Hess’s Law H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. However, we can estimate H using published H values and the properties of enthalpy. © 2009, Prentice-Hall, Inc.

Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” © 2009, Prentice-Hall, Inc.

Enthalpies of Formation An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. © 2009, Prentice-Hall, Inc.

Standard Enthalpies of Formation Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). © 2009, Prentice-Hall, Inc.

Writing enthalpy of formation equations: CO2(g) C(s) + O2(g)  CO2(g) H2O(l) H2(g) + ½ O2(g)  H2O (l)

Calculation of H We can use Hess’s law in this way: H = nHf°products – mHf° reactants where n and m are the stoichiometric coefficients. © 2009, Prentice-Hall, Inc.

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Use Hf° for reactants and products to solve for the enthalpy of this reaction: C3H8(g) Hf° = -104.7 kJ

Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” © 2009, Prentice-Hall, Inc.

Hess’s Law Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products. © 2009, Prentice-Hall, Inc.

Hess’s Law Problem C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) Calculate the enthalpy for the following reaction using the given chemical equations: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) 3C(s) + 4H2(g)  C3H8(g) ΔH1 = - 104.7 kJ C(s) + O2(g)  CO2(g) ΔH2 = -393.5 kJ 2H2(g) + O2(g)  2H2O(l) ΔH3 = -571.7 kJ

Hess’s Law and enthalpy of reaction Steps for solving using Hess’s Law: Set up equations so reactants and products are on proper side of the yield arrow. Use multipliers to get the correct amount of reactants and products (includes ΔHrxn) Make sure other substances that are not part of the overall equation cancel out. Add the sum of the ΔH for the contributing reactions.

Rearranged equations & adjust with multipliers C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) C3H8(g)  3C(s) + 4H2(g) ΔH1 = + 104.7 kJ 3[C(s) + O2(g)  CO2(g)] ΔH2 = -393.5 kJx3 2[2H2(g) + O2(g)  2H2O(l)] ΔH3 = -571.7 kJx2

Answer C3H8(g)  3C(s) + 4H2(g) ΔH1 = + 104.7 kJ 3C(s) + 3O2(g)  3CO2(g) ΔH2 = - 1180 4H2(g) + 2O2(g)  4H2O(l) ΔH3 = - 1143 C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) ΔH = -2219.2 kJ

Hess’s law problem: Given data: 4CuO(s)  2Cu2O(s) + O2(g) ΔH = 288 kJ Cu2O(s)  Cu(s) + CuO(s) ΔH = 11 kJ Write the standard formation reaction for CuO(s) Calculate the standard enthalpy of formation for CuO(s) using Hess’s Law.