1. Thermochemistry

2. Energy is the capacity to do work Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom 6.1

3. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Energy Changes in Chemical Reactions Temperature is a measure of the thermal energy. 90 0 C 40 0 C greater thermal energy 6.2

4. Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open mass & energyExchange: closed energy isolated nothing 6.2

5. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy energy + 2HgO (s) 2Hg (l) + O 2 (g) 6.2 energy + H 2 O (s) H 2 O (l)

6. Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. energy, pressure, volume, temperature 6.3  E = E final - E initial  P = P final - P initial  V = V final - V initial  T = T final - T initial

7. First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.  E system +  E surroundings = 0 or  E system = -  E surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! 6.3 Chemical energy lost by combustion = Energy gained by the surroundings system surroundings

8. Another form of the first law for  E system 6.3  E = q + w  E is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -P  V when a gas expands against a constant external pressure

9. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.  H = H (products) – H (reactants)  H = heat given off or absorbed during a reaction at constant pressure H products < H reactants  H < 0 H products > H reactants  H > 0 6.4

10. Thermochemical Equations H 2 O (s) H 2 O (l)  H = 6.01 kJ Is  H negative or positive? System absorbs heat Endothermic  H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm. 6.4

11. Thermochemical Equations CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = -890.4 kJ Is  H negative or positive? System gives off heat Exothermic  H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm. 6.4

12. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m x s Heat (q) absorbed or released: q = m x s x  t q = C x  t  t = t final - t initial 6.5

13. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (  H 0 ) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (  H 0 ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f The standard enthalpy of formation of any element in its most stable form is zero.  H 0 (O 2 ) = 0 f  H 0 (O 3 ) = 142 kJ/mol f  H 0 (C, graphite) = 0 f  H 0 (C, diamond) = 1.90 kJ/mol f 6.6

15. The standard enthalpy of reaction (  H 0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD H0H0 rxn d  H 0 (D) f c  H 0 (C) f = [+] - b  H 0 (B) f a  H 0 (A) f [+] H0H0 rxn n  H 0 (products) f =  m  H 0 (reactants) f  - 6.6 (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

16. 16 Chemical Equilibrium Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

17. 17 Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) H 2 O (g) 2NO 2 (g) NO 2

18. 18 N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium

19. 19 constant

20. 20 N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] aA + bB cC + dD K = [C] c [D] d [A] a [B] b Law of Mass Action

21. 21 K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will K = [C] c [D] d [A] a [B] b aA + bB cC + dD

22. 22 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P 2 N2O4N2O4 P aA (g) + bB (g) cC (g) + dD (g) K p = K c (RT)  n  n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b) In most cases K c  K p

23. 23 Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = ′ [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] = constant K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] =K c [H 2 O] ′ General practice not to include units for the equilibrium constant.

24. 24 The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT)  n  n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7

25. 25 The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = 0.400 atm and P NO = 0.270 atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO 2 2 2 POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

26. 26 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. [CaO][CO 2 ] [CaCO 3 ] K c = ′ [CaCO 3 ] [CaO] K c x ′

27. 27 P CO 2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO

28. 28 Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate K p and K c for the reaction? NH 4 HS (s) NH 3 (g) + H 2 S (g) K p = P NH 3 H2SH2S P= 0.265 x 0.265 = 0.0702 K p = K c (RT)  n K c = K p (RT) -  n  n = 2 – 0 = 2 T = 295 K K c = 0.0702 x (0.0821 x 295) -2 = 1.20 x 10 -4

29. 29 A + B C + D C + D E + F A + B E + F K c = ′ [C][D] [A][B] K c = ′ ′ [E][F] [C][D] [E][F] [A][B] K c = KcKc ′ KcKc ′′ KcKc K c = KcKc ′′ KcKc ′ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

30. 30 N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] 2NO 2 (g) N 2 O 4 (g) K = [N 2 O 4 ] [NO 2 ] 2 ′ = 1 K = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

31. 31 Writing Equilibrium Constant Expressions 1.The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. 2.The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. 3.The equilibrium constant is a dimensionless quantity. 4.In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. 5.If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

32. 32 Chemical Kinetics and Chemical Equilibrium A + 2B AB 2 kfkf krkr rate f = k f [A][B] 2 rate r = k r [AB 2 ] Equilibrium rate f = rate r k f [A][B] 2 = k r [AB 2 ] kfkf krkr [AB 2 ] [A][B] 2 =K c = In terms of chemical kinetics, the equilibrium constant of a reaction can be expressed as a ratio of the rate constants of the forward and reverse reactions.

33. 33 Predicting the Direction of a Reaction The equilibrium constant Kc for the formation of hydrogen iodide from molecular hydrogen and molecular iodine in the gas phase is 54.3 at 430°C. Suppose that in a certain experiment we place 0.243 mole of H2, 0.146 mole of I2, and 1.98 moles of HI all in a 1.00-L container at 430°C. Will there be a net reaction to form more H2 and I2 or more HI? Inserting the starting concentrations in the equilibrium constant expression, we write

34. 34 The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium

35. 35 Calculating Equilibrium Concentrations 1.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3.Having solved for x, calculate the equilibrium concentrations of all species.

36. 36 At 1280 o C the equilibrium constant (K c ) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x

37. 37 K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M

38. 38 If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Châtelier’s Principle Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress

39. 39 Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left aA + bB cC + dD Add Remove

40. 40 Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas

41. 41 Le Châtelier’s Principle Changes in Temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases colder hotter N 2 O 4 (g) 2NO 2 (g)

42. 42 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

43. 43 Chemistry In Action Life at High Altitudes and Hemoglobin Production K c = [HbO 2 ] [Hb][O 2 ] Hb (aq) + O 2 (aq) HbO 2 (aq) Studies show that long-time residents of high-altitude areas have high hemoglobin levels in their blood—sometimes as much as 50 percent more than individuals living at sea level!

44. 44 Chemistry In Action: The Haber Process N 2 (g) + 3H 2 (g) 2NH 3 (g)  H 0 = -92.6 kJ/mol

45. 45 Le Châtelier’s Principle - Summary ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyes*no Volumeyes*no Temperatureyes Catalystno *Dependent on relative moles of gaseous reactants and products