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
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)}  +  CuSO_{4(aq)}  à  Fe_{2}SO_{4(aq)}  +  Cu_{(s)}  
Blue solution  Colourless solution 
SO_{2}Cl_{2(g)}  à  SO_{2(g)}  +  Cl_{2(g)} 
Red tangent is at t = 0 (initial rate): rate = 0.50  0.00 = 1.5 x 10^{4} mol dm^{3} s^{1} 3300  0 

Green tangent is at t = 3000: rate = 0.38  0.14 = 6.0 x 10^{5} mol dm^{3} s^{1} 4000  0 
Question 1 P 115
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} 
To look at how the order effects the rate we'll keep the numbers simple.
Lets make:
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} 
Remember  a reaction usually has more than one reactant:
A  +  B  +  C  à  Products 
Rate  a 
[A] ^{0} 
Rate  a 
[B] ^{ 1} 
Rate  a 
[C] ^{ 2} 
Combining these 3 give:
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  = 

k  = 

k  = 




mol dm^{3} 

mol dm^{3}  
k  =  s^{1}  k  =  s^{1}  k  =  s^{1}  
mol dm^{3}  mol dm^{3}  
k  =  dm^{3} mol^{1} s^{1}  k  =  dm^{3} mol^{1} s^{1} 
Questions 1  3 P117
and / or carry out method 2 at home
Concentration  time graphs:
Half  life:
Concentration  time graph for a first order reaction:
2N_{2}O_{(g)}  à  2N_{2(g)}  +  O_{2(g)} 
Other concentration  time graphs:
Zero order  First order  Second order 



Qu 1  2 P 119
Qu 1 P159
Orders from rate  concentration graphs
Zero order  First order  Second order 
Horizontal straight line 
Proportional straight line 
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:
Na_{2}S_{2}O_{3(aq)}  +  2HCl_{(aq)}  à  2NaCl_{(aq)}  +  S_{(s)}  +  SO_{2(aq)}  +  H_{2}O_{(l)} 
Na_{2}S_{2}O_{3} 
HCl 

Proportional straight line 1st order wrt [Na_{2}S_{2}O_{3}] 
Horizontal straight line zero order wrt [HCl] 
Rate  =  k [Na_{2}S_{2}O_{3}]^{1} [HCl]^{0} 
Rate  =  k [Na_{2}S_{2}O_{3}] 
Questions 1,2 P 121
Initial rates and the rate constant
Determination of orders by inspection:
Example:
Consider the reaction:
2NO_{(g)} 
+  O_{2(g)}  à  4NO_{2} 
Experiment  Initial concentration [NO] (mol dm^{3})  Initial concentration [O_{2}] (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 O_{2}:
Rate  =  k [NO]^{2 } [O_{2}]^{1} 
Rate  =  k [NO]^{2 } [O_{2}]^{1}  Rate  =  k [NO]^{2 } [O_{2}]^{1}  
k  =  Rate  k  =  Rate  
[NO]^{2 } [O_{2}]^{1}  [NO]^{2 } [O_{2}]^{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})^{2} 


k  =  s^{1}  
(mol dm^{3})^{2}  
k  =  dm^{6} mol^{2} s^{1} 
The rate constant, k:
Concentration  
rate constant 
The effect of temperature on the rate constant
Questions 12 P123
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 
NO_{2(g)}  +  CO_{(g)}  à  NO_{(g)}  +  CO_{2(g)} 
Second order wrt NO_{2(g)}  
Zero order wrt CO_{(g)} 
First step 
NO_{2}  +  NO_{2}  à  Slow  Rate Determining Step 
First step  NO_{2}  +  NO_{2}  à  Slow  Rate Determining Step  
Second step  +  CO  à  +  CO_{2}  Fast  
Overall  NO_{2}  +  CO  à  NO  +  CO_{2} 
First step  NO_{2}  +  NO_{2}  à  Slow  Rate Determining Step  
Second step  +  CO  à  NO_{2}  +  CO_{2}  Fast  
Overall  NO_{2}  +  CO  à  NO  +  CO_{2} 
First step  NO_{2}  +  NO_{2}  à  NO_{3}  +  Slow  Rate Determining Step  
Second step  NO_{3}  +  CO  à  NO_{2}  +  CO_{2}  Fast 
Overall  NO_{2}  +  CO  à  NO  +  CO_{2} 
First step  NO_{2}  +  NO_{2}  à  NO_{3}  +  NO  Slow  Rate Determining Step 
Second step  NO_{3}  +  CO  à  NO_{2}  +  CO_{2}  Fast 
Overall  NO_{2}  +  CO  à  NO  +  CO_{2} 
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).