1.09 Rate equations

• what is the generalised rate equation?

  r= k [A]m[B]n

  K= rate constant

  [A]= concentration of A (unit moldm-3)

  m, n are called reaction orders
• what are orders usually?

  orders are usually intergers 0,1,2
• what does 0 mean?

  0 means the reaction is zero order with respect to that reactant
• what does 1 mean?

  first order
• what does 2 mean?

  second order
• how are the orders worked out?

  the orders arent the same as the stoichiometric coefficients in the balanced equation- worked out experimentally
• how is the total order for a reaction worked out?

  the total order for a reaction is worked out by adding all the individual orders together (m+n)
• what is the happens for zero order reaction?

  for 0 order: the concentration of A has no effect on the rate equation

  r= k [A]0

  
• what happens for first order reaction?

  for 1 order: the rate of reaction is directly proportional to the concentration of A

  r= k [A]1

  
• what happens for second order?


  the rate of reaction is proportional to the concentration of A squared

  r= k[A]2

  
• what is 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 afixed temperature

  3- the value of K refers to a specific temperature and it increases if we increase temperature
• for 1st order overall reaction what is the unit for K?

  s-1
• for second order overall reaction what is the unit of K?

  mol-1dm3s-1
• for a third overall reaction what is the unit of K?

  mol-2dm6s-1
• 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?

  k= rate/ [A] [B]2

  k= s-1/moldm-3 moldm-3

  k= s-1/mol2dm-6

  unit of k= mol-2dm-6s-1
• what is a continuous rate method?

  when we follow one experiment over time recording the change in concentration we call it a continuous rate method

  the gradient represents the rate of reaction

  the reaction is fastest at the start where the gradient is the steepest

  
• when does the rate drop?

  the rate drops as the reactants start to get used up and their concentration drops

  the graph (continuous rate) will eventually become horizontal and the gradient becomes 0 which represents the reaction having stopped
• when will the measurement of change in volume of gas work?

  this works if there is a change in the number of moles of gas in a reaction

  using a gas syringe is a common way of following this
• what is a gas syringe work well for?

  gas syringe works well for measuring continuous rate but a typical gas syringe only measures 100ml of gas so you dont want a reaction to produce more than this volume

  quantities of reactants needs to be added carefully
• what is the initial rate?

  initial rate is the rate at the start of the reaction, where it is fastest

  it can then be calculated from the gradient of a continuous monitoring concentration vs time graph at time=0
• why is the measure of initial rate preferable?

  we know the concentrations at the start of the reaction
• what is the method for measuring the initial rate?

  - measure 50cm3 of the 1.0moldm-3 hydrochloric acid and add to conical flask

  - set up the gas syringe in the stand

  - weigh 0.20g of magnesium

  - add the magnesium ribbon to the conical flask, place the bung formly into the top of the flsk and start the timer

  record the volume of hydrogen gas collected every 15 seconds for 3 mins

  
• what is the equation for working out the gradient?

  change in y/change in x
• what happens in reactions where there are several reactants?

  in reactions where there are several reactants, if the concentration of one of the reactant is kept in a large excess then that reactant will appear not to affect the rate and will be pseudo-zero order

  this is because its concentration stays virtually constant and doesnt affect rate
• the higher the concentration/ temperature/ surface area.....

  the faster the rate (steeper the gradient)
• what happens if the magnesium or marble chips is in excess of the acid?

  if the magnesium or marble chips is in excess of the acid, then the final volume of gas produced will be proportional to the amount of moles of acid
• what will different volumes of the same initial concentration ave?

  different volumes of the same initial concentration will have the same initial rate (if other conditions are the same) but will end at different amounts
• how can the initial rate be calculated?

  the initial rate can be calculated from taking the gradient of a continuous monitoring concentration vs time graph at time= 0

  initial rate can also be calculated from clock reactions where the time taken to reach a fixed concentration is measured
• what is a common clock reaction?

  hydrogen peroxide reacts with iodide ions to form iodine

  the thiosulfate ion then immediately reacts with iodine formed in the second reaction:

  H2O2 (aq) + 2H+ + 2I- makes I2 (aq) + 2H2O (l)

  2S2O32- (aq) + I2 (aq) makes 2I- (aq) + S4O6 2- (aq)

  when the I2 produced has reacted with all of the limited amount of thiosulfate ions present, excess I2 remains in solution

  reaction with the starch then suddenly foms a dark blue-black colour

  a series of experuments is carried out, in which the concentration of iodide ions is varied, while keeping the concentration of all of the other reagents the same

  - in each experiment the time taken (t) for the reaction mixture to turn blue is measured
• in clock reactions, what are there?

  in clock reactions there are often two successive reactions

  the end point is achieved when one limited reactant runs out, resulting in a sudden colour change

  by repeating the experiment several times, varying the concentration of a reactant e.g. I-, (keeping the other reactants at constant concentration) you can determine the order of reaction with respect to that reactant

  the initial rate of reaction can be represented as (1/t)
• how to work out the orders 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 + nlog [Y]

  Y intercept= log k

  Y= mx+c

  a graph of log rate vs log [Y] will yield a straight line where the gradient is equal to the order n

  in this experiment high concentration with quick times will have the biggest percentage errors

  
• deduce the rate equation for the following equation

  A+B+2C makes D+ 2E

  using the initial rate data from a table

  in order to calculate the order for a paricular reactant it is easiest to compare two experiments where only the reactant is being changed

  if concentration is doubled and rates stays the same order=0

  if concentration is doubled and rate doubles order= 1

  if concentration 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. therfore the order with respect to reactant A is first order

  - for reactant B compae between experiement 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 second 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 0 order

  - the overall rate equation is r= k[A] [B]2

  the reaction is 3rd overall order and the unit of the rate constant= mol-2dm6s-1
• how to work out the orders when 2 reactant concentrations are changed simultaneously?

  in some equations it is possible to compare between 2 experiments where only one reactant has its initial concentration changed

  if, however, reactants are changed then the effect of both individual changes in concentration are multiplied together to give an overall change on rate
• what does this rate equation describe:

  r= k [A] [B]2

  if the [A] is x2 then the rate would be x2

  if [B] is x3 the rate woukd be x3^2= x9

  if these changes happened at the same time then the rate would be x2x9= x18
• how to calculate the value for K?

  rearrnge the rate equation to give k

  r=k [x] [Y]2

  k= r/[x] [y]2

  values forr , [x] and [y] given in the table

  k= 2.4 x 10-6/ 0.2 x 0.2^2

  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
• what happens for 0 order reactants?

  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

  the gradient is also the value of the rate constant
• what happens to the rate constant if temperature is increased?

  increasing the temperature increases the value of the rate constant k

  increasing temperature increases the rate constant k

  relationship is given by the arrhenius equation
• what is the arrhenius equation?

  k=Ae^(-Ea/RT)

  A= arrhenius constant

  R= gas constant

  EA= activation energy
• a reaction carried out at 35 degrees C has a value of k= 4.26 x 10-8 s-1

  the activation energy EA= 95.8 KJmol-1

  the gas constant R= 8.31 jk-1mol-1

  calculate a valye for Arrhenius constant, A, for the reaction using equation k= Ae ^-EA/RT


  k= Ae ^-EA/RT

  rearrange to A= K/e ^-EA/RT

  = 4.26 x 10-8/ e^-95800 /(8.31/308)

  = 4.26 x 10-8/ e^-37.4

  = 4.26 x 10-8/5.55 x 10-17

  = 7.67 x 10^8 s-1