Δ H Every chemical reaction and change of physical state releases or absorbs heat.

1. Enthalpy ΔH 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.

2. Thermochemistry Study of heat changes that accompany chemical reactions and phase changes Thermochemical equations are used to represent these reactions 4Fe(s) + 3 O2(g) → 2Fe2O3(s) + 1625kJ 27kJ + NH4NO3(s) → NH4+(aq) + NO3-(aq) First equation – exothermic- heat pack Second equation – endothermic – cold pack

3. 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

4. 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

5. We measure change in H (heat content) Exothermic reaction a downhill change Endothermic reaction an uphill energy change Enthalpy ΔH = H2 – H1 Products – reactants

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

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

8. 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.

9. 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 ΔHr C(cr) + O2 → CO2 + energy (393.5kJ) ∆Hr = -393.5kJ

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

11. Elements in the Natural State ΔHºf = 0 ELEMENTS ONLY

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

13. Enthalpy of Formation • ΔHf represents the production of one mole from ... • its free elements in their standard states. (units = kJ/mole)

14. ΔHr = ΣΔHºf (products) - ΣΔHºf (reactants) Σ (Sigma) is the symbol used to indicate summation. It means to ADD all the values of ΔH for all the products and subtract ... the sum of all the ΔH of the reactants Calculation of Enthalpy of Reaction

15. Calculate the enthalpy change in the following chemical reaction carbon monoxide + oxygen → carbon dioxide 1) Write a balanced equation 2CO (g) + O2→ 2CO2 EXAMPLE 1

16. 2) Calculate ΔHf products 2 moles CO2 x (-393.5 kJ) = -787.0 kJ 3) Calculate Δ Hf reactants 2 moles CO + 1 mole O2 = 2(-110.5 kJ) + 1(0 kJ) =-221 kJ ΔHr =SΔHf (products) -SΔHf (reactants) ΔHr = (-787.0 kJ)-(-221. kJ) = -566.0 kJ This means the 566.0 kJ are released in the reaction 2CO (g) + O2→ 2CO2

17. Practice • Compute ΔHºr for the following reactions: 2NO(g) + O2(g) → 2NO2(g) -114.14kJ __FeO(cr)+ O2(g) → __ Fe2O3(cr) • -560.4kJ 4 2

18. Solving for change in enthalpy we get ΔHr = ΣΔHºf (products) - ΣΔHºf (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. Enthalpy

19. products Exothermic Reactions • Δ Hr= 75 kJ -115 kJ • = - 40 kJ • Negative Δ Hr for EXOTHERMIC reactants 115 kJ Enthalpy (kJ) 75 kJ

20. products Endothermic Reactions 247 kJ • Δ H = 247 kJ - 122 kJ • = 125 kJ • Positive Δ Hr for ENDOTHERMIC Enthalpy (kJ) 122 kJ reactants

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

23. 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.

24. Hess’s Law N2 + 2O2 ↔ 2NO2 41kJ N2O4 ↔ 2NO2 35kJ N2 + 2O2 ↔ N2O4 Reverse the second reaction to get:

25. Hess’s Law N2 + 2O2 ↔ 2NO2ΔH= 41kJ 2NO2 ↔N2O4ΔH= 35kJ N2 + 2O2 ↔ N2O4

26. Hess’s Law N2 + 2O2 ↔ 2NO2ΔH= 41kJ 2NO2 ↔N2O4ΔH= 35kJ N2 + 2O2 ↔ N2O4 Reversing the second reaction reverses the sign of the enthalpy of the reaction

27. Hess’s Law N2 + 2O2 ↔ 2NO2ΔH= 41kJ 2NO2 ↔N2O4ΔH=-35kJ N2 + 2O2 ↔ N2O4 Now add the two

28. Hess’s Law N2 + 2O2 ↔ 2NO2ΔH= 41kJ 2NO2 ↔N2O4ΔH=-35kJ N2 + 2O2 ↔ N2O46kJ

29. End

30. 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.

31. Increase in disorder; • Decrease in disorder; ∆S > 0 ∆S < 0

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

33. Exergonic reactions (spontaneous) • Endergonic reactions (nonspontaneous)

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

35. 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)

36. GIBBS FREE ENERGY CALCULATIONS • ΔH values are relative; free elements are considered to have • change in enthalpy is found by;

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

38. 1. Organize the data you will use from the appropriate table. • 2. Multiply each ΔGf0 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 ΔGf0 .

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

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