1. Unit 8 Chemical Kinetics & Thermodynamics

2. Chemical Kinetics Chemical kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. Rate is how much a quantity changes in a given period of time.

3. Collision Theory says that in order for reactions to occur between substances, particles must collide. For a collision to be effective, particles must: 1.Collide with sufficient energy. 2.Have a favorable orientation. Collision Theory

4. Effect of Orientation on a Collision

5. Smaller pieces = larger surface area The rate of a chemical reaction depends on the number of collisions between particles. Rate-influencing factors are: Nature of Reactants Surface Area Temperature Concentration Presence of Catalysts Rate-Influencing Factors Visual Concept

6. Effect of Concentration The rate of a reaction often depends on the concentration of one or more of the reactants. Greater concentration of reactant particles = more collisions = increase in reaction rate.

7. Rate Law Expressions Rate law equations are based on the Molarity concentrations of the reactant(s). (Indicate Molarity concentrations with brackets.) The Rate (R) involves the product of the concentrations of the individual reactants. Experiments have proven that coefficients of reactants have an exponential effect on R. Ex: For the equation A + 2 B → C R = k [A][B] 2 (k = rate law constant)

8. Rate Law Expressions Sample Problem For the following single-step chemical reaction: 2SO 2 + O 2 → 2 SO 3 1.Write the rate law expression. R = k [SO 2 ] 2 [O 2 ] 2.What happens to the Rate if the [SO 2 ] is cut by ½? Rate is 4x slower 3.What happens to the Rate if the [O 2 ] is tripled? Rate is 3x greater

9. Reaction Mechanism Chemical reactions often occur as a series of consecutive steps (this is called a reaction mechanism.) Products in an early step that are reactants in a later step are called intermediates. Example: Overall reaction: H 2(g) + 2 ICl (g)  2 HCl (g) + I 2(g) Mechanism: 1)H 2(g) + ICl (g)  HCl (g) + HI (g) 2)HI (g) + ICl (g)  HCl (g) + I 2(g) intermediate

10. Rate-Determining Step One of the steps will always occur at a slower rate than the others. The rate-determining step is the slowest step (bottleneck) in a reaction mechanism. ONLY the rate-determining step is used for rate law expressions.

11. Rate-Determining Step Sample Problem Equation:2NO 2 + F 2 → 2 NO 2 F Step 1:NO 2 + F 2 → NO 2 F + F“slow” Step 2:F + NO 2 → NO 2 F“fast” 1.Write the rate law expression for this reaction. R = k [NO 2 ] [F 2 ] 2.Given the following data, calculate the [NO 2 ] used: R = 0.00905 M/s, k = 1.44 x 10 -3 1 / M ● s, [F 2 ] = 1.11 M [NO 2 ] = 5.66 M [NO 2 ] = R k [F 2 ] = 0.00905 M/s (1.44 x 10 -3 1 / M ● s )(1.11 M) Rate-determining step

12. Getting a Reaction Started Energy is needed to overcome the repulsion between molecules and get a reaction started. Just as a ball cannot get over a hill without added energy, a reaction cannot occur unless the molecules have enough energy to get over this initial energy barrier.

13. Activation Energy Activation energy (E a ) is the minimum amount of energy required to get a reaction started.

14. Exothermic Reactions In an exothermic reaction, energy is released. Therefore, the energy of the products must be less than the energy of the reactants. The great majority of chemical reactions in nature are exothermic.

15. Endothermic Reactions In an endothermic reaction, energy is absorbed. Therefore, the energy of the products must be greater than the energy of the reactants.

16. Catalysts Catalysts are substances which affect the rate of a reaction without being consumed. Catalysts increase the rate of a reaction by changing the reaction mechanism and decreasing the activation energy.

17. Enzymes Due to the complexity of organic molecules, most biological reactions need a catalyst to proceed at a reasonable rate. Protein molecules that catalyze biological reactions are called enzymes. Enzymes pull the reactants onto an active site that orients them for the reaction.

18. Rate Order Experiments have shown that there are 3 ways that a reactant’s concentration can affect the overall rate of the reaction: Rate Order 0 ([X] 0 ) - The concentration of a reactant has no effect on reaction rate. Rate Order 1 ([X] 1 ) - The concentration of a reactant has a direct effect on reaction rate. Rate Order 2 ([X] 2 ) - The concentration of a reactant has an exponential effect on reaction rate. The order of a reaction can be determined only by experiment.

19. Determining the Order of a Reaction [A] (M) Rate (M/s) 0.100.015 0.200.030 0.400.060 x2 [A] (M) Rate (M/s) 0.100.015 0.200.015 0.400.015 x2 [A] (M) Rate (M/s) 0.100.015 0.200.060 0.400.240 x1 x4 When [A] doubles, the rate doubles When [A] doubles, rate doesn’t change When [A] doubles, the rate quadruples Rate = k[A] 1 First Order Rate = k[A] 0 = k Zero Order Rate = k[A] 2 Second Order

20. Rate Order of a Reaction Sample Problem From the data above, determine the following: 1.Rate order of each reactant. second order in [NO 2 ] and zero order in [CO] 2.Overall rate order of the reaction. 2 + 0 = 2 second order reaction 3.The rate law for the reaction. Rate = k [NO 2 ] 2 [NO 2 ] (M)[CO] (M)Rate (M/s) 0.10 0.0021 0.200.100.0082 0.20 0.0083 0.400.100.033 x2x4 constant x2 x1

21. Finding the Rate Constant Sample Problem Using the data above, and the rate law we just found, (R = k [NO 2 ] 2 ), determine the following: 1.Value of k. 2.Reaction rate when the concentration of [NO 2 ] is 0.50 M and [CO] is 1.00 M. [NO 2 ] (M)[CO] (M)Rate (M/s) 0.10 0.0021 0.200.100.0082 0.20 0.0083 0.400.100.033 k = R [NO 2 ] 2 = 0.0021 M/s (0.10 M) 2 = 0.21 M -1 s -1 R = k [NO 2 ] 2 = (0.21 M -1 s -1 ) (0.50 M) 2 = 0.053 M/s Use data from any one line of chart to find k

22. Thermodynamics Thermodynamics is the study of energy transfer in reactions.

23. Energy Energy is the ability to do work. Some forms of energy are: Mechanical (kinetic and potential) Electrical Heat or thermal Light or radiant Nuclear Chemical Energy is commonly measured in joules. 1 joule is the amount of energy needed to move a 1 kg mass at a speed of 1 m/s. (1 J = 1 )

24. First Law of Thermodynamics The First Law of Thermodynamics is the Law of Conservation of Energy which states that energy cannot be created or destroyed. However, energy can be transferred between objects, or between forms. (Ex: heat → light → sound)

25. Energy Exchange Conservation of Energy requires that the total energy change in a system and its surroundings must be zero. Energy is exchanged between system and surroundings through either heat exchange or work being done. Heat is the exchange of thermal energy between a system and its surroundings.

26. Enthalpy the Enthalpy change (  H) of a reaction is the heat exchanged in a reaction under conditions of constant pressure.  H for a reaction is equal to the difference between the  H of formation of the products and the reactants.  H rxn =  n  H f (products) -  n  H f (reactants)  means sum n is the coefficient Elements in their standard state have a  H f of zero.

27. Calculating Enthalpy of Reaction Sample Problem Calculate the Enthalpy Change in the Reaction: 2C 2 H 2 (g) + 5O 2 (g)  4CO 2 (g) + 2H 2 O(l) Reactant/ Product  H f (kJ/mol) C2H2(g)C2H2(g)+227.4 CO 2 (g)-393.5 H 2 O(l)-285.8  H rxn =  n  H f (products) -  n  H f (reactants)  H rxn = [(4  H CO2 + 2  H H2O ) – (2  H C2H2 + 5  H O2 )]  H rxn = [(4(-393.5) + 2(-285.8)) – (2(+227.4) + 5(0))]  H rxn = -2145.6 kJ – 454.8 kJ  H rxn = -2600. kJ

28. Stoichiometry Involving ∆H The amount of heat generated or absorbed during a reaction depends on the amount of the substances reacting. Example: C 3 H 3 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g) ∆H = -2044 kJ Means 2044 kJ of heat is given off when 1 mole of C 3 H 3 reacts, or when 3 moles of CO 2 is produced. These relationships can be used as ratios in stoichiometric conversions.

29. Ammonia reacts with oxygen as follows: 4NH 3 (g) + 5O 2 (g)→ 4NO(g) + 6 H 2 O(g) ∆H = -906 kJ Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH 3. mol NH 3 Stoichiometry Involving ∆H Sample Problem 155 g NH 3 g NH 3 -2070 kJ NH 3 = 17.0 1 4 kJ -906

30. Hess’s Law Hess’s Law – the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process. Possible steps in Hess’s Law: 1.Reverse equation/ change sign on ΔH. 2.Multiply or Divide coefficients/multiply or divide ΔH.

31. Hess’s Law Sample Problem Calculate the enthalpy of formation for CH 4 : C(s) + 2H 2 (g) → CH 4 (g) ∆H f = ? The component reactions are: C(s) + O 2 (g) → CO 2 (g)∆H c = -393.5 kJ H 2 (g) + ½O 2 (g) → H 2 O(l) ∆H c = -285.8 kJ CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ∆H c = -890.8 kJ CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ∆H c = +890.8 kJ - 2 moles of H 2 are used to make CH 4, so multiply the 2 nd equation by 2 (including ∆H.) -CH 4 is on the products side, not the reactants side, so reverse the 3 rd reaction and change the sign on ∆H. -Cancel unwanted terms and add the ∆H’s. 22 -571.6 kJ -74.3 kJ

32. Sign of ∆H In an exothermic reaction, the energy of the products is less than the energy of the reactants, so ∆H is negative. In an endothermic reaction, the energy of the products is greater than the energy of the reactants, so ∆H is positive. Exothermic ∆H < 0 Endothermic ∆H > 0

33. Spontaneous Processes A spontaneous process is one that occurs without ongoing outside intervention. A nonspontaneous process is not impossible, but it requires energy input. if a process is spontaneous in one direction, it must be nonspontaneous in the other.

34. Enthalpy and Reaction Tendency Most spontaneous reactions are exothermic (∆H is negative). Products have less energy than reactants. Less energy = greater stability. But not all spontaneous reactions are exothermic. Ex: The melting of ice is endothermic and spontaneous (at >0 o C). Enthalpy must not be the sole criteria for spontaneity.

35. Entropy Entropy (S) is a measure of the degree of disorder in a system. Changes that increase entropy: Physical state (phase) changes (solid < liquid < gas) A larger number of moles of products than reactants Increase in temperature Solids dissociating into ions upon dissolving Expansion (greater volume) of gases

36. Second Law of Thermodynamics The Second Law of Thermodynamics states that for any spontaneous process, the entropy of the universe increases (  S univ is positive). The entropy of the system could decrease as long as the entropy of the surroundings increases by a greater amount. (  S univ =  S sys +  S surr )

37. Temperature Dependence of  S The increase in  S surr often comes from the heat released in an exothermic reaction. the amount the entropy of the surroundings changes depends on the temperature it is at originally. the higher the original temperature, the less effect addition or removal of heat has.

38. Gibbs Free Energy The total amount of energy available in the system to do work on the surroundings is called the Gibbs Free Energy (  G).  G can be calculated using the following equation (T is the Kelvin temperature): Systems tend toward lower Gibbs free energy (lower chemical potential.) A negative  G = a spontaneous reaction. A positive  G = a nonspontaneous reaction.  G =  H – T  S

39. Gibbs Free Energy Sample Problem Given:CCl 4(g)  C (s, graphite) + 2 Cl 2(g)  H = +95.7 kJ and  S = +142.2 J/K at 25°C. 1.Calculate  G and determine if it is spontaneous:  G =  H – T  S  G = 95.7 x 10 3 J – (298K) (142.2 J/K)  G = +5.33 x 10 4 J 2.Determine at what temperature (if any) the reaction becomes spontaneous:  G =  H – T  S < 0 95.7 x 10 3 J – T (142.2 J/K) < 0 95.7 x 10 3 J < T (142.2 J/K) Nonspontaneous T > 673 K Spontaneous at:

40. Effect on Spontaneity When  H and  S have opposite signs, spontaneity does not depend on the temperature, but when they have the same sign it does.  G =  H – T  S

41. Third Law of Thermodynamics The Third Law of Thermodynamics states that for a perfect crystal at absolute zero, absolute entropy (S) = 0. So every substance that is not a perfect crystal at absolute zero has some energy from entropy. Perfect crystals never occur in practice; imperfections just get "frozen in" at low temperatures. Scientists have achieved temperatures extremely close to absolute zero.