kJ mol-1 2 2 - 572 1

Using enthalpy cycles instead… • counter-clockwise reaction energy (-3,222) has to equal clockwise reaction energy (-3,267 + ΔH) • answer is + 45 KJ/mol

clockwise needs to equal counter-clockwise • -109 = 52.2 + (-92.3) + ΔH • ΔH =- 68.9 kJ mol-1

Bond Enthalpies. Topic 5.4 • can be used to calculate the enthalpy change for a chemical reaction if we know the energy necessary to break or form bonds in the gaseous state • breaking bonds • energy is required so enthalpy is positive (endothermic) • the molecule was stable so energy was necessary to break apart the molecule • forming bonds • energy is released so enthalpy is negative (exothermic) • the new molecule is more stable than the individual atoms so energy is released

a molecule with strong chemical bonds generally has less tendency to undergo chemical change than does one with weak bonds • SiO bonds are among the strongest ones that silicon forms • it is not surprising that SiO2 and other substances containing SiO bonds (silicates) are so common • it is estimated that over 90 percent of Earth's crust is composed of SiO2 and silicates

we use average bond enthalpies • again, in the gaseous state • different amount of energy can be required to break the same bond • example- methane, CH4 • if you took methane to pieces, one hydrogen at a time, it needs a different amount of energy to break each of the four C-H bonds • every time you break a hydrogen off the carbon, the environment of those left behind changes, and the strength of the remaining bonds is affected • therefore, the 412 kJ mol-1 needed to break C-H is just an average value, may vary, and is not very accurate

The average bond enthalpies for several types of chemical bonds are shown in the table below: 20

Bond Enthalpy Calculations Example 1: Calculate the enthalpy change for thereaction. Is it endo or exothermic? N2 + 3 H2 2 NH3 • Bonds broken • 1 N N = 945 kJ • 3 H-H 3(435) = 1305 kJ • Total = 2250 kJ • Bonds formed • 2x3 = 6 N-H: 6 (390) = - 2340 kJ • Net enthalpy change • = (+ 2250) + (- 2340) = - 90 kJ (exothermic) H-H

Example 2 energy 2H2 + O2 2H2O course of reaction

Working out ∆H Show all the bonds in the reactants H―H + O=O energy H―H 2H2O course of reaction

O O H H H H Working out ∆H Show all the bonds in the products H―H + O=O energy H―H course of reaction

O O H H H H Working out ∆H Show the bond energies for all the bonds 436 + O=O energy 436 course of reaction

O O H H H H Working out ∆H Show the bond energies for all the bonds 436 + 498 energy 436 course of reaction

O H H Working out ∆H Show the bond energies for all the bonds 436 + 498 energy 436 464 + 464 course of reaction

Working out ∆H Show the bond energies for all the bonds 436 + 498 energy 436 464 + 464 464 + 464 course of reaction

Working out ∆H Add the reactants’ bond energies together energy 1370 464 + 464 464 + 464 course of reaction

Working out ∆H Add the products’ bond energies together 1370 energy 1856 course of reaction

Working out ∆H ∆H = energy in ― energy out 1370 1856 + - energy 1370 1856 course of reaction

Working out ∆H ∆H = energy in ― energy out 1370 1856 - 486 + - 1370 energy 1856 course of reaction

Working out ∆H ∆H = energy in ― energy out ∆H = -486 exothermic 1370 energy 1856 course of reaction

Topic 5.3 and 5.4 Hess’s Law and Bond Enthalpies