Rate of reaction

  Collision theory
  Effect of temperature
  Effect of concentration
  Effect of pressure
  Activation energy
  Catalysis

Rate of reaction

                        The rate of a reaction is defined as the change in concentration of a reactant or product in a given time

  Rate = Change in concentration   Units: mole dm-3 = mol dm-3 s-1  
      Time     s      

Representing concentration

Measuring rates

  SLOWS DOWN As fewer collisions take place
  STOPS When one reactant has been used up
  y/x Concentration  /  time
  Gradient Rate

A)  Measuring the decrease in concentration of a reactant:

A)  Measuring the increase in concentration of a product:

Qu 1 - 3   P113

Measuring reaction rates

Obtaining data for a concentration - time graph:

Acid - Base reactions Gas production Visible changes
Titrations

Change in volume

(gas collection)

Precipitation (colorimeter)
pH meter Loss in mass Colour changes (colorimeter)

Using a colorimeter

  Fe(s) + CuSO4(aq) Fe2SO4(aq) + Cu(s)
      Blue solution   Colourless solution    

 

 

 

 

 

Concentration of copper sulphate solution

1M

0.8M

0.6M

0.4M

0.2M

Measuring the rate from a concentration - time graph

  SO2Cl2(g) SO2(g) + Cl2(g)

Red tangent is at t = 0 (initial rate):

rate = 0.50 - 0.00 =   1.5 x 10-4 mol dm3 s-1

              3300 - 0

Green tangent is at t = 3000:

rate = 0.38 - 0.14 =   6.0 x 10-5 mol dm3 s-1

              4000 - 0

Question 1  P 115

Orders and the rate equation

Orders:

  1)  No effect
  2)  Double the rate
  3)  Quadruple the rate
 

Rate is proportional to the concentration

  Rate a

[A] x

  x

is the order

The rate equation and the rate constant:

 

Rate is proportional to the concentration

  Rate = k

[A] x

 

k = 1

     
  [A] =

1

Doubling the concentration of A from 1M to 2M
  [A] =

2

Zero order - where x = 0:

  Rate = k

[A] x

 
  Rate = 1

[1] 0

Rate = 1
 

Now double the concentration of A

  Rate = 1 [2] 0

Rate = 1

Doubling the concentration has no effect on the rate

First order - where x = 1:

  Rate = k

[A] x

 
  Rate = 1

[1] 1

Rate = 1
 

Now double the concentration of A

  Rate = 1 [2] 1

Rate = 2

Doubling the concentration doubles the rate wrt [A]

Second order - where x = 2:

  Rate = k

[A] x

 
  Rate = 1

[1] 2

Rate = 1
 

Now double the concentration of A

  Rate = 1 [2] 2

Rate = 4

Doubling the concentration quadruples the rate wrt [A]

The rate equation and the rate constant:

 

Rate is proportional to the concentration

  Rate = k

[A] x

  A + B + C Products
  Rate a

[A] 0

  Rate a

[B] 1

  Rate a

[C] 2

  Rate a

[A] 0

[B] 1 [C] 2
  Rate = k

[A] 0

[B] 1 [C] 2
  Rate = k [B] 1 [C] 2

Overall order:

  Rate = k [B] 1 [C] 2

                            Orders of reaction and the rate equation can only be determined experimentally.

Units of rate constants:

  Rate = k  [A] 1   Rate = k  [A] 2     Rate = k  [A]  [B]  
                           
  k = Rate   k = Rate     k = Rate  
      [A] 1       [A] 2         [A]  [B]  
                           
  k = mol dm-3 s-1   k = mol dm-3 s-1   k = mol dm-3 s-1
      mol dm-3       mol dm-3 mol dm-3       mol dm-3 mol dm-3
                           
  k = mol dm-3 s-1   k = mol dm-3 s-1   k = mol dm-3 s-1
      mol dm-3       mol dm-3 mol dm-3       mol dm-3 mol dm-3
                           
  k = s-1   k = s-1     k = s-1  
              mol dm-3         mol dm-3  
                           
          k = dm3 mol-1 s-1     k = dm3 mol-1 s-1  

Questions 1 - 3  P117

PracticalStudents to follow method 1

Video and / or carry out method 2 at home

Half - lives

Concentration - time graphs:

Half - life:

Concentration - time graph for a first order reaction:

  2N2O(g) 2N2(g) + O2(g)    

Other concentration - time graphs:

Zero order First order Second order
  • Concentration decreases at a constant rate - Straight line
  • Remember, this reactant has no effect on the rate but will get used up in the reaction.
  • Concentration halves at equal time intervals - Constant half - lives.
  • As the concentration halves the rate also halves.
  • Concentration decreases rapidly - Half life increases.
  • As the concentration halves, the rate reduces by 1/4 so half lives increase.

Qu 1 - 2  P 119

Qu 1  P159

Orders from rate - concentration graphs

Zero order First order Second order
  • Remember, this reactant has no effect on the rate.
  • The rate continues at a steady pace.

 

 

Horizontal straight line

  • As the concentration doubles the rate also doubles, as the concentration triples, the rate triples.
  • The rate and concentration are in direct proportion to each other.

 

Proportional straight line

  • As the concentration doubles the rate also quadruples, as the concentration triples, the rate increases by 9.
  • The rate increases by the square of the concentration.
  • What if rate - concentration2 was plotted?

Proportional curve

Initial rates:

A)  From concentration - time graphs:

B)  From clock reactions:

  appearance of a precipitate
  disappearance of a solid
  a change in colour

Example - Sodium thiosulphate and hydrochloric acid:

Video

Na2S2O3(aq) + 2HCl(aq) 2NaCl(aq) + S(s) + SO2(aq) + H2O(l)
 

Na2S2O3

HCl

 
 
  • As the concentration doubles the rate also doubles, as the concentration triples, the rate triples.
  • The rate and concentration are in direct proportion to each other.

Proportional straight line

1st order wrt [Na2S2O3]

  • Remember, this reactant has no effect on the rate.
  • The rate continues at a steady pace.

 

Horizontal straight line

zero order wrt [HCl]

       
  Rate = k  [Na2S2O3]1  [HCl]0
  Rate = k  [Na2S2O3]

Questions 1,2  P 121

Initial rates and the rate constant

Determination of orders by inspection:

Example:

2NO(g)

+ O2(g) 4NO2
Experiment Initial concentration [NO] (mol dm-3) Initial concentration [O2] (mol dm-3) Initial rate (mol dm-3 s-1)
1 0.00100 0.00100 1.82  x  10-6
2 0.00100 0.00300 5.46  x  10-6
3 0.00200 0.00100 7.28  x  10-6

Order for NO:

Order for O2:

Rate equation:

 

  Rate = k  [NO]2      [O2]1

Rate constant:

 

  Rate = k  [NO] [O2]1   Rate = k  [NO] [O2]1  
                 
  k = Rate   k = Rate  
      [NO] [O2]1       [NO] [O2]1  
                 
  k = 5.46  x  10-6   k = mol dm-3 s-1
     
(0.00100)2  x  (0.00300)
      (mol dm-3)2 mol dm-3
                 
  k = 1820   k = mol dm-3 s-1
              (mol dm-3)2 mol dm-3
                 
          k = s-1  
              (mol dm-3)2  
                 
          k = dm6 mol-2 s-1  

 

Practical

The rate constant, k:

  Concentration
  rate constant

The effect of temperature on the rate constant

Practical

r     =          k [A]x [B]y

Questions 1-2  P123

Rate - determining step

Analogy - Making ham sandwiches

 

Step 1   Step 2   Step 3
Get out 2 slices of bread   Butter the bread add the ham between the slices and assemble the sandwich   Put the sandwich in a sandwich bag

Predicting reaction mechanisms from rate equations:

NO2(g) + CO(g) NO(g) + CO2(g)
  Second order wrt NO2(g)
  Zero order wrt CO(g)

r     =          k [NO2]2

             

First step

NO2 + NO2       Slow - Rate Determining Step
First step NO2 + NO2       Slow - Rate Determining Step
Second step   + CO   + CO2 Fast
Overall NO2 + CO NO + CO2  
First step NO2 + NO2       Slow - Rate Determining Step
Second step   + CO NO2 + CO2 Fast
Overall NO2 + CO NO + CO2  
First step NO2 + NO2 NO3 +   Slow - Rate Determining Step
Second step NO3 + CO NO2 + CO2 Fast
Overall NO2 + CO NO + CO2  
First step NO2 + NO2 NO3 + NO Slow - Rate Determining Step
Second step NO3 + CO NO2 + CO2 Fast
Overall NO2 + CO NO + CO2  

Summary:

If a reactant appears in the rate equation, then that reactant takes part in the slow step of the reaction (RDS).  If it does not appear in the rate equation then the reactant does not participate in the slow step (RDS).

Qu 1 - 2  P 125  /  Qu 1 - 3  P 159  /  Qu 1 - 2  P161