1. Q-Kinetics Prem Sattsangi Copyright 2009

2. #2-Chemical Kinetics: (p.574) Study of the Rate of a Chemical Process
Reactions happen because,
molecules collide with each other.
Reactions proceed faster,
if the following are increased:
1. Reactant concentration. [Increased concentration,
more collisions per second.]
2. Temperature. [Faster moving molecule,
more collisions per sec.]

3. #3-Average rate (M/s) (p. 576)
Time (s) [C4H9Cl] (M)
Calculate average rate between 100 sec -200 sec
Av. Rate =- {[0.0671]final – [0.0820M]init.}
200 secfinal – 100 secinitial
= 1.49 x 10-4 M/s

4. #4-Rate and Stoichiometry (p. 579)
Reactants  Products
N2(g) + 3H2(g)  2NH3(g)
As reaction proceeds, concentration of:
Reactants decrease (minus sign), -D[R]/Dt
Products increase, D[P]/Dt
To equate the rates, divide by the coefficient, in the balanced equation.

5. #5-Study of Reaction Rate as a function of Concentration at 25oC (p
#5-Study of Reaction Rate as a function of Concentration at 25oC (p. 580) [one reagent a a time]
Reactants Products
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
No. [NH4+] [NO2-] Rate
Constant
x 10-7
x 10-7
Doubling the [NH4+] doubles the rate.
x 10-7
x 10-7
Doubling the [NO2-] doubles the rate.

6. #6-Rate Law Equation (p. 581)
Rate = k [NH4+]1 x [NO2-]1
“k” = Rate constant,
(Value increases at higher temperature.)
Exponents “1” are the order of reaction.
This reaction is First order in NH4+ and
First order in NO2-
Overall reaction order = = 2

7. #7-Figuring out Order of a reaction (p
#7-Figuring out Order of a reaction (p.581) [Doubling Reactant Concentration]
[R]r (order “r”) is related to Rate
Doubling the concentration, “[2R]r”
and its effect on Rate:
Rate Equation “r” Order
UNCHANGED [2R]0 = 1 x Rate “0”
2 x [2R]1 = 2 x Rate “1”
4 x [2R]2 = 4 x Rate “2”
8 x [2R]3 = 8 x Rate “3”

8. #8-Figuring out Order of a reaction [Tripling Reactant Concentration]
[R]r (order “r”) is related to Rate
Tripling the concentration, “[3R]r”
and its effect on Rate:
Rate Equation “r” Order
UNCHANGED [3R]0 = 1 x Rate “0”
3 x [3R]1 = 3 x Rate “1”
9 x [3R]2 = 9 x Rate “2”
27 x [3R]3 = 27 x Rate “3”

9. #9-Orders (0, 1, 2, or 3) Trial [A](M) [B](M) Rate (M/s)
x 10-6
x 10-6
x 10-5
In 1 and 2 [A] is constant, [B] is doubled
Rate remained same, Order in B = 0
In 1 and 3 [B] is constant, [A] is tripled
Rate changed to 9 x or 32.
Order in A = 2

10. #10-Calculating the value of “k” (p. 581) at 25oC
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
No. [NH4+] [NO2-] Rate
x 10-7
Rate = k [NH4+]1 x [NO2-]1
k = __Rate Ms-1___ = __ 5.4 x 10-7 Ms-1___
[NH4+]1 x [NO2-]1 [ M ]1 x [0.200 M]1
= 2.7 x 10-4 M-1s-1

11. #11 Units of “k” [M] in Rate (p. 581) cancels M in the denominator.
Rate law k = Unit
Rate = k[A]1 x [B] __Rate Ms-1___ M-1s-1
[A M]1 x [B M]1
Rate = k[A]1 x [B] __Rate Ms-1___ M-2s-1
[A M]1 x [B M]2
Rate = k[A]2 x [B] __Rate Ms-1___ M-2s-1
[A M]2 x [B M]1
Rate = k[A]2 x [B] __Rate Ms-1___ M-3s-1
[A M]2 x [B M]2

12. #12-Energy of activation
Data: Set Temp(oC) time(s)
D (T’) 23 (296K) (Dt’) 350
G (T) 13 (286K) (Dt) 650 (Average)
Temperatures must be in K
R= 8.31J/K.mol
Ea = __________________ = 43.6 kJ/mol