Introduction



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
/
~
~
~
/
3.1.9.1 Rate equations (A-level only)
The rate of a chemical reaction is related to the concentration of reactants by a rate equation of the form: Rate = k[A]m [B]n where m and n are the orders of reaction with respect to reactants A and B and k is the rate constant.
The orders m and n are restricted to the values 0, 1, and 2.
Students should be able to:
• define the terms order of reaction and rate constant
• perform calculations using the rate equation
Calculating order of reactions using initial rates



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
/
~
~
~
/
3.1.9.2 Determination of rate equation (A-level only)
The rate equation is an experimentally determined relationship.
Students should be able to:
• use concentration–time graphs to deduce the rate of a reaction
• use initial concentration–time data to deduce the initial rate of a reaction
• use rate–concentration data or graphs to deduce the order (0, 1 or 2) with respect to a reactant
• derive the rate equation for a reaction from the orders with respect to each of the reactants
Effect of temperature on rate constants


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
/
~
~
~
/
The rate constant k varies with temperature as shown by the equation:
k = Ae–Ea/RT
where A is a constant, known as the Arrhenius constant, Ea is the activation energy and T is the temperature in K.
Students should be able to:
• explain the qualitative effect of changes in temperature on the rate constant k
• perform calculations using the equation k =Ae–Ea/RT
• understand that the equation k =Ae–Ea/RT can be rearranged into the form ln k = –Ea /RT + ln A and know how to use this rearranged equation with experimental data to plot a straight line graph with slope –Ea /R These equations and the gas constant, R, will be given when required.
Rate equations and reaction mechanisms


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.
/
~
~
~
/
3.1.9.2 Determination of rate equation (A-level only)
The orders with respect to reactants can provide information about the mechanism of a reaction.
Students should be able to:
use the orders with respect to reactants to provide information about the rate determining/limiting step of a reaction.
Required Practical 7: Initial rate method
/
~
~
~
/
3.1.9.2 Determination of rate equation (A-level only)
Required practical 7
Measuring the rate of reaction:
• by an initial rate method
Required Practical 7: Continuous monitoring method.
/
~
~
~
/
3.1.9.2 Determination of rate equation (A-level only)
Required practical 7
Measuring the rate of reaction:
• by a continuous monitoring method.
Credits: Neil Goalby