I. Reaction Energy Section 17-1 and 17-2 Ch. 17 – Reaction Energy and Reaction Kinetics

A. Reviewing… n Heat and Temperature  Temperature  Joule  Calorimeter  Heat

A. Reviewing… n Heat Capacity and Specific Heat Specific Heat – used to compare heat absorption of different materials n Heating and Cooling Curves q = m · ∆H f or v q = Cp · m · ∆T

B. Energy of Reactions n Reaction Pathway n Shows the change in energy during a chemical reaction

B. Energy of Reactions n Exothermic n reaction that releases energy n products have lower PE than reactants 2H 2 (l) + O 2 (l)  2H 2 O(g) + energy energy released

B. Energy of Reactions n Endothermic n reaction that absorbs energy n reactants have lower PE than products 2Al 2 O 3 + energy  4Al + 3O 2 energy absorbed

7 The heat that is released or absorbed in a chemical reaction Equivalent to  H C + O 2 (g)  CO 2 (g) kJ C + O 2 (g)  CO 2 (g)  H = kJ In thermochemical equation, it is important to indicate the physical state H 2 (g) + ½ O 2 (g)  H 2 O(g)  H = kJ H 2 (g) + ½ O 2 (g)  H 2 O(l)  H = kJ B. Energy of Reaction

8 Exothermic this is why gummy bears are widely disliked!! Gummy Bear Sacrifice Gummy Bear Sacrifice Gummy Bear Sacrifice and anotherGummy Bear Sacrifice and another Endothermic Mr. Wizard B. Energy of Reactions

9 The heat content a substance has at a given temperature and pressure n Can’t be measured directly because there is no set starting point n The reactants start with a heat content n The products end up with a heat content ΔH n So we can measure how much enthalpy changes ΔH C. Enthalpy

Enthalpy Change in Reaction n H = total of all forms of energy (bonds, PE, KE, phases, etc.) n ΔH = heat of reaction = change in enthalpy during chemical reaction ΔH > 0 thermodynamically stable - does not spontaneously decompose at room temperature ΔH < 0 thermodynamically unstable - spontaneous decomposition at room temperature -ΔH  favors spontaneous change! (kinetically stable - overall negative energy but imperceptible change due to slowness of reaction)

11 Energy ReactantsProducts  Change is down ΔH is <0 Exothermic

12 Energy ReactantsProducts  Change is up ΔH is > 0 Endothermic

D. Heat of Formation Heat of Formation – energy transferred when one mole of a compound is formed from its elements  ΔH f ° is symbol  depends on temperature, pressure and phase  ΔH f ° - standard heat of and kPa  Appendix A-14 page 902 Stable = large negative Unstable = small negative or positive

D. Heat of Formation 2HgO(s) --> 2Hg(l) + O2(g)ΔH = 182 kJ What’s the standard heat of formation of HgO? (stable or unstable?)

D. Heat of Formation 4Fe(s) + 3O 2 (g) --> 2Fe 2 O 3 (s) ΔH = kJ What is the heat of formation of rust? (stable or unstable ?)

16 The standard heat of formation of an element in it’s standard state is arbitrarily set at “0 kJ” This includes the diatomics! D. Heat of Formation

E. Calculating Heat of Reaction method 1: ΔH = H final - H initial (state function) heat of reaction = ∑ ΔH formation of products - ∑ΔH formation of reactants multiply heat of formation by coefficient if more than one mole being used in balanced equation

18 CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)  H= kJ E. Calculating Heat of Reaction

Cl 2 (g) + 2HBr(g)  2HCl(g) + Br 2 (g) 2NO(g) + O 2 (g)  2NO 2 (g)

20 method 2: Hess’s Law If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. E. Calculating Heat of Reaction

21 If you turn an equation around, you change the sign: H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H= kJ then H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H = kJ If you multiply the equation by a number, you multiply the heat by that number: 2 H 2 O(g)  H 2 (g) + O 2 (g)  H = kJ Or, you can just leave the equation “as is” E. Calculating Heat of Reaction

CuO(s) + H 2 (g)  Cu(s) + H 2 O(g) ΔH = ? Cu(s) + ½ O 2 (g)  CuOΔH 1 = -155 kJ H 2 (g) + ½ O 2  H 2 O(g) ΔH 2 = -242 kJ rearranging: CuO(s) --> Cu(s) + ½ O 2 (g) 155 kJ H 2 (g) + ½ O 2 --> H 2 O(g) -242 kJ ΔH = -87 kJ

E. Calculating Heat of Reaction Calculate the enthalpy for the formation of C 5 H 12. 5C(s) + 6H 2 (g)  C 5 H 12 ΔH = ? C(s) + O 2 (g)  CO 2 (g)ΔH = kJ H 2 (g) + ½ O 2 (g)  H 2 O(l)ΔH = kJ C 5 H 12 (s) + 8O 2 (g)  5CO 2 (g) + 6H 2 O(l)ΔH = kJ

EXAMPLE (a little harder) B 2 O 3 (s) + 3H 2 O(g) → 3O 2 (g) + B 2 H 6 (g) ΔH = kJ H 2 O(l) → H 2 O(g) ΔH = +44 kJ H 2 (g) + ½ O 2 (g) → H 2 O(l) ΔH = -286 kJ 2B(s) + 3H 2 (g) → B 2 H 6 (g) ΔH = +36 kJ 2B(s) + 3/2 O 2 (g) → B 2 O 3 (s) ΔH =?

SOLUTION After the multiplication and reversing of the equations (and their enthalpy changes), the result is: B 2 H 6 (g) + 3O 2 (g) → B 2 O 3 (s) + 3H 2 O(g) ΔH = kJ 3H 2 O(g) → 3H 2 O(l) ΔH = (-44 x 3) = -132 kJ 3H 2 O(l) → 3H 2 (g) + 3/2 O 2 (g) ΔH = (+286 x 3) = +858 kJ 2B(s) + 3H 2 (g) → B 2 H 6 (g) ΔH = +36 kJ Adding these equations and canceling out the common terms on both sides, we get 2B(s) + 3/2 O 2 (g) → B 2 O 3 (s) ΔH = kJ

26 Heat of Combustion The heat from the reaction that completely burns 1 mole of a substance: C + O 2 (g)  CO 2 (g) kJ same as saying: C + O 2 (g)  CO 2 (g)  H = kJ