1. Rate Equations and Order of Reactions14
Rate Equations and Order of Reactions
Increasing concentration of reactants can increase the rate of reaction.
Is there any mathematical relationship between rate of reaction and concentration of reactants?
YES!!!!!!

2. 2NO(g) + 2H2(g)  N2(g) + 2H2O (l)
For the reaction:
2NO(g) + 2H2(g)  N2(g) + 2H2O (l)
Initial concentration / mol dm-3
Initial rate / mol dm-3 s-1 NO H2
0.0250
0.100
2.4  10-6
0.050
1.2  10-6
0.0125
0.6  10-6
Rate a [H2(g)] when [NO(g)] is constant
Rate a [NO2(g)]2 when [H2(g)] is constant
Rate a [NO(g)]2 [H2(g)]
Rate = k[NO(g)]2 [H2(g)]

3. Rate Equations and Order of Reactions
14.1
Rate Equations and Order of Reactions

4. Rate Equation In general, for the reaction: mA + nB  products Rate =
14.1 Rate Equations and Order of Reactions (SB p.25 NB p.8)
Rate Equation
In general, for the reaction:
mA + nB  products
Rate =
= k [A]x [B]y m n
(a rate equation)
k = rate constant
x = order of reaction with respect to reactant A
y = order of reaction with respect to reactant B
Overall order of reaction = x + y

5. Rate Equation mA + nB  products Rate = k [A]x [B]y
14.1 Rate Equations and Order of Reactions (SB p.25)
Rate Equation
mA + nB  products
Rate = k [A]x [B]y
x and y are determined experimentally.
x and y may NOT equal to m and n.
x and y may not be whole numbers!!!
Rate of reaction is independent of products’ concentration

6. Order of Reactions Usually integers (0, 1, 2, … …) If x = 1
14.1 Rate Equations and Order of Reactions (SB p.26)
Order of Reactions
Do Class Examples on p. 8
Usually integers (0, 1, 2, … …)
If x = 1
 first order with respect to reactant A
If y = 2
 second order with respect to reactant B
The overall reaction is a third order reaction.

7. Rate Equation Rate = k [A]x [B]y
14.1 Rate Equations and Order of Reactions (SB p.25 NB p. 8)
Rate Equation
Rate = k [A]x [B]y
log rate = log k + x log[A] + y log[B]
If B is kept large excess, its change in concentration will be negligible when compared with A.
 log[B] ~ constant
 log rate = c’ + x log[A]
 when is log rate plotted against log[A]
 straight line with slope = x

8. Zeroth, First and Second Order Reactions
14.2
Zeroth, First and Second Order Reactions

9. For the following reaction: A  product
[A] mol dm-3
Time (s)
1.00
0.80 10
0.70 20
0.65 30
0.53 40 :
How can we find out the order of reaction with respects to [A]?

10. Zeroth Order Reactions
14.2 Zeroth, First and Second Order Reactions (SB p.27 NB p.10)
Zeroth Order Reactions
A  products Rate = k [A]0 = k
e.g.
CH3COCH3(aq) + I2(aq) + H+(aq)
 CH3COCH2I(aq) + I-(aq) + 2H+(aq)
Rate = k[CH3COCH3(aq)][H+(aq)][I2(aq)]0
The rate of reaction is independent of [I2]!!!
This means if we keep [CH3COCH3(aq)] and [H+(aq)] constant then measure the initial rate with various [I2(aq)], we will get the same initial rate.

11. Zeroth Order Reactions
14.2 Zeroth, First and Second Order Reactions (SB p.27)
Zeroth Order Reactions
[I2(aq)]

12. Zeroth Order Reactions WHY???
14.2 Zeroth, First and Second Order Reactions (SB p.27)
Zeroth Order Reactions
WHY???
The reaction is a multi-step reaction.
Only the slowest step affect the overall reaction rate. A B C D
Iodine is not involved in the slowest step of the reaction.

13. A Plot of [A] against time will give a straight line
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.37)
Zeroth Order Reaction
A Plot of [A] against time will give a straight line
A  product
Rate equation is:
[A] = -kt + [A]0

14. First Order Reaction e.g Decomposition of H2O2(aq) to H2O and O2
14.2 Zeroth, First and Second Order Reactions (SB p.28)
First Order Reaction
e.g Decomposition of H2O2(aq) to H2O and O2
2H2O2(aq)  2H2O(l) + O2(g)
Rate = k [H2O2(aq)]
Though the stoichiometric coefficient of H2O2(aq) is 2
Order of reaction with respect to H2O2(aq) is 1 (determined experimentally)

15. A Plot of ln [A] against time will give a straight line
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.37 NB p. 9))
First Order Reaction
A Plot of ln [A] against time will give a straight line
A  product
Rate equation is:
In A = -k1t + In [A]0

16. Half-life of First Order reaction
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.38)
Half-life of First Order reaction
The time taken for half of the reactant to be converted to the product is known as the half-life of the reaction.
After time t,
[A] = [A]0
t is called the half life of the reaction.

17. Half-life of First Order Reaction
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.38)
Half-life of First Order Reaction
ln[A] = -k1t + ln[A]0
k1t = ln[A]0 - ln[A]
If t = half-life (t0.5), then [A]0 = 2[A]:
k1t0.5 = ln 2
t0.5 = constant

18. Half-life of a First Order Reaction
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.39)
Half-life of a First Order Reaction
For a first order reaction:
Half-life is a constant
Do Class Examples 1, 3 on p.10

19. Second Order Reaction A  product Rate equation is:
14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40 NB p.11)
Second Order Reaction
A plot of 1/[A] vs time will give a straight line
A  product
Rate equation is:
Integrating the above rate equation, obtain:

20. Summary Plot a second graph of vs time
Do Q. 5 on p. 77
Do Q. 14 on p. 80
Q. 13 (more difficult)
Do Q. 3 on NB p.17 Q. 2 on NB p. 21
Summary
Steps to determine the order of reaction
Plot a concentration-time graph
 a straight line  zeroth order
 a curve with constant half life  first order
Confirmation: ln[A] vs time  straight line
Plot a second graph of vs time
 a straight line  second order

21. Determination of Simple Rate Equations by Graphical Method
14.4 – 14.5
Determination of Simple Rate Equations by Graphical Method
Notes p. 18

22. log (rate)-log (concentration) Graph
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.32)
log (rate)-log (concentration) Graph
rate = k[A]n
log (rate) = n log [A] + log k
Plotting log (rate) against log [A], a straight line is obtained for any orders
Slope of straight line = Order of reaction (n)
y-intercept = log k
Do Q. 1 on NB p. 16

23. We need to plot two graphs:
(1) [A] vs time to find the rate at different [A]
(by slope to the curve)
(2) log (rate) vs log[A] to find the order and k

24. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
Decide which curve in the following graph corresponds to
(i) a zeroth order reaction;
(ii) a first order reaction.
(i) (3)
(ii) (2)

25. Isolation technique When two or more substances react:
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.33)
Do Q. 1, 3 on p. 27
Do Q 13 on p.80
Isolation technique
When two or more substances react:
A + B  products
Order of reaction with respect to reactant A can be found by
keeping concentration of B constant (by using much excess B – isolation technique)
Rate = k[A]x[B]y  Rate = k’ [A]x

26. The END

27. Q.13 on p. 80
Products (not reactants)
t(s)
v(cm3) 5 33 10 56 15 73 25 92
3600
110
5400
ln (a)
vol. of N2 (a) in reactant/cm3
110 77 54 37 18