5.1.1 How Fast? Reaction Rates PDF

5.1.1 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....

5.1.1 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. concentration The initial rate is the rate at the start of the reaction where it is fastest. rate = gradient of tangent to curve Reaction rates can be calculated from graphs of concentration of reactants or products, by drawing a tangent to the curve (at different times) and calculating the gradient of the tangent. time Rate Equations The rate equation relates mathematically the rate of reaction to the concentration of the reactants. For the following reaction, aA + bB r is used as symbol for rate products, the generalised rate equation is: r = k[A]m[B]n The unit of r is usually mol dm-3 s-1 The square brackets [A] means m, n are called reaction orders the concentration of A Orders are usually integers 0,1,2 (unit mol dm-3) 0 means the reaction is zero order with respect to that reactant k is called the rate constant 1 means first order 2 means second order The total order for a reaction is worked NOTE: the orders have nothing to do with the stoichiometric out by adding all the individual orders coefficients in the balanced equation. They are worked out together (m+n) experimentally. Calculating orders from initial rate data For zero order: the concentration of A has no effect on the initial 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 For a rate concentration graph to show the Graphs of initial rate against concentration show the order of a particular reactant the different orders. The initial rate may have been concentration of that reactant must be varied calculated from taking gradients from concentration whilst the concentrations of the other /time graphs. reactants should be kept constant. N Goalby chemrevise.org 1  Continuous rate experiments Continuous rate data 0.060 This is data from one experiment where the concentration of a substance is followed throughout the experiment. If half-lives are constant [A] then the order is 1st order This data is processed by plotting the data and calculating 0.030 successive half-lives. The half-life of a first-order reaction 0.015 is independent of the concentration and is constant. 0.0075 If half-lives rapidly increase then the order t½ t½ t½ is 2nd order. Time (min) The rate constant (k) 1. The units of k depend on the overall order of For a 1st order overall reaction the unit of k is reaction. It must be worked out from the rate s-1 equation For a 2nd order overall reaction the unit of k is 2. The value of k is independent of concentration and mol-1dm3s-1 time. It is constant at a fixed temperature. For a 3rd order overall reaction the unit of k is 3. The value of k refers to a specific temperature and mol-2dm6s-1 it increases if we increase temperature Example (first order overall) Rate = k[A][B]0 m = 1 and n = 0 Remember: the values of the reaction orders must - reaction is first order in A and zero order in B be determined from experiment; they cannot be - overall order = 1 + 0 = 1 found by looking at the balanced reaction equation - usually written: Rate = k[A] Calculating units of k 1. Rearrange rate equation 2. Insert units and to give k as subject cancel k = Rate k = mol dm-3s-1 Unit of k = s-1 [A] mol dm-3 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 1. Rearrange rate equation to 2. Insert units and 3. Simplify fraction give k as subject cancel k= s-1 k = Rate k = mol dm-3s-1 Unit of k = mol-2dm6s-1 mol2dm-6 [A][B]2 mol dm-3.(moldm-3)2 N Goalby chemrevise.org 2 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: Deduce the rate equation for the following reaction, A+ B+ 2C  D + 2E, using the initial rate data in the table Experiment [A] [B] [C] Rate In order to calculate the order for a particular mol dm- mol dm-3 mol dm-3 mol dm-3 s-1 reactant it is easiest to compare two 3 experiments where only that reactant is 1 0.1 0.5 0.25 0.1 being changed 2 0.2 0.5 0.25 0.2 If conc is doubled and rate stays the same: order= 0 3 0.1 1.0 0.25 0.4 If conc is doubled and rate doubles: order= 1 4 0.1 0.5 0.5 0.1 If conc is doubled and rate quadruples : order= 2 For reactant A compare between experiments 1 and 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 For reactant C compare between experiments 1 and 4 : 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 The reaction is 3rd order overall and the unit of the The overall rate equation is r = k [A] [B]2 rate constant =mol-2dm6s-1 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 N Goalby chemrevise.org 3 Example Deduce the rate equation for the reaction, between X and Y, using the initial rate data in the table Experiment Initial concentration of Initial concentration Initial X/ mol dm–3 of Y/ mol dm–3 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 overall rate equation is r = k [X] [Y]2 The effect of Y, therefore, on rate is to have quadrupled it. The reaction is 3rd order overall and the unit of Y must be second order 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 k = 2.40 x 10–6 k = 3.0 x 10-4 mol-2dm6s-1 [X] [Y]2 0.2 x 0.22 Remember k is the same for all experiments done at the same temperature. Increasing the temperature increases the value of the rate constant k 4 N Goalby chemrevise.org Effect of temperature on rate constant Increasing the temperature increases the value of the rate constant k 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. temperature The Arrhenius equation can be rearranged 1/T ln (Rate) ln k = constant – EA/(RT) k is proportional to the rate of reaction so ln k can be replaced by ln(rate) Gradient = - EA/ R From plotting a graph of ln(rate) or ln k against 1/T the activation energy can be calculated from measuring EA = - gradient x R the gradient of the line 1/T Example 0.0029 0.003 0.0031 0.0032 0.0033 0.0034 Temperature time t -1.6 T (K) 1/T (s) 1/t Ln (1/t) x1,y1 297.3 0.003364 53 0.018868 -3.9703 -2.1 310.6 0.00322 24 0.041667 -3.1781 ln (Rate) 317.2 0.003153 16 0.0625 -2.7726 -2.6 323.9 0.003087 12 0.083333 -2.4849 -3.1 335.6 0.00298 6 0.166667 -1.7918 -3.6 x2,y2 gradient = y2-y1 The gradient should x2-x1 always be -ve -4.1 In above example gradient =-5680 use a line of best fit use all graph paper EA = - gradient x R (8.31) choose points far apart on the graph to = - -5680 x8.31 calculate the gradient = 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 N Goalby chemrevise.org 5 Techniques to investigate rates of reaction There are several different methods for measuring reactions rates. Some reactions can be measured in several ways Measurement of the change in volume of a gas 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 (CH3)2C=CH2(g) + HI(g) (CH3)3CI(g) measurement markings on the barrel to show measurements HCOOH(aq) + Br2(aq)  2H+(aq) + 2Br - (aq) + CO2(g) can be made. Measurement of change of mass 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) Titrating samples of reaction mixture with acid, alkali, sodium thiosulfate etc 1. Small samples are removed from the reaction mixture 2. quench (which stops the reaction)- can be done by by dilution with water by cooling by adding a reagent that reacts with one of the reactants 3. then titrate with a suitable reagent. HCOOCH3(aq) + NaOH(aq)  HCOONa(aq) + CH3OH(aq) The NaOH could be titrated with an acid BrO3 –(aq) + 5Br –(aq) + 6H+(aq)  3Br2(aq) + 3H2O(l) The H+ could be titrated with an alkali CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + H+(aq) + I–(aq) The I2 could be titrated with sodium thiosulfate Colorimetry. If one of the reactants or products is coloured H2O2(aq) + 2I- (aq) + 2H+(aq)  2H2O(l) + I2(aq) then colorimetry can be used to measure the change in colour of the reacting mixtures The I2 produced is a brown solution Measuring change in electrical conductivity Can be used if there is a change in the number HCOOH(aq) + Br2(aq)  2H+(aq) + 2Br - (aq) + CO2(g) of ions in the reaction mixture Measurement of optical activity. If there is a change in the optical activity through the reaction this could be followed in a CH3CHBrCH3 (l) + OH−(aq)  CH3CH(OH)CH3 (l) + Br−(aq) polarimeter N Goalby chemrevise.org 6 Rate Equations and Mechanisms A mechanism is a series of steps through which the reaction Each step can have a different rate of progresses, often forming intermediate compounds. If all the reaction. The slowest step will control the steps are added together they will add up to the overall overall rate of reaction. The slowest step equation for the reaction 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 2 overall reaction Example 1 A + 2B + C  D + E overall reaction Mechanism A + 2B + C  D + E Step 1 A + B  X + D fast Mechanism Step 2 X+CY slow Step 1 A + B  X + D slow Step 3 Y+BE fast Step 2 X+CY fast Step 3 Y+BE fast r = k [X]1[C]1 r = k [A]1[B]1[C]o The intermediate X is not one of the reactants so must be replaced with the substances that make C is zero order as it appears in the up the intermediate in a previous step mechanism in a fast step after the slow step A+BX+D r = k[A]1[B]1[C]1 Example 3 Example 4 Overall Reaction Using the rate equation rate = k[NO]2[H2] and NO2(g) + CO(g) NO(g) + CO2(g) the overall equation 2NO(g) + 2H2(g) N2(g) + 2H2O(g), the following three-step mechanism for the reaction was Mechanism: suggested. X and Y are intermediate species. Step 1 NO2 + NO2 NO + NO3 slow Step 1 NO + NO  X Step 2 NO3 + CO NO2 + CO2 fast Step 2 X + H2 Y NO3 is a reaction intermediate Step 3 Y + H2  N2 + 2H2O NO2 appears twice in the slow steps so it is second order. CO does not appear in Which one of the three steps is the rate-determining step? the slow step so is zero order. Step 2 – as H2 appears in rate equation and combination of r = k [NO2]2 step 1 and 2 is the ratio that appears in the rate equation. Example 5: SN1 or SN2? You don’t need to learn details of these mechanisms. Remember the nucleophilic substitution reaction of haloalkanes and hydroxide ions. The same reaction can also occur via a different mechanism This is a one step mechanism Overall Reaction H H δ+ δ- - (CH3)3CBr + OH–  (CH3)3COH + Br – H3C C Br H3C C OH + :Br -HO: Mechanism: H H (CH3)3CBr  (CH3)3C+ + Br – slow CH3CH2Br + OH-  CH3CH2OH + Br- slow step (CH3)3C+ + OH–  (CH3)3COH fast This is called SN2. The rate equation is Substitution, Nucleophilic, The rate equation is This is called SN1. r = k [CH3CH2Br] [OH-] 2 molecules in rate Substitution, Nucleophilic, determining step r = k [(CH3)3CBr] 1 molecule in rate determining step N Goalby chemrevise.org 7