OCR A2 F325 5.1.1 Rates of reaction Dr Carl Thirsk

Contents Rate and orders The rate equation Measuring rates and determining orders Half lives Initial rates Effect of temperature Rate equations and mechanisms Selected exam questions

How fast? Grade/Level Success Criteria We will be covering: Rate graphs, orders of reaction and rate equations Determining rate constants and orders using data Grade/Level Success Criteria C Explain and use the term: rate of reaction and order of reaction A/B Deduce reaction rates from initial rate data and concentration–time graphs A*/A Relate rate data to reaction mechanisms

Rate and orders

Starter questions What is meant by rate of reaction? Which factors affect the rate of reaction? How do you measure the rate of reaction?

Defining rate Rate of reaction is the change in concentration of a reactant or product in a given time: Rate= change in concentration (mol dm −3 ) time (s) Rate has units of mol dm-3 s-1 Square brackets, [   ] are used to represent the concentration, mol dm-3 of a reactant or product.

Measuring rate As a reaction proceeds, the reactants get used up. This means that the concentration decreases. This explains why reactions: Slow down: fewer collisions are taking place Stop: when all the reactant has been used up The rate can be determined by measuring the concentration of a reactant or product at regular time intervals.

Measuring rate When a graph of concentration of reactant is plotted vs. time, the gradient of the curve is the rate of reaction. The gradient of a curve is measured by drawing tangents. The initial rate is the rate at the start of the reaction where it is fastest. Reaction rates can be calculated from graphs of concentration of reactants or products as they change over time.

Concentration change of a reactant during a reaction 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= (𝑦2−𝑦1) (𝑥2−𝑥1)

Measuring an initial rate 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= (𝑦2−𝑦1) (𝑥2−𝑥1) 12

Measuring the rate from a concentration-time graph Sulphuryl chloride decomposes producing sulphur dioxide and chlorine: SO2Cl2(g)  SO2(g) + Cl2(g) The concentration of SO2Cl2 was measured and plotted

Time/s 500 1000 2000 3000 4000 [SO2Cl2]/ moldm-3 0.50 0.43 0.37 0.27 0.20 0.15 Suppose we want to find the initial rate (t = 0) and the rate as t = 3000: After t = 0 Rate = Change in [SO2Cl2]/ change in time Rate = (0.5 – 0) / (3300 – 0) = 1.5 x 10-4 mol dm-3 s-1 blue line = experimental data After t = 3000 Rate = Change in [SO2Cl2]/ change in time Rate = (0.38 – 0.14) / (4000 – 0) = 6.0 x 10-5 mol dm-3 s-1

Orders of reaction What determines whether a collision between two particles will result in a reaction? Activation energy - Ea Only particles that have more energy than Ea will react when they collide. If we increase the concentration of a reactant there will be: A larger number of collisions per second A faster reaction By doing a series of experiments we can work out how concentration of each reactant affects the rate of reaction. Mathematically this effect is called the order with respect to a reactant.

Orders of reaction Take the example reaction: A + B  C For reactant A, we can say rate is proportional to the concentration of A: Rate ∝ [A]m where m is the order of reaction with respect to reactant A Order is always defined in relation to concentrations. There are three common reaction orders.

Zero order reactions For the reaction: A + B + C  Products If the order is 0 with respect to reactant A, then; Rate ∝ [A]0 The rate is not affected by the concentration of A. We can increase the concentration of A as much as we like and the rate of reaction will not change. This is because any number raised to the power of 0 becomes 1

First order reactions For the reaction: A + B + C  Products If the order is 1 with respect to reactant B, then; Rate ∝ [B]1 If we double the concentration of B the rate also doubles This is because any number raised to the power of 1 is itself If [B] increases by 2 times, rate increases by 21 times = 2 times If [B] increases by 3 times, rate increases by 31 times = 3 times

Second order reactions For the reaction: A + B + C  Products Second Order: If the order is 2 with respect to reactant C, then; Rate ∝ [C]2 If we double the concentration of C the rate quadruples This is because any number raised to the power of 2 …….behaves likes this: If [C] increases by 2 times, rate increases by 22 times = 4 times If [B] increases by 3 times, rate increases by 32 times = 9 times

Rate ∝ [A]0 , Rate ∝ [B]1 and Rate ∝ [C]2 Combining orders For the reaction: A + B + C  Products Let the orders of reaction with respect to the different reactant be as follows: A = 0, B = 1, and C=2 So the rate is affected by the concentrations of reactant as follows: Rate ∝ [A]0 , Rate ∝ [B]1 and Rate ∝ [C]2 Combining these expressions gives: Rate ∝ [A]0[B]1[C]2

The rate equation

The Rate Equation and The Rate Constant, k For the reaction: A + B + C  Products The rate constant, k, is a proportionality constant that links the rate of reaction with the concentration of reactants. So if: Rate ∝ [A]0[B]1[C]2 We can write a rate equation as follows: Rate = k[A]0[B]1[C]2 Which simplifies to: Rate = k[B]1[C]2 since [A]0 = 1.

Summary For the general reaction: aA + bB →products The generalised rate equation is: r = k[A]m[B]n r is used as the standard symbol for chemical rate r has units mol dm-3s-1 m, n are called reaction orders Orders are usually integers 0, 1 ,2 0 means the reaction is zero order with respect to that reactant 1 means first order 2 means second order

The Overall Order of Reaction The overall order of reaction is the sum of the individual orders. From the example: Rate = k[B]1[C]2 The overall order is 1 + 2 = 3 VERY IMPORTANT: The orders have nothing to do with the stoichiometric coefficients in the balanced equation. Orders can only be worked out experimentally

The Rate Equation and Orders of Reaction Questions The rate of equation for a reaction between P and Q is: rate = k[P]2[Q] How will the rate change if: The concentration of P is doubled The concentration of Q is tripled The concentration of P and Q are both tripled. 2) In a reaction between R, S and T: When the [R] is doubled the rate is unchanged When the [S] is quadrupled the rate increases by 16 times When the [T] is halved the rate is halved Deduce the orders w.r.t R, S and T Hence write down the rate equation for the reaction Rate quadruples Rate triples Rate multiplies by 27 [R]0 [S]2 [T]1 rate = k [S]2[T] = 3rd order

Key points about the rate constant, k The units of k depend on the overall order of reaction and must be worked out from the rate equation. The larger the value of k, the faster the reaction A fast reaction has a large value of k A slow reaction has a small value of k The value of k is independent of concentration and time. It is constant at a fixed temperature. The value of k refers to a specific temperature and it increases if we increase temperature.

Finding k and the units of k The units of the rate constant depend on the overall order in the reaction. We can determine the units of k by substituting units into the rate equation. For a first order reaction, the rate equation will be: rate = k[A]1 The units of this will be: moldm-3s-1 = k x moldm-3 Hence k for a first order reaction has units s-1

Rate constant units For a 1st order overall reaction the unit of k is s-1 For a 2nd order overall reaction the unit of k is mol-1 dm3s-1 For a 3rd order overall reaction the unit of k is mol-2 dm6s-1

The Rate Equation and Orders of Reaction Question The reaction between ozone, O3(g), and ethene, C2H4(g), has the rate equation: rate = k[O3(g)][C2H4(g)] 5 x 10-8 mol dm-3 O3(g) was reacted with 1.0 x 10-8 mol dm-3 C2H4(g). The initial rate was 1.0 x 10-12 mol dm-3 s-1 Calculate the rate constant, k, for this reaction and state the units. k = rate k = 1 x 10-12 = 2000 mol dm3 s-1 [O3][C2H4] (5 x 10-8)(1 x 10-8)

Measuring rates Half life method Initial rates

Continuous rate experiments This is an experiment where the concentration of a substance is followed throughout the experiment. This data can be processed by plotting the data and calculating successive half-lives. The half life is the time taken for the concentration of a reactant to half. The half-life of a first-order reaction is independent of the concentration and is constant. If half-lives rapidly increase then the order is 2nd order.

Measuring half-lives The half life of a reactant is the time it takes for the concentration of that reactant to reduce by half. The shorter the half life, the faster the reaction is progressing For a 1st order reaction: The half-life is constant Half-life is the same regardless of the concentration Here half-life is 53 seconds 2N2O(g)  2N2(g) + O2(g) 1st order w.r.t to N2O at a constant temp

Half-lives for a zero order reaction Concentration decreases at a constant rate. Half-life decreases with time

Half-lives for a first order reaction Concentration halves in equal time intervals Half life is constant

Half-lives for a second order reaction Concentration decreases rapidly then slows down. Half-life increases with time

Using half-lives Questions Hydrogen peroxide, H2O2, reacts in a first order reaction with a half-life of 27 seconds. If the initial concentration of hydrogen peroxide is 1.60 mol dm-3 what is the concentration after 81 seconds? Paracetamol has a biological half-life of 380 s. How long will it take for the level of paracetamol in the body to fall to one 16th of its original value? 0.2 mol dm-3 1520 s or 25.3 min

Determining orders from rate-concentration graphs Orders can also be determined by plotting rate against concentration. Different orders give different shaped graphs. The following graphs are of initial rate against concentration. It is important to understand and learn the shapes of these graphs.

Rate is not affected by changes in conc. Zero order For zero order: the concentration of A has no effect on the rate of reaction r = k[A]0 = k zero-order Rate is not affected by changes in conc.

If [A] is doubled then rate is doubled First order For first order: the rate of reaction is directly proportional to the concentration of A: r = k[A]1 first-order If [A] is doubled then rate is doubled

If [A] is doubled then rate increases by 22 = 4 times Second order For second order: the rate of reaction is proportional to the concentration of A squared: r = k[A]2 second-order If [A] is doubled then rate increases by 22 = 4 times

Relating concentration vs. time and rate vs. concentration graphs Zero order Measure half-lives Measure initial rates

Relating concentration vs. time and rate vs. concentration graphs First order Measure half-lives Measure initial rates

Relating concentration vs. time and rate vs. concentration graphs Second order Measure half-lives Measure initial rates

Determining initial rates The preceding graphs of initial rate against concentration show the different orders. The initial rate may have been calculated from taking gradients from concentration vs. time graphs at t=0s For a rate concentration graph to show the order of a particular reactant, the concentration of that reactant must be varied whilst the concentrations of the other reactants should be kept constant. Another way to determine initial rates is to use clock reactions

Clock reactions Clock reactions produce some kind of distinct change after a certain period of time: appearance of a precipitate disappearance of a solid a change in colour During these types of reactions you are measuring the rate for the initial period of the reaction. This means that there will not be a significant change in concentration so we can assume that the rate is just 1/t. In clock reactions, experiments are repeated with different concentrations of one reactant, then again with another reactant. Graphs are plotted for each reactant to determine the order.

Sodium thiosulphate and hydrochloric acid reaction The reaction is: Na2S2O3(aq) + 2HCl(aq)2NaCl(aq) + S(s) + SO2(aq) + H2O(l) The experiment is repeated with different [HCl] and then again with different [Na2S2O3] Graphs are plotted for each The reaction produces sulfur, which is formed as a suspension This causes the solution opacity to increase The rate of reaction is measured by seeing how long it takes for a cross placed under the beaker to vanish

Na2S2O3(aq) + 2HCl(aq)  2NaCl(aq) + S(s) + SO2(aq) + H2O(l) Can you write the rate equation for this reaction? Results Na2S2O3(aq) + 2HCl(aq)  2NaCl(aq) + S(s) + SO2(aq) + H2O(l) Rate = k [Na2S2O3]1 [HCl]0 Rate = k [Na2S2O3] Na2S2O3 HCl As the concentration doubles the rate also doubles, as the concentration triples, the rate triples. The rate and concentration are in direct proportion to each other. This reactant has no effect on the rate. The rate continues at a steady pace.

Working out orders from experimental initial rate data Normally, to work out the rate equation we do a series of experiments where the initial concentrations of reactants are changed (one at a time) and measure the initial rate each time. In questions, this data is normally presented in a table. To calculate the order for a particular reactant, compare two experiments where only that reactant is being changed If conc. is doubled and rate stays the same: order= 0 w.r.t that reactant If conc. is doubled and rate doubles: order= 1 If conc. is doubled and rate quadruples: order= 2

Worked example Work out the rate equation for the following reaction, A + B + 2C  D + 2E, using the initial rate data in the table: For reactant A compare between experiments 1 and 2 For reactant B compare between experiments 1 and 3 For reactant C compare between experiments 1 and 4

Worked example The overall rate equation is r = k[A][B]2 The reaction is 3rd order overall and the unit of the rate constant = mol-2dm6s-1

Reactant concentrations are changed simultaneously In most questions, it is possible to compare between two experiments where only one reactant has its initial concentration changed. If, however, both reactants are changed then the effect of both individual changes on concentration are multiplied together to give the overall effect on rate.

Reactant concentrations are changed simultaneously In a reaction where the rate equation is: r = k [A] [B]2 If the [A] is x2, that rate would x2 If the [B] is x3, that rate would x32= x9 If these changes happened at the same time then the rate would x2 x9 = x18 In these questions, you will have to consider the combined effect different reactants have on the rate constant to figure out the orders

Worked example Work out the rate equation for the reaction, between X and Y, using the initial rate data in the table: For reactant X, compare between experiments 1 and 2 For reactant X as the concentration doubles (Y staying constant) so does the rate Therefore the order with respect to reactant X is first order

Worked example Work out the rate equation for the reaction, between X and Y, using the initial rate data in the table: Comparing between experiments 2 and 3: Both X and Y double and the rate goes up by 8 We know X is first order so that will have doubled rate The effect of Y, therefore, on rate is to have quadrupled it. Y must be second order

Worked example Work out the rate equation for the reaction, between X and Y, using the initial rate data in the table: The overall rate equation is r = k [X] [Y]2 The reaction is 3rd order overall and the unit of the rate constant =mol-2dm6s-1.

Calculating a value for k using initial rate data To do this, choose any one of the experiments and put the values into the rate equation that has been rearranged to give k. Using experiment 3: r = k [X][Y]2, k = r / [X][Y]2 k = 2.40 x 10–6 / 0.2 x 0.22 = 3.0 x 10-4 mol-2dm6s-1

Effect of temperature

The Effect of Temperature on k The key factor in the rate of reaction is the number of collisions that exceed the activation energy, Ea. Rate increases with temperature by much more than can be explained just from an increased frequency of collisions. rate = k [NO]2[O2] In the above reaction, the rate increases sharply with increasing T even if the concentrations are unchanged. Obviously the temperature cannot change the concentrations of the reactants, so this shows that k must change with T. Increasing T speeds up most reactions by increasing the value of k For many reactions the rate doubles as T increases by 10˚C, demonstrating that a greater number of particles are above Ea

The Effect of Temperature on k Rate of reaction can be measured at different temperatures and k calculated for each temperature. The rate constant, k, increases with increasing temperature Increasing temperature increases the rate constant k. The relationship is given by the Arrhenius equation k = Ae-(Ea/RT) where A is a constant, R is the gas constant and Ea is activation energy. You do not need to know this equation, just be aware that the relationship between k and T is not directly proportional but is as shown on the graph

Rate equations and mechanisms Rate determining step

Rate equations and mechanisms A mechanism is a series of steps through which the reaction progresses, often involving intermediate compounds. If all the steps are added together they will add up to the overall equation for the reaction. Each step can have a different rate of reaction. The slowest step will control the overall rate of reaction. The slowest step is called the rate-determining step (r.d.s.)

Rate equations and mechanisms The molecularity (number of moles of each substance) of the molecules in the slowest step will be the same as the order of reaction for each substance. r=k[A]m If m=0, the reaction is zero order w.r.t A and A is not involved in the r.d.s. A must be involved in a fast step. If m=1, the reaction is first order w.r.t A and 1 mole of A is involved in the r.d.s. of the mechanism. If m=2, the reaction is second order w.r.t A and 2 moles of A feature in the r.d.s of the mechanism

Predicting mechanisms from rate equations NO2(g) + CO(g)NO(g) + CO2(g) The overall balanced equation tells us that: 1 mol NO2(g) reacts with 1 mol CO(g) to produce 1 mol NO(g) and 1 mol CO2(g) The overall equation does not tell you anything about the mechanism. To do this we need to carry out some rate experiments. If these experiments are carried out the rate equation is found to be: rate = k[NO2]2 What are the orders of reaction with respect to the two reactants? [NO2] = second order [CO] = zero order

Rate-Determining Step and Molecularity NO2(g) + CO(g ) NO(g) + CO2(g) r = k[NO2]2 If a reactant appears in the rate equation then it must participate in the r.d.s. The order with respect to that reactant tells you how many particles of that reactant are involved in the r.d.s. (i.e. its molecularity) What possible mechanism is consistent with both the rate equation and the overall equation?

Rate-Determining Step and Molecularity Overall Reaction: NO2(g) + CO(g) → NO(g) + CO2(g) Mechanism: Step 1: NO2 + NO2→ NO + NO3 slow Step 2: NO3 + CO → NO2 + CO2 fast NO3 is a reaction intermediate. NO2 appears twice in the slow step so it is second order. CO does not appear in the slow step so is zero order. r = k[NO2]2

Example 1 Overall reaction: A + 2B + C  D + E Mechanism: Step 1: A + B  X + D slow Step 2: X + C  Y fast Step 3: Y+BE fast C is zero order as it appears in the mechanism after the slow step r = k[A][B]

Example 2 Overall reaction: A + 2B + C  D + E Mechanism: Step 1: A + B  X + D fast Step 2: X + C  Y slow Step 3: Y + BE fast We can write from this mechanism: r = k[X][C] However, X is an intermediate. We must therefore replace it in the rate equation with the substances it was formed from in the step A + B  X + D r = k[A][B][C]

Example 3 rate equation tell us the slow step (r.d.s.) will involve two molecules of NO and one molecule of H2 Using the rate equation rate = k[NO]2[H2] and the overall equation 2NO(g) + 2H2(g)N2(g) + 2H2O(g), the following three-step mechanism for the reaction was suggested. X and Y are intermediate species. Step 1: NO + NO  X Step 2: X + H2  Y Step 3: Y + H2  N2 + 2H2O Which one of the three steps is the r.d.s? Step 2. H2 appears in the rate equation and combining steps 1 and 2 gives the correct molecularity

Example 4 Suggest a rate equation for the reaction whose mechanism is shown below: CH3CH2Br + OH-  CH3CH2OH + Br- slow The rate equation is r = k[CH3CH2Br] [OH-] This is called an SN2 reaction. Substitution, Nucleophilic, 2 molecules in the r.d.s

The same reaction can also occur via a different mechanism: Overall Reaction: (CH3)3CBr + OH–  (CH3)3COH + Br – Mechanism: (CH3)3CBr  (CH3)3C+ + Br- slow (CH3)3C+ + OH–  (CH3)3COH fast Write the rate equation and draw the ‘curly arrows’ mechanism for this reaction What would this reaction be called if the other was SN2? r = k[(CH3)3CBr)] SN1

C3H6O + NaBH4 + H2O  C3H8O + NaBH3.OH Example 5 Carbonyls can be reduced by hydrides, such as sodium tetrahydroborate, NaBH4 An example reaction is: C3H6O + NaBH4 + H2O  C3H8O + NaBH3.OH A student suggests the rate equation for this reaction is: r = k[H2O][C3H6O] Using your knowledge of the mechanism for this reaction, state whether or not you agree with this and why.

Questions 1) The following reaction is first order w.r.t. H2 and first order w.r.t ICl. H2(g) + 2ICl(g)  I2(g) + 2HCl(g) a) Write the rate equation for this reaction. b) The mechanism for this reaction consists of two steps. i) Identify the molecules that are in the r.d.s. Justify you answer. ii) A chemist suggested the following mechanism: 2ICl(g) I2+Cl2 slow H2(g) + Cl2(g)  2HCl(g) fast Suggest, with reasons, whether this mechanism is likely to be correct. [5 marks overall]

Questions 2) The reaction between HBr and oxygen gas occurs rapidly at 700 K. It can be represented by the equation 4HBr + O2  2H2O + 2Br2 The rate equation was found by experiment to be: r = k[HBr][O2] a) Explain why the reaction cannot be a one-step reaction. b) Each of the 4 steps in this reaction involves 1 molecule of HBr. Two of the steps are the same. The r.d.s is the first one and results in the formation of HBrO2. Write equations for the full set of 4 reactions. Hint: HBrO is another intermediate.

Solution to 2b The rate equation shows that one molecule of HBr is involved in the r.d.s. However, there are 4 molecules in the balanced equations. 3 other molecules of HBr must be involved in other (fast) steps, so the reaction cannot just be one-step. Overall mechanism: Step 1: HBr + O2  HBrO2 slow (r.d.s.) Step 2: HBr + HBrO2  2HBrO fast Step 3: HBr + HBrO  H2O + Br2 fast Step 4: HBr + HBrO  H2O + Br2 fast

Further questions

Further questions Orders and rates for simultaneous reactions Answers

Selected exam questions

June 2010

Jan 2012

June 2013