Rate Equations (Kinetics A Level)

The rate equation relates mathematically the rate of reaction to the conc of the reactants. - For the following reaction, - aA + bB → products, the generalised rate equation is: -> r = k[A]^m[B]^n • r is rate; units (mol dm-3 s-1) • [square brackets] is conc (mol dm-3) • k is rate constant .. m, n are called reaction orders >Orders are usually integers 0,1,2 > O means reaction is zero order with respect to that > reactant 1 means first order > 2 means second order > The total order for a reaction is worked by adding all the individual order together (m+n)

For zero order: the concentration of A has no effect on rate of reaction r=k[A]º=k For first order: rate of reaction is directly proportional to conc of A r = k[A]^1 For second order: rate of reaction is proportional to conc of A squared r = k[A]^2 diagram1

• The units of k depend on the overall order of reaction. >must be worked out from rate equation • The value of k is independent of concentration and time. It is constant at a fixed temperature • The value of k refers to a specific temperature and increases if we increase temperature . for a 1st order overall reaction the unit of k is s-1 For 2nd order overall reaction the unit of k is mol-1 dm3 s-1 For 3rd order overall reacion for the unit of k is mol-2 dm6 s-1

Example 1st order: Rate = k[A][B]0 - overall order m=1 and n=0 - 0 + 1 =1 so overall is first order - first orders written as rate=k[A] STo find units - rearrange to make k subject K = rate/[A] - insert units to cancel - mol dm-3 s-1 / mol dm-3 - = s-1

Continuous Monitoring - When we follow one experiment over time recording change in conc - we call it a continuous rate method. • The gradient represents rate of reaction. • The reaction is fastest at start where gradient is steepest. • The rate drops as the reactants start to get used up; their conc drops. • The graph will eventually become horizontal and the gradient • becomes zero which represents the reaction having stopped.

This works if theres a change in number of moles of gas in reaction. - Using a gas syringe is a common way of following this. - works well for measuring continuous rate but a typical gas syringe only measures 100ml of gas - so you don’t want a reaction to produce more than this volume. - - Quantities of reactants need to be calculated carefully. Mg + 2HCl → MgCl +H2 Eg of Typical Method • Measure 50 cm’ of the hydrochloric acid;add to conical flask. • Set up gas syringe in stand : Weigh 0.20 g of magnesium, • Add magnesium ribbon to the conical flask, place bung firmly into top of the flask and start the timer. • • Record volume of hydrogen gas collected every 15 seconds for 3 minutes. .. • Large excess of reactants - In reactions with several reactants, if the conc of one of the reactant is kept in a large excess - then reactant wont affect rate and will be pseudo-zero order. - its conc stays virtually constant and does not affect rate. • The initial rate - rate at the start of the reaction, where it is fastest. - calculated from the gradient of a continuous monitoring conc vs time graph at time = zero. - A measure of initial rate is preferable as we know the concs at the start of the reaction. diagram 2

• The higher the conc/ temp/ SA, the faster the rate (steeper the gradient) • If magnesium or marble chips is in excess of acid, then • final vol of gas produced will be proportional to amt of moles of acid. .. analysing diagram 3 Different vols of same initial concs will have same initial rate (if other conditions are same) >but will end at different amounts Need to calculate and compare initial moles of reactants > to distinguish between diff finishing volumes.

• The initial rate can be calculated from taking gradient • of a continuous monitoring conc vs time graph at time = zero • • Initial rate can also be calculated from clock reactions where time taken to reach a fixed conc is measured. Initial rate represented as 1/t .. • In clock reactions there are often two successive reactions. • end points achieved when one limited reactant runs out, resulting in sudden colour change. • By repeating experiment several times, varying conc of a reactant e.g. I- (keeping other reactants constant conc) • you can determine order of reaction with respect to that reactant

• Hydrogen peroxide reacts w iodide ions to form iodine. • thiosulfate ion then immediately reacts iodine formed, in second reaction as shown below. H202(aq) + 2H+(aq) + 2l-(aq) → I2(aq) + 2H20(l) 2S2032-(aq) + I2(aq) → 2I-(aq) + S4062-(aq) • When the I2 produced has reacted with all of limited amount of thiosulfate ions present, • ## excess I2 remains in solution. • Reaction with starch then suddenly forms a dark blue-black colour. • A series of experiments is carried out, in which the conc of iodide ions is varied, • while keeping concs of all of the other reagents the same. • In each experiment, time taken (t) for reaction mixture to blue is measured.

Working order out graphically - in an experiment where conc of one reagent is changed and reaction rate measured, - its possible to find order graphically - Log both sides of equation Log rate = log k + n log [Y] Y= c + mx A graph of log rate vs log [Y] will yield a straight line where the gradient is equal to the order n

To calculate order for a particular reactant, its easiest to - compare two experiments where only that reactant is being changed. - If conc is doubled and rate stays same: order= 0 - If conc is doubled and rate doubles: order= 1 - If conc is doubled and rate quadruples : order= 2 Add orders of all reactants in equation to find equation’s order

• in most questions its possible to compare 2 experiments where • ## only one reactant has its initial conc changed • if two reactants are changed then the effect of both individual changes • of conc are multiplied to give overall change in rate

Using the inital rate data tables, choose any one of the experiments - and substitute values into rate equation and rearrange to give k Remember k is same for all experiments done at same temp. > Increasing the temperature increases the value of the rate constant k

For zero order reactants, rate stays constant as the reactant is used up. > means the conc of that reactant has no effect on rate. > Rate = k [A]º so rate = k If the graph is directly proportional, gradient = value of k constant

Increasing temp increases value of the rate constant k - the relationship is given by Arrhenius equation (k = Ae^-Ea/RT - where A is Arrhenius constant, - R is the gas constant, - and Ea is activation energy diagram 5

k = Ae-Ea/RT The Arrhenius equation is usually rearranged to Ln k= Ln A -Ea/(RT) > shd be able to rearrange and substitute values into both these equations. Units: -Temp uses the unit K -R = 8.31 J mol-1K-1 -Activation energy will need to be in J mol-1 to match the units of R -The unit of Arrhenius constant A will be same as the unit of rate constant diagram 6 for examples

Using the rearranged version Ink=In A - Ea/(RT) k is proportional to 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 - gradient of the line .. Eg. - in a graph: y is ln(rate) and x is 1/T - gradient = Ea / R - Ea = gradient x R (rearranged) » gradient is always negative diagram 7 for example

• A mechanism is a series of steps through which reaction progresses, • often forming intermediate compounds. -> If all steps added together theyll add up to overall equation for reaction • Each step can have a diff rate of reaction. • slowest step will control overall rate of reaction. • The slowest step is rate-determining step. • The molecularity (number of moles of each substance) of molecules • in the slowest step will be same as order of reaction for each substance. -> e.g. 0 moles of A in slow step wd mean A is zero order. 1 mole of A in the slow step would mean A is first order diagram 8 lots of examples