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

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5.1.1 How fast?

(b) deduction of: (i) orders from experimental data (ii) a rate equation from orders of the form: rate = k[A]m[B]n, where m and n are 0, 1 or 2 M0.2 Learners are expected to interpret initial rates data to determine orders with respect to reactants. Integrated forms of rate equations are not required. PAG10 HSW8 Use of rate equations. (c) calculation of the rate constant, k, and related quantities, from a rate equation including determination of units M0.0, M0.1, M0.4, M1.1, M2.2, M2.3, M2.4