2. ΔHΔH Every chemical reaction and change of physical state releases or absorbs heat. Goal – to determine whether heat is absorbed or released during a chemical reaction, therefore determine if exothermic or endothermic.

3. Thermochemistry Study of heat changes that accompany chemical reactions and phase changes Thermochemical equations are used to represent these reactions 4Fe(s) + 3 O 2 (g) → 2Fe 2 O 3 (s) + 1625kJ 27kJ + NH 4 NO 3 (s) → NH 4 + (aq) + NO 3 - (aq) First equation – exothermic- heat pack Second equation – endothermic – cold pack

4. Thermochemistry Terms System – specific part of the universe that contains reaction or process you want to study Surroundings – everything in universe other than system Universe = system + surroundings

5. Enthalpy Impossible to know the total heat content of a system because it depends upon many factors Chemists are interested in changes in energy during reactions For many reactions, energy lost or gained can be measured by calorimeter at constant pressure Enthalpy(H) – heat content of a system at constant pressure

6. We measure change in H (heat content) Exothermic reaction a downhill change Endothermic reaction an uphill energy change ΔH = H 2 – H 1 Products – reactants

7. Exothermic Reactions reactants products Enthalpy (kJ) Is ΔH positive or negative? Why? Products - Reactants!

8. Endothermic Reactions Enthalpy (kJ) reactants products ΔH is a positive #

9. Why is there a change in Enthalpy? due to energy required to break the bonds in the reactants and energy produced by forming the chemical bonds in the products.

10. A balanced equation can represent the energy absorbed or released Energy change for the reaction is called the Enthalpy of Reaction and is represented by Δ H r C (cr) + O 2 → CO 2 + energy (393.5kJ) ∆Hr = -393.5kJ

11. Standard States  H = change in enthalpy “ ° ” = Standard state enthalpy at 298.15K (25° C degrees) and 101.325 kilopascals (pressure) (No longer STP!) “f” = formation ΔHº f

12. Elements in the Natural State ELEMENTS ONLY ΔHº f = 0

13. ALL OTHER COMPOUNDS Enthalpy of formation found in standard tables Table C-13 pg921 ΔHº f

14. Enthalpy of Formation Δ H f represents the production of one mole from... its free elements in their standard states. (units = kJ/mole)

15. Calculation of Enthalpy of Reaction ΔH r = ΣΔHº f (products) - ΣΔHº f (reactants) Σ ( Sigma) is the symbol used to indicate summation. It means to ADD all the values of Δ H for all the products and subtract... the sum of all the Δ H of the reactants

16. EXAMPLE 1 Calculate the enthalpy change in the following chemical reaction carbon monoxide + oxygen → carbon dioxide 1) Write a balanced equation 2CO (g) + O 2 → 2CO 2

17. 2) Calculate Δ H f products 2 moles CO 2 x (-393.5 kJ) = -787.0 kJ 3) Calculate Δ H f reactants 2 moles CO + 1 mole O 2 = 2(-110.5 kJ) + 1(0 kJ) =-221 kJ Δ H r =  Δ H f ( products ) -  Δ H f ( reactants) Δ H r = (-787.0 kJ) - (-221. kJ) = -566.0 kJ This means the 566.0 kJ are released in the reaction

18. Practice Compute ΔHº r for the following reactions: 2NO (g) + O 2(g) → 2NO 2(g) -114.14kJ __FeO (cr) + O 2(g) → __ Fe 2 O 3(cr) -560.4kJ 42

19. Solving for change in enthalpy we get ΔH r = ΣΔHº f (products) - ΣΔHº f (reactants) Knowing the enthalpy of formation of each reactant and product, we can calculate the amount of energy produced or absorbed and predict whether a reaction will be exothermic or endothermic.

20. Exothermic Reactions reactants products Enthalpy (kJ) 115 kJ 75 kJ Δ H r = 75 kJ -115 kJ = - 40 kJ Negative Δ H r for EXOTHERMIC

21. Δ H = 247 kJ - 122 kJ = 125 kJ Positive Δ H r for ENDOTHERMIC Endothermic Reactions Enthalpy (kJ) reactants products 247 kJ 122 kJ

22. Spontaneous or Nonspontaneous That is the Question ? Suppose Δ H is negative Δ H f products < Δ H f reactants It’s EXOTHERMIC When Δ H r is negative reaction tends to be SPONTANEOUS Spontaneous means that it will occur without any outside influence How about Δ H positive?

24. Hess’s Law the enthalpy change for a reaction is the sum of the enthalpy changes for a series of reactions that add up to the overall reaction.

25. Hess’s Law N 2 + 2O 2 ↔ 2NO 2 41kJ N 2 O 4 ↔ 2NO 2 35kJ N 2 + 2O 2 ↔ N 2 O 4 Reverse the second reaction to get:

26. Hess’s Law N 2 + 2O 2 ↔ 2NO 2 ΔH= 41kJ 2NO 2 ↔ N 2 O 4 ΔH= 35kJ N 2 + 2O 2 ↔ N 2 O 4

27. Hess’s Law N 2 + 2O 2 ↔ 2NO 2 ΔH=41kJ 2NO 2 ↔ N 2 O 4 ΔH= 35kJ N 2 + 2O 2 ↔ N 2 O 4 Reversing the second reaction reverses the sign of the enthalpy of the reaction

28. Hess’s Law N 2 + 2O 2 ↔ 2NO 2 ΔH=41kJ 2NO 2 ↔ N 2 O 4 ΔH=-35kJ N 2 + 2O 2 ↔ N 2 O 4 Now add the two

29. Hess’s Law N 2 + 2O 2 ↔ 2NO 2 ΔH=41kJ 2NO 2 ↔ N 2 O 4 ΔH=-35kJ N 2 + 2O 2 ↔ N 2 O 4 6kJ

30. End

31. Entropy (S) is the measure of the degree of disorder in a system. A spontaneous process is one that occurs in a system left to itself. No external action is needed to make it happen.

32. Increase in disorder; Decrease in disorder; ∆S > 0 ∆S < 0

33. GIBBS FREE ENERGY Gibbs free energy indicates whether a reaction will occur or not.

34. Exergonic reactions (spontaneous) Endergonic reactions (nonspontaneous)

35. Endothermic reactions occur spontaneously when T Δ S is large. The thermodynamic definition of a system in equilibrium is; when Δ H and Δ S have the same sign, and there is some temperature at which Δ H and T Δ S are numerically equal.

36. A spontaneous reaction proceeds toward equilibrium. Chemical potential energy, G, is least at equilibrium. Enthalpy, entropy, and free energy depend on temperature. ( We will only work at 298.15 K and 100.00 kPa-standard states)

37. GIBBS FREE ENERGY CALCULATIONS Δ H values are relative; free elements are considered to have change in enthalpy is found by;

38. entropy changes and Gibbs free energy is computes as follows;

39. 1. Organize the data you will use from the appropriate table. 2. Multiply each Δ G f 0 by the number of moles from the balanced equation. (Always make sure the chemical equation is balanced). Substitute these values into the equation used to determine Δ G f 0.

40. The Gibbs free energy decreases in a spontaneous reaction because the system is changing to a more stable state.

41. To find 1. Multiply each by the number of moles from the balanced equation. 2. Substitute these values into the equation.