Rate Equations

The rate equation relates mathematically the rate of reaction to the concentration of the reactants. For the following reaction, aA + bB  products, the generalised rate equation is: r = k[A]m[B]n r is used as symbol for rate The unit of r is usually mol dm-3 s -1 The square brackets [A] means the concentration of A (unit mol dm-3 ) k is called the rate constant 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 NOTE: the orders have nothing to do with the stoichiometric coefficients in the balanced equation. They are worked out experimentally The total order for a reaction is worked out by adding all the individual orders together (m+n) For zero order: the concentration of A has no effect on the rate of reaction r = k[A]0 = k For first order: the rate of reaction is directly proportional to the concentration of A r = k[A]1 For second order: the rate of reaction is proportional to the concentration of A squared r = k[A]2 The rate constant (k) 1. The units of k depend on the overall order of reaction. It must be worked out from the rate equation 2. The value of k is independent of concentration and time. It is constant at a fixed temperature. 3. The value of k refers to a specific temperature and it increases if we increase temperature 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-1dm3s -1 For a 3rd order overall reaction the unit of k is mol-2dm6s -1 Example (first order overall) Rate = k[A][B]0 m = 1 and n = 0 – reaction is first order in A and zero order in B – overall order = 1 + 0 = 1 – usually written: Rate = k[A] Remember: the values of the reaction orders must be determined from experiment; they cannot be found by looking at the balanced reaction equation Calculating units of k 1. Rearrange rate equation to give k as subject k = Rate [A] 2. Insert units and cancel k = mol dm-3s -1 mol dm-3 Unit of k = s-1 N Goalby chemrevise.org 1 Example: Write rate equation for reaction between A and B where A is 1st order and B is 2nd order. r = k[A][B]2 overall order is 3 Calculate the unit of k Unit of k = mol-2dm6s -1. 1. Rearrange rate equation to give k as subject k = Rate [A][B]2 2. Insert units and cancel k = mol dm-3s -1 mol dm-3 .(moldm-3 )2 3. Simplify fraction k = s -1 mol2dm-6

Working out rate order graphically In an experiment where the concentration of one of the reagents is changed and the reaction rate measured it is possible to calculate the order graphically Rate = k [Y]n Log both sides of equation Log rate = log k + n log [Y] A graph of log rate vs log [Y] will yield a straight line where the gradient is equal to the order n Taking rate equation Y = c + m x In this experiment high concentrations with quick times will have the biggest percentage errors. 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. Working out orders when two 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 effect on rate. 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 x2x9= x 18 Example work out the rate equation for the reaction, between X and Y, using the initial rate data in the table Experiment Initial concentration of X/ mol dm–3 Initial concentration of Y/ mol dm–3 Initial rate/ mol dm–3 s –1 1 0.05 0.1 0.15 x 10–6 2 0.10 0.1 0.30 x 10–6 3 0.20 0.2 2.40 x 10–6 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 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 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 5 N Goalby chemrevise.org Calculating a value for k using initial rate data Using the above example, 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 k = 3.0 x 10-4 mol-2dm6s -1 Remember k is the same for all experiments done at the same temperature. Increasing the temperature increases the value of the rate constant k For zero order reactants, the rate stays constant as the reactant is used up. This means the concentration of that reactant has no effect on rate. Rate = k [A]0 so rate = k As the rate is the gradient of the graph on the right, the gradient is also the value of the rate constant. [A] Time (min) Example: work out the rate equation for the following reaction, A+ B+ 2C  D + 2E, using the initial rate data in the table Experiment [A] mol dm- 3 [B] mol dm-3 [C] mol dm-3 Rate mol dm-3 s -1 1 0.1 0.5 0.25 0.1 2 0.2 0.5 0.25 0.2 3 0.1 1.0 0.25 0.4 4 0.1 0.5 0.5 0.1 In order to calculate the order for a particular reactant it is easiest to compare two experiments where only that reactant is being changed For reactant A compare between experiments 1 and 2 If conc is doubled and rate stays the same: order= 0 If conc is doubled and rate doubles: order= 1 If conc is doubled and rate quadruples : order= 2 For reactant A as the concentration doubles (B and C staying constant) so does the rate. Therefore the order with respect to reactant A is first order For reactant B compare between experiments 1 and 3 : As the concentration of B doubles (A and C staying constant) the rate quadruples. Therefore the order with respect to B is 2nd order As the concentration of C doubles (A and B staying constant) the rate stays the same. Therefore the order with respect to C is zero order For reactant C compare between experiments 1 and 4 : 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 T

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 gas constant and Ea is activation energy. Increasing the temperature increases the value of the rate constant k The Arrhenius equation can be rearranged ln k = constant – Ea/(RT) k is proportional to the rate of reaction so ln k can be replaced by ln(rate) From plotting a graph of ln(rate) or ln k against 1/T the activation energy can be calculated from measuring the gradient of the line ln (Rate) 1/T Gradient = – Ea/ R Ea = – gradient x R Effect of Temperature on Rate Constant: The Arrhenius Equation -4.1 -3.6 -3.1 -2.6 -2.1 -1.6 0.0029 0.003 0.0031 0.0032 0.0033 0.0034 ln (Rate) 1/T use a line of best fit use all graph paper choose points far apart on the graph to calculate the gradient

Rate Equations and Mechanisms A mechanism is a series of steps through which the reaction progresses, often forming 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. 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. e.g. 0 moles of A in slow step would mean A is zero order. 1 mole of A in the slow step would mean A is first order 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 r = k [A]1 [B]1 [C]o r = k [X]1 [C]1 The intermediate X is not one of the reactants so must be replaced with the substances that make up the intermediate in a previous step A + B  X + D r = k[A]1 [B]1 [C]1 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 Example 2 Example 3 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 r = k [NO2 ]2 Example 5: SN1 or SN2? You don’t need to remember the details here. Remember the nucleophilic substitution reaction of haloalkanes and hydroxide ions. This is a one step mechanism CH3CH2Br + OH-  CH3CH2OH + Br- slow step The rate equation is r = k [CH3CH2Br] [OH-] 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 The rate equation is r = k [(CH3 )3CBr] This is called SN2. Substitution, Nucleophilic, 2 molecules in rate determining step This is called SN1. Substitution, Nucleophilic, 1 molecule in rate determining step C is zero order as it appears in the mechanism in a fast step after the slow step NO2 appears twice in the slow steps so it is second order. CO does not appear in the slow step so is zero order. 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 rate-determining step? Example 4 Step 2 – as H2 appears in rate equation and combination of step 1 and 2 is the ratio that appears in the rate equation.

A-level Chemistry exemplar for required practical No. 7 – part a Measuring the rate of reaction by an initial rate method: An ‘Iodine Clock’ experiment: To investigate the reaction of iodide(V) ions with hydrogen peroxide in acidic solution and to determine the order of the reaction with respect to iodide ions. Student sheet The ‘Iodine Clock’ experiment can be used to determine the effect of a change in concentration of iodide ions on the reaction between hydrogen peroxide and iodide ions. Introduction Hydrogen peroxide reacts with iodide ions to form iodine and the thiosulfate ion immediately reacts with iodine as shown below. H2O2(aq) + 2H+(aq) + 2I– (aq) → I2(aq) + 2H2O(l) 2S2O3 2– (aq) + I2(aq) → 2I– (aq) + S4O6 2– (aq) When the I2 produced has reacted with all of the limited amount of thiosulfate ions present, excess I2 remains in solution. Reaction with the starch then forms a dark blue-black colour. By varying the concentration of I – , you can determine the order of reaction with respect to I – ions. Requirements You are provided with the following:  0.25 mol dm–3 dilute sulfuric acid  0.10 mol dm–3 potassium iodide solution  0.05 mol dm–3 sodium thiosulfate solution (in a shared burette)  0.10 mol dm–3 hydrogen peroxide solution (in a shared burette)  starch solution  50 cm3 burette  funnel suitable for filling a burette  stand and clamp  white tile  plastic dropping pipette  25 cm3 measuring cylinder  50 cm 3 measuring cylinder  100 cm3 beaker  250 cm3 beaker  stirrer  stopwatch  paper towels to dry beakers  plentiful supply of distilled or deionised water. Suggested method Experiment 1 a) Rinse a 50 cm3 burette with potassium iodide solution. Fill the burette with potassium iodide solution. b) Transfer 10.0 cm3 of hydrogen peroxide solution from the shared burette provided to a clean, dry 100 cm3 beaker. You will use this in step (h). c) Use a 50 cm3 measuring cylinder to add 25 cm3 of sulfuric acid to a clean, dry 250 cm3 beaker. d) Use a 25 cm3 measuring cylinder to add 20 cm3 of distilled or deionised water into the 250 cm3 beaker. e) Use a plastic dropping pipette to add about 1 cm3 of starch solution to this beaker. f) Use your burette to add 5.0 cm3 of potassium iodide solution to the mixture in the 250 cm3 beaker. g) Finally, add 5.0 cm3 of sodium thiosulfate solution from the shared burette provided to the mixture in the 250 cm3 beaker. Make sure this sodium thiosulfate solution is added last. h) Stir the mixture in the 250 cm3 beaker. Pour the hydrogen peroxide solution from the 100 cm3 beaker into the 250 cm3 beaker and immediately start the timer. Stir the mixture. i) Stop the timer when the mixture in the 250 cm3 beaker turns blue-black. Record the time to an appropriate precision in a table of your own design. This experiment could take several minutes. j) Rinse the 250 cm3 beaker with distilled or deionised water and dry it with a paper towel. Experiments 2–5 k) Repeat steps (b) to (j) in four further experiments using the volumes shown in the following table. Volumes of solutions added to 250 cm3 beaker Volume in 100 cm3 beaker Experiment Sulfuric acid 0.25 M / cm3 Starch / cm3 Water / cm3 Potassium iodide 0.10 M / cm3 Sodium thiosulfate 0.05 M / cm 3 Hydrogen peroxide 0.10 M / cm3

Continuous Monitoring When we follow one experiment over time recording the change in concentration we call it a continuous rate method. The gradient represents the rate of reaction. The reaction is fastest at the start where the gradient is steepest. The rate drops as the reactants start to get used up and their concentration drops. The graph will eventual become horizontal and the gradient becomes zero which represents the reaction having stopped. time concentration Measurement of the change in volume of a gas Mg + HCl  MgCl2 +H2 This works if there is a change in the number of moles of gas in the reaction. Using a gas syringe is a common way of following this. It works quite well for measuring continuous rate but a typical gas syringe only measures 100ml of gas so you don’t want at reaction to produce more than this volume. Quantities of reactants need to be calculated carefully. gas syringe Typical Method • Measure 50 cm3 of the 1.0 mol dm–3 hydrochloric acid and add to conical flask. • Set up the gas syringe in the stand • Weigh 0.20 g of magnesium. • Add the magnesium ribbon to the conical flask, place the bung firmly into the top of the flask and start the timer. • Record the volume of hydrogen gas collected every 15 seconds for 3 minutes. The initial rate is the rate at the start of the reaction, where it is fastest. It is obtained by taking the gradient of a continuous monitoring conc vs time graph at time = zero. A measure of initial rate is preferable as we know the concentrations at the start of the reaction Large Excess of reactants In reactions where there are several reactants, if the concentration of one of the reactant is kept in a large excess then that reactant will appear not to affect rate and will be pseudo-zero order . This is because its concentration stays virtually constant and does not affect rate.A-level Chemistry exemplar for required practical No. 7 – part b Measuring the rate of reaction by a continuous monitoring method: The reaction between magnesium and hydrochloric acid Student sheet Requirements You are provided with the following:  magnesium ribbon  0.8 mol dm–3 hydrochloric acid  50 cm3 measuring cylinder  100 cm3 conical flask  rubber bung and delivery tube to fit conical flask  100 cm3 gas syringe OR trough/plastic container with 100 cm3 measuring cylinder  stand, boss and clamp  stopwatch or timer  distilled or deionised water. Suggested method a) Measure 50 cm3 of the 0.8 mol dm–3 hydrochloric acid and add to conical flask. b) Set up the gas syringe in the stand (or alternative gas collection method as shown by your teacher). Using a gas syringe. Using a trough c) Add one 6 cm strip of magnesium ribbon to the conical flask, place the bung firmly into the top of the flask and start the timer. d) Record the volume of hydrogen gas collected every 15 seconds for 2.5 minutes. Repeat steps (a) to (d) using 0.4 mol dm–3 hydrochloric acid, made by mixing 25 cm3 of the 0.8 mol dm–3 hydrochloric acid with 25 cm3 of distilled or deionised water. Analysis a) Plot a graph of volume of hydrogen produced on the y-axis against time in seconds for each hydrochloric acid concentration. Draw a line of best fit. b) Draw a tangent to each line of best fit at time, t = 0 s c) Calculate the gradient of each tangent in order to deduce the rate of each reaction. d) Compare the two rate values obtained. Sample results The following table is a sample results table using results from the trial of this experiment. Volume of gas collected / cm3 Time / s 0.4 mol dm–3 hydrochloric acid 0.8 mol dm–3 hydrochloric acid 0 0.0 0.0 15 4.0 15.0 30 6.0 29.0 45 9.0 43.0 60 17.0 54.0 75 20.0 66.0 90 22.0 75.0 105 24.0 84.0 120 27.0 91.0 135 30.0 95.0 150 32.0 96.0 These were obtained using 0.8 mol dm–3 hydrochloric acid and 0.4 mol dm–3 hydrochloric acid with 6 cm strips of magnesium ribbon.