Chapter 17: Thermochemistry Calculating Heats of reaction

Learning Targets You will determine the heat of a reaction using Hess’s law and standard heat of formations. You will use bond energies to calculate the energy involved in a chemical reaction.

Hess’s Law Allows us to calculate the heat of reaction indirectly when it cannot be directly measured. States that 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. You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Hess’s Law Although the enthalpy change for this reaction cannot be measured directly, you can use Hess’s law to find the enthalpy change for the conversion of reactants to products by using the corresponding reactions. Can multiply, divide or reverse corresponding reactions. Whatever you do to the reaction you MUST do to the ∆ H. You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Book Example C(s, diamond) → C(s, graphite) a. C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ b. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ c. CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ If you add equations b and c, you get the equation for the conversion of diamond to graphite. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ CO2(g) → C(s, graphite) + O2(g) ΔH = 393.5 kJ C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Calculate the ΔH for the following reaction: Example #1 4 CuO (s)  2 Cu2O (s) + O2 (g) ΔH = 288 kJ Cu2O (s)  Cu (s) + CuO (s) ΔH = 11 kJ Calculate the ΔH for the following reaction: 2 Cu (s) + O2 (g)  2 CuO (s) -310 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Calculate ΔH for the reaction Example #2 From the following information S (s) + 3 2 O2 (g)  SO3 (g) ΔH = -395.2 kJ 2 SO2 (g) + O2 (g)  2 SO3 (g) ΔH = -198.2 kJ Calculate ΔH for the reaction S (s) + O2 (g)  SO2 (g) -296.1 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Calculate ΔH for the reaction Try on your own #1 Given the following data: C2H2 (g) + 5 2 O2 (g)  2 CO2 (g) + H2O (l) ΔH = - 1300. kJ C (s) + O2 (g)  CO2 (g) ΔH = - 394 kJ H2 (g) + 1 2 O2 (g)  H2O (l) ΔH = - 286 kJ Calculate ΔH for the reaction 2 C (s) + H2 (g)  C2H2 (g) 226 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Calculate ΔH for the reaction Try on your own #2 Given the following data: 2 O3 (g)  3 O2 (g) ΔH = - 427 kJ O2 (g)  2 O (g) ΔH = + 495 kJ NO (g) + O3 (g)  NO2 (g) + O2 (g) ΔH = - 199 kJ Calculate ΔH for the reaction NO (g) + O (g)  NO2 (g) -233 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Standard Heats of Formation Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. For a reaction that occurs at standard conditions the standard heat of reaction is the difference between the standard heats of formation of all the reactants and products. You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Example What is the standard heat of reaction (ΔH°) for the reaction of CO(g) with O2(g) to form CO2(g)? 2CO(g) + O2(g) → 2CO2(g) ΔHf°(reactants) = 2 mol CO(g)  ΔHf°CO(g) + 1 mol O2(g)  ΔHf°O2(g) ΔHf°(products) = 2 mol CO2(g)  ΔHf°CO2(g) ΔH° = ΔHf°(products) – ΔHf°(reactants) = (–787.0 kJ) – (–221.0 kJ) = –566.0 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Example #1 Calculate the standard heat of reaction for the following reaction: 2 SO2 (g) + O2 (g)  2 SO3 (g) -197.8 kJ ∆H° = ∆Hf°products − ∆Hf° reactants ∆H° = (2∙−395.7 kJ mol ) - [ 2 ∙−296.8 kJ mol + 0.0 kj/mol ] [ 2 ∙−296.8 kJ mol + 0.0 kj/mol ] ∆H° = -197.8 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Calculate ∆Hf° for the following reaction: Example #2 ∆H° = ∆Hf°products − ∆Hf° reactants ∆H° = 4 ·90.37 kJ + 6 ·(−241.8 kJ mol 4 ·90.37 kJ + 6 ·(−241.8 kJ mol −[ 4·−46.19 kJ mol + (5·(6 ·0.0 kJ mol )] ∆H° = - 9.046 x 102 kJ Calculate ∆Hf° for the following reaction: 4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g) - 9.046 x 102 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

CaCO3 (s)  CaO (s) + CO2 (g) Try on your own (2) What is the standard heat of reaction (∆H°) for the decomposition of hydrogen peroxide? 2 H2O2 (l)  2 H2O (l) + O2 (g) -1.960 x 102 kJ Use standard heats of formation (∆Hf°) to calculate the change in enthalpy for this reaction. CaCO3 (s)  CaO (s) + CO2 (g) 178.4 kJ You will determine the heat of a reaction using Hess’s law and standard heat of formations.

Bond Energy Bond energy is the amount of energy required to break apart a mole of molecules into its component atoms. When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, there will always be an accompanying energy change. ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed) You will use bond energies to calculate the energy involved in a chemical reaction.

Example #1 Reactants = broken bonds and Products = formed bonds Reactants: 2 x 436 kJ/mol = 872 kJ 1 x 499 kJ/mol = 499 kJ Products: 4 x 463 kJ/mol = 1852 kJ ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed) ΔH = (872 kJ + 499 kJ) – (1852 kJ) ΔH = - 481 kJ Find ∆H° for the following reaction given the following bond energies: 2 H2 (g) + O2 (g)  2 H2O (g) ∆H° = - 481 kJ You will use bond energies to calculate the energy involved in a chemical reaction.

CH3CH2CH3 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g) Example #2 The complete combustion of propane can be represented by the following equation: CH3CH2CH3 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g) Find ∆H° given the following bond energies: - 2054 kJ/mol Reactants = broken bonds and Products = formed bonds Reactants: 8 x 413 kJ/mol = 3304 kJ 5 x 498 kJ/mol = 2490 kJ 2 x 347 kJ/mol = 694 kJ Products: 6 x 805 kJ/mol = 4830 kJ 8 x 464 kJ/mol = 3712 kJ ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed) ΔH = (3304 kJ + 2490 kJ + 694 kJ) – (4830 kJ + 3712 kJ) ΔH = - 2054 kJ Bond Type Average bond enthalpy (kJ/mol) C-H + 413 C-C + 347 O=O +498 C=O +805 H-O +464 You will use bond energies to calculate the energy involved in a chemical reaction.

Try on your own (2) Calculate the heat of reaction for H2 + Cl2  2 HCl - 185 kJ Using the bond enthalpies provided, calculate the heat of reaction, ΔH, for: C2H4(g) + H2(g) → C2H6(g) - 124 kJ Bond Energies H – H 436 kJ/mol Cl – Cl 243 kJ/mol H - Cl 432 kJ/mol Bond Enthalpies C – H 413 kJ C = C 614 kJ C – C 348 kJ H – H 436 kJ You will use bond energies to calculate the energy involved in a chemical reaction.