How fast?

The rate of reaction is defined as the change in concentration of a substance in unit time Its usual unit is mol dm-3s -1 When a graph of concentration of reactant is plotted vs time, the gradient of the curve is the rate of reaction. 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 Initial rate = gradient of tangent time concentration 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) Calculating orders from initial rate data 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 initial Graphs of initial rate against concentration show the different orders. The initial rate may have been calculated from taking gradients from concentration /time graphs 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. 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 2 0.060 0.030 0.015 0.0075 t ½ t ½ t ½ Time (min) [A] Continuous rate data This is data from one experiment where the concentration of a substance is followed throughout the experiment. Continuous rate experiments This data is processed by plotting the data and calculating successive half-lives. If half-lives are constant then the order is 1st order 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 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 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. This data is normally presented in a table. 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 N Goalby chemrevise.org 3 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 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

temperature k 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 E. -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 Temperature T (K) 1/T time t (s) 1/t Ln (1/t) 297.3 0.003364 53 0.018868 -3.9703 310.6 0.00322 24 0.041667 -3.1781 317.2 0.003153 16 0.0625 -2.7726 323.9 0.003087 12 0.083333 -2.4849 335.6 0.00298 6 0.166667 -1.7918 x1 ,y1 x2 ,y2 gradient = y2 -y1 x2 -x1 In above example gradient =-5680 The gradient should always be -ve Ea = – gradient x R (8.31) = – -5680 x8.31 = 47200 J mol-1 The unit of Ea using this equation will be J mol-1 . Convert into kJ mol-1 by dividing 1000 Ea = +47.2 kJ mol-1 E

Techniques to investigate rates of reaction measurement of the change in volume of a gas Titrating samples of reaction mixture with acid, alkali, sodium thiosulphate etc Colorimetry. Measurement of optical activity. Measurement of change of mass Measuring change in electrical conductivity H2O2 (aq) + 2I- (aq) + 2H+ (aq)  2H2O(l) + I2 (aq) HCOOCH3 (aq) + NaOH(aq)  HCOONa(aq) + CH3OH(aq) (CH3 )2C=CH2 (g) + HI(g) (CH3 )3CI(g) BrO3 – (aq) + 5Br – (aq) + 6H+ (aq)  3Br2 (aq) + 3H2O(l) HCOOH(aq) + Br2 (aq)  2H+ (aq) + 2Br – (aq) + CO2 (g) HCOOH(aq) + Br2 (aq)  2H+ (aq) + 2Br – (aq) + CO2 (g) There are several different methods for measuring reactions rates. Some reactions can be measured in several ways 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. If drawing a gas syringe make sure you draw it with some measurement markings on the barrel to show measurements can be made. This works if there is a gas produced which is allowed to escape. Works better with heavy gases such as CO2 HCOOH(aq) + Br2 (aq)  2H+ (aq) + 2Br – (aq) + CO2 (g) CH3COCH3 (aq) + I2 (aq) → CH3COCH2 I(aq) + H+ (aq) + I– (aq) Small samples are removed from the reaction mixture, quenched (which stops the reaction) and the titrated with a suitable reagent. The NaOH could be titrated with an acid The H+ could be titrated with an alkali The I2 could be titrated with sodium thiosulphate If one of the reactants or products is coloured then colorimetry can be used to measure the change in colour of the reacting mixtures The I2 produced is a brown solution Can be used if there is a change in the number of ions in the reaction mixture If there is a change in the optical activity through the reaction this could be followed in a polarimeter CH3CHBrCH3 (l) + OH− (aq)  CH3CH(OH)CH3 (l) + Br− (aq)

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.