1. DP Chemistry R. Slider

2. Rate Equation Recall that the rate of a reaction is a measure of the change in concentration of a reactant, R, (or product, P) over time. The rate of disappearance of the reactants is equal to the rate of appearance of the products for a 1:1 molar ratio. This can be seen graphically above Notice the initial rate is measured from the graph Units for rate

3. Measuring reaction rates The rate of reaction changes as the reaction proceeds. This can be seen by the change in the gradient of the curves on the graph. We can collect information about rate at different times by measuring the gradient at different points on the graph. Notice in the above graph, the formation of HI is initially occurring at a faster rate than the disappearance of reactants as seen by the steeper gradient. Can you explain why? H 2(g) + I 2(g)  2HI (g)

4. Reaction Order When the concentration of a particular reactant is changed, the rate of reaction may also change This relates to reaction order which can only be determined experimentally Scenario 1: First order Doubling [A] doubles the rate of reaction. This means: Rate α [A] Or Rate = k[A] where k is the rate constant which is dependent upon temperature and use of a catalyst. This reaction is “first order with respect to A” Scenario 2: Second order Doubling [A] increases the rate 4 times. This means: Rate α [A] 2 Or Rate = k[A] 2 This reaction is “second order with respect to A” Consider the following reaction for which we have a measured initial rate for A and B: A + B  products We can alter concentrations and measure the change in rate

5. Reaction Order When the concentration of a particular reactant is changed, the rate of reaction may also change This relates to reaction order which can only be determined experimentally Scenario 3: Zero order Doubling [A] does not change the rate of reaction. This means: Rate α [A] 0 Or Rate = k[A] 0 This reaction is “zero order with respect to A” Consider the following reaction for which we have a measured initial rate: A + B  products We can alter concentrations and measure the change in rate Scenario 4: Overall reaction order Doubling [A] increases the rate 4 times and doubling [B] doubles the rate. This means: Rate α [A] 2 [B] Or Rate = k[A] 2 [B] This reaction is “second order with respect to A and first order with respect to B” Overall reaction order = 2 + 1 = 3 This is the rate expression for the reaction

6. Order of Reaction with respect to reactant, R Index value (value of m) Rate equation Experimental observation Zero0Rate=k[R] 0 No change in rate with a change in [R] One1Rate=k[R] 1 Rate and [R] are directly proportional – as [R] is doubled or tripled, so is the rate Two2Rate=k[R] 2 Rate is directly proportional to [R] 2 – doubling [R] will increase the rate fourfold Reaction Order Summary For a general reaction: R  P, Rate α [R] m

7. Rate Constant (k) What units? The units of the rate constant is dependent upon the overall rate expression. To determine the units for the rate constant, simply solve for k and derive the units. Source: Chemistry for use with the IB Diploma Program, Derry et. al. A small value for k is an indication of a slow rate of reaction whereas a large value is indicative of a fast reaction rate.

8. Rate expression So, now we know the rate expression has a general form that looks something like this: For the reaction A + B  products the rate expression looks like: Rate = k [A] m [B] n Order of reaction with respect to A rate in mol dm -3 s -1 Order of reaction with respect to B rate constant concentrations in mol dm -3 Be sure to practise solving problems involving this rate expression to solve for: Rate Rate Rate constant Rate constant Unknown concentration Unknown concentration Be sure to practise solving problems involving this rate expression to solve for: Rate Rate Rate constant Rate constant Unknown concentration Unknown concentration

9. Determining the rate expression Procedure: 1.Measure the initial rate for a series of reactant concentrations 2.Change one of the concentrations keeping the other constant and measure the rate again 3.Change the concentration of the one previously kept constant, and measure the initial rate again. 4.Repeat this procedure until all reactants have been changed and enough data is obtained Experiment [A] / mol dm -3 [B] / mol dm -3 [C] / mol dm -3 Rate/ mol dm -3 s -1 10.1 6.2 x 10 -4 20.10.20.11.2 x 10 -3 30.1 0.26.2 x 10 -4 40.20.10.22.5 x 10 -3 You try: Look at the data to the right and determine the rate expression and the rate constant for a reaction that has 3 reactants, A, B and C.

10. Determining the rate expression Procedure: 1.Measure the initial rate for a series of reactant concentrations 2.Change one of the concentrations keeping the other constant and measure the rate again 3.Change the concentration of the one previously kept constant, and measure the initial rate again. 4.Repeat this procedure until all reactants have been changed and enough data is obtained Experiment [A] / mol dm -3 [B] / mol dm -3 [C] / mol dm -3 Rate/ mol dm -3 s -1 10.1 6.2 x 10 -4 20.10.20.11.2 x 10 -3 30.1 0.26.2 x 10 -4 40.20.10.22.5 x 10 -3 You try: Look at the data to the right and determine the rate expression and the rate constant for a reaction that has 3 reactants, A, B and C. Exp’t 2: Doubles [B], which doubles the rate Exp’t 3: Doubles [C], which has no effect on the rate Exp’t 4: Doubles [A], which quadruples the rate Therefore, A is second order, B is first order and C is zero order and the rate expression is: Rate = k[A] 2 [B] (note [C] 0 = 1 so is not included in the rate expression) Solving for k, 6.2 x 10 -4 = k [0.1] 2 [0.1] 1 k = 6.2 x 10 -1 dm 6 mol -2 s -1

11. Determining reaction order Experiment We need to be able to measure the change in concentration of either a reactant or a product such as a gas being produced. Then we can change the concentration of reactants one by one to determine how they affect the reaction rate. Concentration (vol here) vs. Time graph This graph allows us to calculate instantaneous rate information. The gradient equals the rate. Rate vs. Concentration graph (not shown) Plotting rates vs. changes in concentration allows us to easily determine the order of the reaction by analysing their shape. Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htmhttp://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm

12. Reaction order graphs - zero Rate vs. Concentration For a zero order reaction, rate is constant with changes in concentration A  products, rate = k Concentration vs. Time Because the rate is constant, [A] will decrease by the same amount every second and the gradient is constant and negative (-k) Also, the time it takes for half of the reactants to disappear (1/2 life), decreases with reduced concentration Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htmhttp://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm

13. Reaction order graphs - first Rate vs. Concentration For a first order reaction, the rate increases in proportion to changes in concentration A  products, rate = k[A] Concentration vs. Time The rate will decrease every second and the gradient will become less negative as [A] decreases Also, the 1/2 life remains constant with reduction in concentration Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htmhttp://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm

14. Reaction order graphs - second Rate vs. Concentration For a second order reaction, the rate increases exponentially with increases in concentration (like y = x 2 ) A + B  products, if [A] = [B] rate = k[A] 2 Concentration vs. Time The rate will decrease every second and the gradient will become less negative, but more dramatically than first order reactions Also, the 1/2 life increases with reduction in concentration Source: http://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htmhttp://ibchem.com/IB/ibnotes/full/kin_htm/16.1.htm

15. The reaction between NO (g) and Cl 2(g) has been studied at 50 °C, recording the initial rate of formation of NOCl (g) for the initial concentrations of reactants as shown in the table. NO (g) + ½Cl 2(g)  NOCl (g) Exercise [NO (g) ] (mol dm -3 ) [Cl 2(g) ] (mol dm -3 ) Initial Rate (mol dm -3 s -1 ) 0.250 1.43 x 10 -6 0.2500.5002.86 x 10 -6 0.500 11.4 x 10 -6 1. The order of reaction with respect to NO (g) is: 2.The order of reaction with respect to Cl 2(g) is: 3.The overall order of reaction is: 4.The value of the rate constant, k, at 50 °C is: mol -2 dm 6 s -1 5. The rate of formation of NOCl when [NO (g) ] and [Cl 2(g) ] are both equal to 0.110 mol dm -3 is: 6. The rate at which NO is reacting, at the instant when Cl 2 is reacting at the rate of 2.21 x 10 -7 mol dm -3 s -1 is: 7. The rate at which NOCl is forming, at the instant when Cl 2 is reacting at the rate of 2.21 x 10 -7 mol dm -3 s -1 is:

16. The reaction between NO (g) and Cl 2(g) has been studied at 50 °C, recording the initial rate of formation of NOCl (g) for the initial concentrations of reactants as shown in the table. NO (g) + ½Cl 2(g)  NOCl (g) Exercise answers [NO (g) ] (mol dm -3 ) [Cl 2(g) ] (mol dm -3 ) Initial Rate (mol dm -3 s -1 ) 0.250 1.43 x 10 -6 0.2500.5002.86 x 10 -6 0.500 11.4 x 10 -6 1. The order of reaction with respect to NO (g) is: 2 2.The order of reaction with respect to Cl 2(g) is: 1 3.The overall order of reaction is: 3 4.The value of the rate constant, k, at 50 °C is: 9.17 x 10 -5 mol -2 dm 6 s -1 5. The rate of formation of NOCl when [NO (g) ] and [Cl 2(g) ] are both equal to 0.110 mol dm -3 is: 1.22 x 10 -7 mol dm -3 s -1 6. The rate at which NO is reacting, at the instant when Cl 2 is reacting at the rate of 2.21 x 10 -7 mol dm -3 s -1 is: 4.42 x 10 -7 mol dm -3 s -1 (half of Cl2 as seen in the reaction ratio) 7. The rate at which NOCl is forming, at the instant when Cl 2 is reacting at the rate of 2.21 x 10 -7 mol dm -3 s -1 is: 4.42 x 10 -7 mol dm -3 s -1 (same as the disappearance of NO in #6)

17. Graphing Exercise Plot this data to determine the order of the reaction: 2NOBr(g)  2NO(g) + Br2 (g)

18. Reaction Mechanisms Most reactions occur in more than one step because it is rare that more than two individual particles will simultaneously collide and successfully react. Source: http://www.talktalk.co.uk/reference/encyclopaedia/hutchinson/m0030471.html http://www.talktalk.co.uk/reference/encyclopaedia/hutchinson/m0030471.html intermediate Reactions often go through intermediate species, which means a reaction may go through multiple steps before reaching the final products. reaction mechanisms. It is important to note that there may be more than one possible mechanism. These possible multi-step pathways are known as reaction mechanisms. It is important to note that there may be more than one possible mechanism. elementary step or elementary process. Each step in the reaction mechanism is known as the elementary step or elementary process.

19. Molecularity This is a description of each elementary step, indicating the number of reacting particles. Quite simply: Number of reacting particles Molecularity 1Unimolecular 2Bimolecular 3Termolecular ExampleRate/order CuCO 3  CO 2 + CuOk[CuCO 3 ]/first NO 2 + CO  NO + CO 2 k[NO 2 ][CO]/second 2CO + O 2  2CO 2 k[CO] 2 [O 2 ]/third Unimolecular and bimolecular are by far the most common. Single step termolecular reactions are very rare and no examples of higher molecularity are known.

20. Reaction Mechanism example 1 Consider this reaction: 2 NO(g) + O 2 → 2 NO 2 This reaction does not occur in a single step, however, but rather through two steps. Step 1: 2 NO → N 2 O 2 Step 2: N 2 O 2 + O 2 → 2 NO 2 Overall, Step 1: 2 NO → N 2 O 2 Step 2: N 2 O 2 + O 2 → 2 NO 2 Overall: 2 NO(g) + O 2 → 2 NO 2

21. Reaction Mechanism example 2 This is a reaction between 2-bromo-2- methylpropane and the hydroxide ions from sodium hydroxide solution: Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.htmlhttp://www.chemguide.co.uk/physical/basicrates/ordermech.html This shows a two-step mechanism. The first step shows electrons being transferred to the Br forming two ions. This is slow due to strong bonds between the carbon and bromine. The second step is likely to be fast due to the strong attraction between the positive carbon and negative hydroxide. The slowest step is known as the rate determining step because the rate can only be as fast as the slowest step. Notice that adding the two elementary steps together gives the overall balanced equation for the reaction

22. Reaction Mechanism example 2 This is a reaction between 2-bromo-2- methylpropane and the hydroxide ions from sodium hydroxide solution: Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.htmlhttp://www.chemguide.co.uk/physical/basicrates/ordermech.html Experimentally: We find that the overall rate expression is: Rate = k[(CH3)3CBr] Notice that the [OH-] has no effect on the rate. This supports the assumption that the first step is slow. If the second step were also slow, increasing the [OH-] would have an effect on the rate. The rate determining step must have a rate expression that matches the rate expression of the overall reaction. The molecularity of this step is equal to the overall order of reaction.

23. Reaction Mechanism example 3 This is a reaction between bromo-ethane and the hydroxide ions. Similar species, but different mechanism. Source: http://www.chemguide.co.uk/physical/basicrates/ordermech.htmlhttp://www.chemguide.co.uk/physical/basicrates/ordermech.html Experimentally: We find that the overall rate expression is: Rate = k[(CH3)3CBr][OH-] Notice that in this example the [OH-] has an effect on the rate. This supports the assumption that the reaction occurs in one step. The reaction occurs all at once due to the partial positive charge on the carbon atom

24. Exercise Step 1:2 NO → N 2 O 2 Step 2:N 2 O 2 + H 2 → N 2 O + H 2 O Step 3:N 2 O + H 2 → N 2 + H 2 O For this reaction find the following: the overall balanced equation any reaction intermediates

25. Exercise answers Step 1:2 NO → N 2 O 2 Step 2:N 2 O 2 + H 2 → N 2 O + H 2 O Step 3:N 2 O + H 2 → N 2 + H 2 O For this reaction find the following: the overall balanced equation any reaction intermediates Net Reaction:2 NO + 2 H 2 → N 2 + 2 H 2 O To identify the reaction intermediates, look for substances that first appear on the product side of the equation, but then appear in the next step as a reactant. In this example there are two reaction intermediates - N 2 O 2 and N 2 O

26. If the reaction 2 NO 2 + F 2  2 NO 2 F follows the mechanism, Step 1. NO 2 + F 2  NO 2 F + F (slow) Step 2. NO 2 + F  NO 2 F (fast) Work out the rate expression What is the order of the overall reaction Exercise

27. If the reaction 2 NO 2 + F 2  2 NO 2 F follows the mechanism, Step 1. NO 2 + F 2  NO 2 F + F (slow) Step 2. NO 2 + F  NO 2 F (fast) Rate = k[NO 2 ][F 2 ]..because the rate expression of the slowest step is the same as the overall rate expression. Therefore the overall order of reaction is 2. Exercise answers

28. Activation Energy Recall that the activation energy, E a, is the energy required for particles to react. We have also discussed that raising the temperature increases the number of particles that are able to react as seen in the Maxwell- Boltzmann distribution curves We have also stated that the rate constant, k, is affected by temperature and catalysts. So, how can we quantify this?

29. Activation Energy Now, all we need to do is understand this proportionality…

30. Arrhenius Equation Therefore, since f α k… rate constant Arrhenius constant Mathematical quantity (2.718) Activation energy Gas constant (8.314 J K -1 mol -1 ) Temperature in Kelvin Note: the Arrhenius constant, A, is based on the probability of correct molecular orientation and frequency of collisions. It is virtually constant over a wide temperature range for a particular reaction. It is also called the frequency factor

31. Arrhenius Equation The negative sign on the exponent means that: As Ea increases, k decreases As T increases, k increases So… The rate constant will increase with increasing temperature and decrease with decreasing temperature rate constant Arrhenius constant Mathematical quantity (2.303) Activation energy Gas constant (8.314 J K -1 mol -1 ) Temperature in Kelvin

32. Arrhenius Equation – integrated form The Arrhenius Equation is also sometimes written in the so-called integrated form. This is just a rearrangement of the equation to help us to: solve for temperature determine Ea using a graph Where, ln is the natural logarithm (your scientific calculator will have a button) Also, no need to memorize either of these equations. They are found in your Chemistry Data Booklet.

33. What happens to the fraction of particles that are able to react when we raise the temperature from 20 0 C to 30 0 C, assuming the activation energy is 50 kJ mol -1 ? Exercise

34. 1. 2. = 1.21 x 10 -9 = 2.38 x 10 -9 This shows that an increase in 10 0 C has effectively doubled the fraction of particles that can react.

35. Determining Ea using a graph We can use the integrated form of the Arrhenius Equation to determine the activation energy. Recall, This can be rewritten in the form of an equation for a straight line, y = mx + b

36. Determining E a using a graph So, a graph of ln k vs. 1/T will give a straight line that will allow us to determine the value of E a for a reaction from the gradient. Now, practice some of these problems…