The document discusses the kinetics of a second-order reaction with the rate law expressed as rate = k [a]², where 'a' is the reactant and 'b' is the product. It provides concentration data over time, demonstrating how concentrations of 'a' decrease while 'b' increases, and highlights the effects of different reaction rate constants on reaction speed. It also details the graphical analysis, showing that the reciprocal of concentration vs time produces a linear relationship, reinforcing the second-order kinetics model.

1. Rate Law: Second-Order Reaction
2A → B
Rate Law:
Rate = k x [A]2
B appears at same rate that 2A's
disappear

2. Time (s) [A] [B] K 0.05 M/s
0 1.50 0.00
Reaction
Begins:
Initial
Concentration
Of A is 1.50 M
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Concentration v Time
A rearranging to B
[A]
[B]
time
Concentration(M)

3. In first second,
0.11 molar
Reduction in [A]
And 0.06 molar
Increase in [B]
Time [A] [B]
0 1.50 0.00
1 1.39 0.06
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Second Order Reaction
Rate = k [A]²
[A]
[B]
time (s)
[A](M)

4. In second
second, rate is
Somewhat
Lower because
[A] is lower
Time [A] [B]
0 1.50 0.00
1 1.39 0.06
2 1.29 0.10
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Second Order Reaction
Rate = k[A]²
time (s)
[A](M)

5. After 40 seconds it is easy to
see the steep decay and
product curves.
Time [A] [B]
0 1.50 0.00
5 1.07 0.21
10 0.84 0.33
15 0.69 0.41
20 0.58 0.46
25 0.51 0.50
30 0.45 0.53
35 0.40 0.55
40 0.37 0.57
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Second Order Reaction
Rate = k[A]²
[A]
[B]
time (s)
[A](M)

6. Over the same time period,
plotting the natural log of the
concentration v time does
NOT give a straight line.
Time ln[A]
0 0.405
5 0.068
10 -0.174
15 -0.371
20 -0.545
25 -0.673
30 -0.799
35 -0.916
40 -0.994
0 5 10 15 20 25 30 35 40 45
-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
f(x) = -0.034x + 0.229
R² = 0.958
Second order reaction
Ln[A] v time
ln[A]
Linear (ln[A])
time (s)
Ln[A]

7. However, as suggested by
our analysis, the reciprocal
of [A] v. time is a linear
function, with k as the
coefficient.
Tim
e
1/[A]
0 0.667
5 0.935
10 1.190
15 1.449
20 1.724
25 1.961
30 2.222
35 2.500
40 2.703
0 5 10 15 20 25 30 35 40 45
0
0.5
1
1.5
2
2.5
3
f(x) = 0.051x + 0.678
R² = 1.000
Second Order Reaction
Reciprocal of [A] v time
1/[A]
Linear (1/[A])
time (s)
1/[A](M¯¹)
K = 0.05

8. This is the effect of changing
the rate constant to 0.01 M/s.
The reaction is much
slower.
Lower Rate Constant
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Second Order Reaction
Rate = k[A]²
[A]
[B]
time (s)
[A](M)

9. This is the effect of changing
the rate constant to 0.10 M/s.
The reaction is much faster.
Higher Rate Constant
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Second Order Reaction
Rate = k[A]²
[A]
[B]
time (s)
[A](M)

10. This is the effect of changing
the rate constant to 0.10 M/s.
The reaction is much faster.
Higher Rate Constant
0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Second Order Reaction
Rate = k[A]²
[A]
[B]
time (s)
[A](M)