2. Energy = force x distance (Joules) In chemical reactions, we need energy usually in the form of heat. Energy is absorbed to break the bonds of the reactants and energy is given out when new bonds are formed in the products. Exothermic reactionsEndothermic reactions

3.  Heat is the energy transferred between objects that are at different temperatures.  The amount of heat transferred depends on the amount of the substance. ◦ Energy is measured in units called joules (J).

4.  Temperature is a measure of “hotness” of a substance and represent the average kinetic energy of the particles in a substance.  It does not depend on the amount of the substance. Do both beakers contain the same amount of heat?

5.  All chemical reactions are accompanied by some form of energy change  ExothermicEnergy is given out  Endothermic Energy is absorbed

6.  Enthalpy (H) is the internal energy stored in the reactants.  The absolute value of enthalpy of reactants or products cannot be found but we can only measure the difference between them.  We measure the enthalpy change, ∆ H i.e. the amount of heat released or absorbed when a chemical reaction occurs at constant pressure, measured in kilojoules per mole (kJmol -1 ). ∆ H = H (products) – H (reactants)

7. The standard conditions for enthalpy changes are :  Temperature of 298K or 25 0 C  Pressure of 100kPa (or 1 atm)  Concentration of 1 moldm -3 for all solutions  All substances in their standard states  in [kJmol -1 ]

8. Enthalpy Energy of products & reactants  In reaction A + B  C + D, the enthalpy of the products (C&D) is < the enthalpy of the reactants (A&B). Energy change : ∆ H = H (products) - H (reactants) ∆ H is negative since H (products) < H (reactants)  Heat is lost to the surrounding  Temperature increases Map

9.  Combustion reactions ◦ CH₄ (g) + 2O₂ (g)  CO₂ (g) + 2H₂O (l)  Neutralization (acid + base) ◦ NaOH(aq) + HCl(aq)  NaCl(aq) + H₂O(l)  Respiration ◦ C₆H₁₂O₆ (aq) + 6O₂ (g)  6CO₂ (g) + 6H₂O (l)

11. Enthalpy Energy of products & reactants  In reaction A + B  C + D, the enthalpy of the products (C&D) is > the enthalpy of the reactants (A&B). Energy change : ∆ H = H (products) - H (reactants) ∆ H is positive since H (products) > H (reactants)  Heat is gained from the surrounding  Temperature decreases

12.  Thermal decomposition  CaCO₃ (s)  CaO (s) + CO ₂ (g)  Photosynthesis  6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)

14.  Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin.  Unit : Jg -1 0 C -1 The specific heat capacity of alminium is 0.90 Jg -1 0 C -1. If 0.90J of energy is put into 1g of aluminium, the temperature will be raised by 1 0 C. Calculating heat absorbed and released q = c × m × ΔT q = heat change (absorbed or released) [ J ] c = specific heat capacity of substance m = mass of substance in grams ΔT = change in temperature in Celsius

15.  Heat given off by a process is measured through the temperture change in another substance (usually water).  Due to the law of conservation of energy, any energy given off in a process must be absorbed by something else  Assumption : energy given out will be absorbed by the water and cause a temperature change.  calculate the heat through the equation Q = mc ΔT

16. 1. How much heat is required to increase the temperature of 20 grams of nickel (specific heat capacity 440Jkg -1 0 C -1 ) from 50 0 C to 70 0 C? 2. If 500J of heat is added to 100.0g of samples of each of the substances below, which will have the largest temperature increase? SubstanceSpecific heat capacity / Jg -1 K 1 gold0.129 silver0.237 Copper0.385 water4.18

17. (1) Std enthalpy change of neutralisation - Overall enthalpy change (heat produced) when 1 mole of water is formed when an acid reacts with an alkali under std condition. E.g. HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) ΔH n θ =-394 kJmol -1 - Std enthalpy change of solution - Overall enthalpy change (heat produced or absorbed) when 1 mole of a solute is dissolved completely in excess solvent. E.g. NaOH(s) + (aq)  NaOH(aq) ΔH s θ =-394 kJmol -1

18. (3) Std enthalpy change of combustion - Overall enthalpy change (heat produced) when 1 mole of the substance is completely burnt in oxygen under std condition. C(s) + O 2 (g)  CO 2 (g) ΔH c θ =-394 kJmol -1 (4) Std enthalpy heat of formation - Overall enthalpy change (heat produced or absorbed) when 1 mole of substance is formed from its elements in their standard states. 2C(g) + 3H 2 (g) + ½O 2 (g)  C 2 H 5 OH(l) ΔH f θ =-277 kJmol -1

19. Example: Which of these are correct? (1) 2Fe(s) + O 2 (g)  Fe 2 O 3 (s) (2) 2C(s) + O 2 (g)  2CO(g) (3) C(s) + O 2 (g)  CO 2 (g) (4) C 2 H 2 (g) + 2H 2 (g)  C 2 H 6 (g) (5) N 2 (g) + 3H 2 (g)  2NH 3 (g)

20.  The standard enthalpy change of combustion for a substance is the enthalpy change (heat produced) when 1 mole of a pure substance is completely burnt in excess oxygen under standard conditions.  Example, CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) ΔH Ɵ c =-698 kJmol -1  The heat given out is used to heat another substance,e.g. water with a known specific heat capacity.  The experiment set-up can be used to determine the enthalpy change when 1 mole of a liquid is burnt. Example : refer to page 185

21. For exothermic reaction, heat produced by reaction is assumed to be absorbed by water, so temp of water increases. Heat change of reaction = - heat change of water = - m H2O x c H2O xΔT H2O For endothermic reaction, heat absorbed by reaction is assumed to be taken from the water, so temp of water decreases. Heat change of reaction = heat change of water = m H2O x c H2O xΔT H2O

22. The following measurements are taken:  Mass of cold water (g)  Temperature rise of the water ( 0 C)  The loss of mass of the fuel (g) We know that it takes 4.18J of energy to raise the temperature of 1g of water by 1 0 C. This is called the specific heat capacity of water, c, and has a value of 4.18Jg -1 K -1. Hence, energy transferred can be calculated using: Energy transfer = mcΔT (joules)  Enthalpy transfer = m x 4.18 x Δ T / No. of moles

23. Given that: Vol of water = 100 cm 3 Temp rise = 34.5 0 C Mass of methanol burned = 0.75g Specific heat capacity of water = 4.18 Jg -10 C -1 Calculate the molar enthalpy change of the combustion of methanol. What is the big assumption made with this type of experiment?

24.  Loss of heat to the surroundings (exothermic reaction); absorption of heat from the surroundings (endothermic reaction). This can be reduced by insulating the calorimeter.  Using incorrect specific heat capacity in the calculation of heat change. If copper can is used, the s.h.c. of copper must be accounted for.  Others include – e.g incomplete combustion. Some of the ethanol could be used to produce CO & soot & water (less heat is given out) Use bomb calorimeter – heavily insulated & substance is ignited electronically with good supply of oxygen

25.  If 1g of methanol is burned to heat 100g of water, raising its temperature by 42K, calculate the enthalpy change when 1 mole of methanol is burnt. Note: Specific heat capacity of water is 4.18 Jg -1 0 C -1 Practice questions page 187 #1-4

26. Enthalpy change of neutralisation (ΔH n )  The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of water molecules is formed when an acid reacts with an alkali under standard conditions. Example, NaOH(g) + HCl(g)  NaCl(g) + H 2 O(l) ΔH Ɵ =-57 kJmol -1  The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water.  Reaction between strong acid and strong base involves H + (aq) + OH - (aq)  H 2 O(l) ΔH Ɵ =-57 kJmol -1 For sulfuric acid, the enthalpy of neutralisation equation is ½ H 2 SO 4 (aq) + KOH(aq)  ½K 2 SO 4 (aq) + H 2 O(l) ΔH Ɵ =-57 kJmol -1 - Example : refer to page 188

27. For neutralisation between a weak acid, a weak base or both, the enthalpy of neutraisation will be smaller than -57 kJmol -1 (less exothermic) CH 3 COOH(aq) + NaOH(aq)  CH 3 COONa(aq) + H 2 O(l) ΔH Ɵ =-55.2 kJmol -1  Some of the energy released is used to ionise the acid.

28.  200.0cm 3 of 0.150 M HCl is mixed with 100.0cm 3 of 0.350 M NaOH. The temperature rose by 1.36 0 C. If both solutions were originally at the same temp, calculate the enthalpy change of neutralisation. Assume that the density of the solution is 1 gcm -3 and the specific heat capacity is 4.18J Jg -1 0 C -1. -56.8kJmol-1

29. The experimental change of neutralisation is -56.8 kJmol -1 The accepted literature value is -57.2 kJmol -1 (1) Heat loss to the environment. (2)Assumptions that (a)the denisty of NaOH and HCl solutions are the same as water. (b)the specific heat capacity of the mixture are the same as that of water

30. When 3 g of sodium carbonate are added to 50 cm 3 of 1.0 M HCl, the temperature rises from 22.0 °C to 28.5°C. Calculate the enthalpy change for the reaction. Assume that the density of the solution is 1 gcm -3 and the specific heat capacity is 4.18J Jg -1 0 C -1. Example : refer to page 189 dissolving ammonium chloride

31. The experimental change of solution is +13.8 kJmol -1 The accepted literature value is 15.2 kJmol -1 (1) Absorption of h eat from the environment. (2)Assumptions that the specific heat capacity of the solution is the same as that of water (3)The mass of ammonium chloride is not taken into consideration when working out the heat energy released.

32. 100.0 cm 3 of 0.100 mol dm -3 copper II sulphate solution is placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction. First step Make sure you understand the graph. Extrapolate to determine the change in temperature. The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used? Enthalpy changes of reaction in solution

33. Determine the limiting reactant Calculate Q Calculate the enthalpy for the reaction. 100.0 cm 3 of 0.100 mol dm -3 copper II sulphate solution is placed in a styrofoam cup. 1.30 g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction.

34. N 2(g) + 3H 2(g)  2NH 3(g) Δ H Ɵ = -92 kJmol -1 What does it mean? The enthalpy change of reaction is -92 kJmol -1 #

35. Fats store plenty of energy compared to sugar. The body would store up to 67.5g of sugar complexes for the energy equivalent of 10g of fat. Calculate the energy in which a fat is converted to sugar: C 18 H 36 O 2 (s) + 8O 2 (g)  3C 6 H 12 O 6 (s) given C 18 H 36 O 2 (s) + 26O 2 (g)  18CO 2 (g) +H 2 O(l) ΔH 1 =-11407kJ C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) +6H 2 O(l) ΔH 2 =-2800kJ

36.  Hess's law can be applied to calculate enthalpies of reactions that are difficult to measure  Enthalpy is a state function  It depends only upon the initial and final state of the reactants/products and not on the specific pathway taken to get from the reactants to the products 

37. Hess’s Law states that the enthalpy change for any chemical reaction is independent of the pathway provided the starting and final conditions, and reactants and products are the same.

38. Consider these thermo-chemial equations: (1) NaOH(s) + (aq)  NaOH(aq) ΔH Ɵ 1 =-43kJmol -1 (2) NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l) ΔH Ɵ 2 =-57kJmol -1 Calculate the enthaply change for (3) NaOH(s) + HCl(aq)  NaCl(aq) + H 2 O(l)

39. (1) Using cycles (2) Manipulating equations (3) Equations (4) Enthalpy level diagrams

40. Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide. NaOH(aq) NaOH(s) NaCl(aq) + H 2 O(l) + HCl(aq) 1. Indirect path: NaOH(s) + (aq)  NaOH(aq) ΔH Ɵ 1 =-43kJmol -1 2. NaOH(aq) + HCl (aq)  NaCl(aq) + H 2 O(l) ΔH Ɵ 2 =-57kJmol -1 3. NaOH(s) + HCl (aq)  NaCl(aq) + H 2 O(l). ΔH2ΔH2 ΔH1ΔH1 Indirect path Direct path + HCl(aq) + H 2 O(l)

41. Calculate the enthalpy change for the thermal decomposition of calcium carbonate. CaCO 3 (s)  CaO(s) + CO 2 (g) CaCO 3 (s) +2HCl(aq)  CaCl 2 (aq) + H 2 O(l) + CO 2 (g) ΔH Ɵ 1 =-17 kJmol -1 CaO(s) +2HCl(aq)  CaCl 2 (aq) + H 2 O(l) ΔH Ɵ 1 =-195kJmol -1

42. Calculate the enthalpy change for the combustion of carbon monoxide to form carbon dioxide. C(s) + O 2 (g)  CO 2 (g) ΔH Ɵ =-394 kJmol -1 2C(s) + O 2 (g)  2CO(g) ΔH Ɵ = -222kJmol -1 2CO(g) + O 2 (g)  2CO 2 (g)

43. From the following data at 25 0 C and 1 atmosphere pressure: Eqn 1: 2CO 2 (g)  2CO(g) + O 2 (g) ΔH Ɵ =566 kJmol -1 Eqn 2: 3CO(g) + O 3 (g)  3CO 2 (g) ΔH Ɵ =-992 kJmol -1 Calculate the enthalpy change calculated for the conversion of oxygen to 1 mole of ozone,i.e. for the reaction O 2 (g)  O 3 (g)

44. Calculate the enthalpy change for the conversion of graphite to diamond under standard thermodynamic conditions. C (s,graphite) + O 2 (g)  CO 2 (g) ΔH Ɵ = -393 kJmol -1 C (s, diamond) + O 2 (g)  CO 2 (g) ΔH Ɵ = -395 kJmol -1

45. Std enthalpy heat of formation - Overall enthalpy change (heat produced or absorbed) when 1 mole of substance is formed from its elements in their standard states. 2C(g) + 3H 2 (g) + ½O 2 (g)  C 2 H 5 OH(l) ΔH f θ =-277 kJmol -1  It is zero for elements.  It can be positive or negative.  It is illustrated by equations which must balance to produce 1 mole of the substance.

46.  The standard enthalpy of formation of strontium chloride is the enthalpy change for which of the following reactions? A. Sr(g) + Cl 2 (g)  SrCl 2 (s) B. Sr(s) + Cl 2 (g)  SrCl 2 (s) C. Sr 2+ (g) + 2Cl - (g)  SrCl 2 (s) D. Sr 2+ (aq) + 2Cl - (aq)  SrCl 2 (s)

47.  The standard enthalpy of formation of magnesium bromide is the enthalpy change for which of the following reactions? A. Mg 2+ (g) + 2Br - (g)  Mg 2+ (Br - ) 2 (s) B. Mg 2+ (g) + 2Br - (g)  Mg 2+ (Br - ) 2 (g) C. Mg(s) + Br 2 (g)  Mg 2+ (Br - ) 2 (s) D. Mg(s) + Br 2 (l)  Mg 2+ (Br - ) 2 (s)

48.  The standard enthalpy of combustion of hydrogen is -286kJmol -1  What is the standard enthalpy of formation of water ?

49.  The enthalpy change of reaction = total enthalpy change of formation of product – total change of formation of reactant  H rxn =

50.  Iron(III)oxide can be reduced to iron using hydrogen. Fe 2 O 3 (s) + 3H 2 (g)  2Fe(s) + 3H 2 O(g) Calculate the standard enthalpy change, ΔH θ SubstanceΔHf θ / kJmol -1 Fe 2 O 3 (s)- 822 H 2 (g)0 Fe(s)0 H 2 O(g)- 242

51. The equation for the decomposition of ammonium dichromate is (NH 4 ) 2 Cr 2 O 7 (s)  N 2 (g) + 4H 2 O(l) + Cr 2 O 3 (s) Consider the following data at 298K. Calculate the standard enthalpy change, ΔH θ SubstanceΔHf θ / kJmol -1 (NH 4 ) 2 Cr 2 O 7 (s)- 1806 N 2 (g)0 H 2 O(l)- 286 Cr 2 O 3 (s)- 1140

52.  Find  H rxn for the following reaction  Ca(OH) 2 (s) + CO 2 (g)  H 2 O(g) + CaCO 3 (s) Reactants HfHf Products HfHf Ca(OH) 2 -986.1 kJ/molH 2 O(g)-241.8 kJ/mol CO 2 -393.5 kJ/molCaCO 3 (s)-1206.9 kJ/mol

53.  Find the standard enthalpy change of formation of Na 2 O(s) given ΔH f θ [Na 2 O 2 (s)] = -505kJmol -1 and ΔH θ for Na 2 O(s) + ½O 2 (g)  Na 2 O 2 (s) is -88kJmol -1

54. Why is a Reaction Exothermic or Endothermic? Remember: Breaking bonds requires energy (Endothermic) while making bonds releases energy (Exothermic).

55.  More energy released from new bonds forming  Less energy required for breaking old bonds  For a reaction to be exothermic, the energy released due to bond formation must be greater than the energy required to break bonds.

56.  More energy required for breaking old bonds  Less energy released from making new bonds  For a reaction to be endothermic, the energy required to break bonds is greater than the energy released from bond formation.

57.  The standard enthalpy required when 1 mole of a particular covalent bond in a gaseous molecule is broken to form gaseous atoms.  The reaction is endothermic.

58.  Also called mean bond enthalpy.  It is the average value of the bond dissociation energy of a particular type of covalent bond, in a range of different gaseous compounds.

59.  Bonds can have different values, even the same types of bonds in the same molecule can have different values.  For H 2 O:  H-O-H (g)  H-O (g) + H (g) ΔH θ diss =+493KJ/mol  H-O (g)  H (g) + O (g) ΔH θ diss = +424KJ/mol

60.  Total amount of energy required to break both O-H bonds: 493 + 424 = 917KJmol -1  Average bond enthalpy of the O-H bond in a water molecule: 917/2 = 459KJmol -1 E.g. C-H bond enthalpy is based on the ave bond energies in CH 4, alkanes and other hydrocarbons.

61. Bond Ave bond enthalpy, ΔH Ɵ (Kjmol -1 ) Bond length (nm) H-H4360.07 C-C3480.15 C-H4120.11 O-H4630.10 N-H3880.10 N-N1630.15 C=C6120.13 O=O4960.12 C Ξ C 8370.12 NΞNNΞN 944 Refer to page 201

62.  Average bond enthalpies are useful for working out overall enthalpy changes of reactions.  To do this we can use the following equation:

63. What Determines A.B.E’s??  Can you find 2 trends in the table?? BondA.B.E (kJ/mol) Bond Length (nm) C-H+4130.108 O-H+4640.096 C-O+3580.143 C-C+3470.154 C +6120.134 C +8380.120 C O+8050.116 O +4980.121 N +9450.110

64. When a hydrocarbon e.g. methane (CH 4 ) burns, CH 4 + O 2  CO 2 + H 2 O What happens?

65. C H H H H + O O O O C H H H H O O O O C H H H H O O OO ENERGY Enthalpy Level (KJ) Progress of Reaction 2 O=O Bond Breaking Bond Forming 4 C-H 4 H-O2 C=O CH 4 + 2O 2  CO 2 + 2H 2 O

66. C H H H H + O O O O C H H H H O O OO Why is this an exothermic reaction (produces heat)?

67. Bond Ave Bond Enthalpy (kJ/mol) C-H412 H-O463 O=O496 C=O743 CH 4 + 2O 2  CO 2 + 2H 2 O Energy absorbed when bonds are broken = (4 x C-H + 2 x O=O) Energy given out when bonds are formed = ( 2 x C=O + 4 x H-O) = 4 x 412 + 2 x 496 = 2640 kJ/mol C H H H H + O O O O C H H H H O O OO Break  Form = 2 x 803 + 4 x 464 = 3338 kJ/mol

68. Energy absorbed when bonds are broken (a) = 2640 kJ/mol Energy released when bonds are formed (b) = 3338 kJ/mol Enthalpy change, ΔH = ∑(bonds broken) - ∑(bonds made) = a + (-b) = 2640 – 3338 = -698 kJ/mol Why is this an exothermic reaction (produces heat)?

69. 2H 2(g) + O2 (g)  2H 2 O (g) BondBond Energy (KJ/mol) H-H436 O=O499 O-H463

70.  We have to figure out which bonds are broken. 2 H-H = 436 x 2 = 872KJ/mol 1 O=O = 499 x 1 = 499KJ/mol 872 + 499 = 1371JK/mol  And which bonds are formed. 2 H 2 O = H-O-H H-O-H = 463 x 4 = 1852kJ/mol

71.  Now we can substitute the values given into the equation: ΔH θ = 1371 – 1852 = -481KJ/mol

72. What can be said about the hydrogenation reaction of ethene? H H C=C (g) + H-H (g)  H-C-C-H (g) H H HH HH

73. What can be said about the combustion of hydrazine in oxygen? H H N-N (g) + O=O (g)  N ΞN (g) + 2 O (g) H H HH

74. Example Calculate the mean Cl-F bond enthalpy given that Cl 2(g) + 3F 2(g)  2ClF 3(g) ΔH Ɵ = -164 kJmol -1 Bond enthalpy for Cl-Cl = 242 kJmol -1 and F-F = 158 kJmol -1

75.  Bond dissociation enthalpies vary slightly between molecules and even within molecules so enthalpy changes calculated using mean bond enthalpies may not always agree with experimental values.

76. Standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from the element under standard conditions. Example C(s)  C(g) Calculate the enthalpy change for the process 3 C (s) + 4H 2(g)  C 3 H 8(g) ΔH Ɵ = -164 kJmol -1 Bond enthalpy for C-H = 412 kJmol -1, H-H = 436 kJmol -1 and C-C = 348 kJmol -1 (ΔH Ɵ at ) ΔH Ɵ at = 715 kJmol -1 Practice questions page 206 #10,12,13

77. Dissolving sugar. Sugar molecules are dispersed throughout the solution and are moving around. More disordered or random. Other examples: melting ice H 2 O(s)  H 2 O(l) evaporating water H 2 O(l)  H 2 O(g)

78. Increasing entropy Entropy (S) : amount of disorder Unit : JK -1 mol -1 S Ɵ : standard entropy Δ S Ɵ : entropy change If Δ S Ɵ > 0 => increase in entropy => increase in disorder E.g. H 2 O(l )  H 2 O(g) Δ S Ɵ =+119 JK -1 mol -1 If Δ S Ɵ decrease in entropy => decrease in disorder E.g. N H 3 (g ) + HCl(g)  NH 4 Cl(s) Δ S Ɵ = - 285 JK -1 mol -1

79. What are the factors that affect ENTROPY?

80. (1) State of matter ◦ Gas particle motion is more random in a gas ◦ Liquid particle motion is less random than in a liquid than a gas but more than a solid ◦ Solid particle motion is restricted. ◦ Examples  (changing state) H 2 O(l)  H 2 O(g)  (changing state) H 2 O(s)  H 2 O(l) ΔS(gas) > ΔS(liquid) > ΔS(solid)

81. (2) Temperature ◦ Comparing two gasses, one at 20  C and one at 80  C ◦ Molecules in the 80  C gas have more kinetic energy, they are moving more and colliding more

82. (3) The number of molecules ◦ More molecules means more possible positions relative to the other molecules  (more moles and change of state) Li 2 CO 3 (s)  Li 2 O(s) + CO 2 (g)  (more moles) MgSO 4  8H 2 O  Mg 2+ (aq) + SO 4 2- (aq) + 8H 2 O(l) (4) More complex molecules have higher entropy values

83. ReactionEntropy (increase/ decrease) ΔS Ɵ ( + / - ) Explanation N 2 (g) + 3H 2 (g)  2NH 3 (g)4 moles of gas to 2 moles of gas CaCO 3 (s)  CaO(s) + CO 2 (g)1 mole of solid to 1 mole of solid + 1 mole of gas CH 4 (g) + 2O 2 (g)  CO 2 (g) +2H 2 O(l)3 moles of gas to 1 mole of gas C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)2 moles of gas to 1 mole of gas Is there an increase or decrease in disorder of the system? Is there an increase or decrease in the no. of moles of gas? For reaction where the no. of moles of gas is the same on both sides, the ΔS = 0. E.g. F 2 (g) + Cl 2 (g)  2ClF(g) Practice Qn 24from pg 230 (textbook decrease- increase+ decrease- -

84. Which of the following reactions has the largest ΔS value?  CO 2 (g) + 3H 2 (g)  CH 3 OH(g) + H 2 O(g)  2Al(s) + 3S(s)  Al 2 S 3 (s)  CH 4 (g) + H 2 O(l)  3H 2 (g) + CO(g)  2S(s) + 3O 2 (g)  2SO 3 (g)

85. Entropy change = total entropy of products – total entropy of reactants ΔS Ɵ = ∑ ΔS Ɵ products - ∑ ΔS Ɵ reactants E.g. Calculate the standard entropy change for the reaction CH 4 (g) + 2O 2(g)  CO 2(g) + 2H 2 O (l) Use standard entropy from pg 230 (textbook)

86.  Some reactions are spontaneous because they give off energy in the form of heat (H < 0).  Others are spontaneous because they lead to an increase in the disorder of the system (S > 0).  Calculations of H and S can be used to probe the driving force behind a particular reaction.

87.  Spontaneous reaction – one that occurs without any outside influence (no input of energy)  A spontaneous reaction does not have to happen quickly.  E.g. 4Na(s) + O 2(g)  2Na 2 O (s) can happen by itself.

88. Calculate H and S for the following reaction and decide in which direction each of these factors will drive the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information: Compound H f o (kJ/mol) S°(J/mol-K) N 2 (g) 0 191.61 H 2 (g) 0 130.68 NH 3 (g) -46.11 192.45

89. What happens when one of the potential driving forces behind a chemical reaction is favorable and the other is not? We can answer this question by defining a new quantity known as the Gibbs free energy (G) of the system, which reflects the balance between these forces. ΔG= ΔH – TΔS ΔG : free energy change ΔG Ɵ : standard free energy change

90.  Temperature, T should be in Kelvin, K 0°C = 273K (273.15K)  Check the units of ΔS, entropy is often given in JK –1 mol –1 but must be converted to kJK –1 mol –1 ΔG ө = ΔH ө – TΔS ө  ө = standard conditions, 25°C (298K) and 1 atm (101.3 kPa)

91.  measures the balance between the two driving forces that determine whether a reaction is spontaneous.  the enthalpy and entropy terms have different sign conventions.  When heat is released in a chemical reaction, the surrounding is hotter and particles move around more => entropy increases. FavourableUnfavourable ΔH Ɵ < 0ΔH Ɵ > 0 ΔS Ɵ > 0ΔS Ɵ < 0

92.  Because of the way the free energy of the system is defined, G o is negative for any reaction for which H o is negative and S o is positive.  For a Favorable, or spontaneous reactions: G o < 0

93.  When a reaction is favored by both enthalpy (H o 0), there is no need to calculate the value of G o to decide whether the reaction should proceed.  Similarly, for reactions favored by neither enthalpy (H o > 0) nor entropy (S o < 0). Free energy calculations become important for reactions favored by only one of these factors.

94. Given that the changes in enthalpy and entropy are -139 kJ and 277 J/K respectively for the reaction given below, calculate the change in Gibbs energy. Then state whether the reaction is spontaneous at 25  C C 6 H 12 O 6 (aq)  2C 2 H 5 OH(aq) + 2CO 2 (g)

95. Calculate H and S for the following reaction: NH 4 NO 3 (s) + H 2 O(l)  NH 4 + (aq) + NO 3 - (aq) Use the results of this calculation to determine the value of G o for this reaction at 25 o C, and explain why NH 4 NO 3 spontaneously dissolves is water at room temperature. Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information: Compound H f o (kJ/mol) S°(J/mol-K) NH 4 NO 3 (s) -365.56 151.08 NH 4 + (aq) -132.51 113.4 NO 3 - (aq) -205.0 146.4

96.  The balance between the contributions from the enthalpy and entropy terms to the free energy of a reaction depends on the temperature at which the reaction is run.  Predict whether the following reaction is spontaneous at 25 0 C: N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

97.  The equation suggests that the entropy term will become more important as the temperature increases. G o = H o - TS o  Since the entropy term is unfavorable, the reaction should become less favorable as the temperature increases.  Predict whether the following reaction is still spontaneous at 500C: N 2 (g) + 3 H 2 (g) 2 NH 3 (g)  Assume that the values of H o and S used in the previous example are still valid at this temperature.

98.  What does the value of G o tell us about the following reaction? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G o = -32.96 kJ  The value of G o for a reaction ◦ measures the difference between the free energies of the reactants and products when all components of the reaction are present at standard- state conditions. ◦ describes this reaction only when all three components are present at 1 atm pressure.  The fact that G o is negative for this reaction at 25 o C means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium.  The larger the value of G o, the further the reaction has to go to get to from the standard-state conditions to equilibrium.

99. ΔG Ɵ = ∑ ΔG f Ɵ products - ∑ ΔG f Ɵ reactants standard free energy of formation : free energy change for the formation of 1 mole of substance from its elements in their standard states & under standard conditions. Calculate ΔG Ɵ for the reaction CaCO 3 (s)  CaO(s) + CO 2 (g) given that Substance ΔG f Ɵ /kJmol -1 CaCO 3 - 1129 CaO- 604 CO 2 - 395 Practice Qn from pg 235 (textbook

100. For a spontaneous reaction ΔG is negative (–) For a non–spontaneous reaction ΔG is positive (+) ΔH : Enthalpy Change ΔS : Entropy Change