2. Hess’s Law and Enthalpies of Formation Sections 5.6-5.7

3. Objectives Establish standard values for enthalpy changes in chemical reactions Use these standards to calculate enthalpy changes for reactions

4. PA Academic Standards 3.4.12 Apply and analyze energy sources and conversions and their relationship to heat and temperature. Determine the heat involved in illustrative chemical reactions. Evaluate mathematical formulas that calculate the efficiency of specific chemical and mechanical systems.

5. Key Terms Enthalpy State Function Hess’s Law Enthalpy of formation Standard enthalpy change Standard enthalpy of formation

6. Hess’s Law  H’s have been measured and recorded for many reactions We can estimate  H for a given reaction from the published  H of other reactions Forgo calorimetric measurements

7. Hess’s Law Hess’s law : “If 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.”

8. 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. The “in between” does not matter. Thus the steps are additive.

9. Hess’s Law

10. Example C + O 2  CO 2 Publications tell us that this can occur as 2 steps C + ½O 2  CO  H  = – 110.5 kJ CO + ½O 2  CO 2  H  = – 283.0 kJ C + CO + O 2  CO + CO 2  H  = – 393.5 kJ C + O 2  CO 2  H  = – 393.5 kJ Hess’s law allows us to add equations. We add all reactants, products, &  H  values

11. Tabulated Enthalpies Experimental data is grouped according to the type of process Examples: Enthalpies of vaporization Enthalpies of fusion Enthalpies of combustion Enthalpies of formation

12. Enthalpies of Formation  H f Enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Subscript f indicates substance has been formed

13. Magnitude of Enthalpy Temperature Pressure State (gas, liquid, solid) of reactants and products In order to compare enthalpies, a set of conditions for tabulation must be defined = standard state

14. Standard State Standard enthalpies of formation,  H f, are measured under standard conditions (25°C and 1.00 atm pressure). 

15. Standard Enthalpy of Change Enthalpy change when all reactants and products are in their standard states ΔH°

16. Standard Enthalpy of Formation Change in enthalpy for the reaction that forms one mole of the compound from its elements with all substances in their standard states Usually reported at 298 K ΔH° f

17. Standard Enthalpy of Formation Standard enthalpy of formation of the most stable form of any element is zero because there is no formation reaction needed when element is in its standard state

18. SAMPLE EXERCISE 5.10 Identifying Equations Associated with Enthalpies of Formation For which of the following reactions would the enthalpy change represent a standard enthalpy of formation? For those where it does not, what changes would need to be made in the reaction conditions? Solution Analyze: The standard enthalpy of formation is represented by a reaction in which each reactant is an element in its standard state and the product is one mole of the compound. Plan: To solve these problems, we need to examine each equation to determine, first of all, whether the reaction is one in which a substance is formed from the elements. Next, we need to determine whether the reactant elements are in their standard states. Solve: In (a) Na 2 O is formed from the elements sodium and oxygen in their proper states, a solid and O 2 gas, respectively. Therefore, the enthalpy change for reaction (a) corresponds to a standard enthalpy of formation. In (b) potassium is given as a liquid. It must be changed to the solid form, its standard state at room temperature. Furthermore, two moles of product are formed, so the enthalpy change for the reaction as written is twice the standard enthalpy of formation of KCl(s). The proper equation for the formation reaction is

19. PRACTICE EXERCISE Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl 4 ). SAMPLE EXERCISE 5.10 continued Reaction (c) does not form a substance from its elements. Instead, a substance decomposes to its elements, so this reaction must be reversed. Next, the element carbon is given as diamond, whereas graphite is the lowest- energy solid form of carbon at room temperature and 1 atm pressure. The equation that correctly represents the enthalpy of formation of glucose from its elements is

20. Calculation of  H Imagine this as occurring in 3 steps: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g)  3 C + 4 H 2 (g) 3 C + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l)

21. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H The sum of these equations is:

22. Calculation of  H We can use Hess’s law in this way:  H =  n  H f(products) -  m  H f(reactants) where n and m are the stoichiometric coefficients. 

23. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H  H= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ