second orderreactiongraph
TRANSCRIPT
Rate Law: Second-Order Reaction
2A → BRate Law:
Rate = k x [A]2
B appears at same rate that 2A's disappear
Time (s) [A] [B] K 0.05 M/s
0 1.50 0.00
ReactionBegins:InitialConcentrationOf A is 1.50 M
0 5 10 15 20 25 30 35 400
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
Co
nce
ntr
atio
n (
M)
In first second,0.11 molarReduction in [A]And 0.06 molarIncrease in [B]
Time [A] [B]
0 1.50 0.00
1 1.39 0.06
0 5 10 15 20 25 30 35 400
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)
In second second, rate isSomewhatLower 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 35 400
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)
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 0.570 5 10 15 20 25 30 35 40
00.20.40.60.8
11.21.41.6
Second Order Reaction
Rate = k[A]²
[A]
[B]
time (s)
[A] (
M)
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 + 0.229R² = 0.958
Second order reaction
Ln[A] v time
ln[A]
Linear (ln[A])
time (s)
Ln
[A]
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 40 450
0.5
1
1.5
2
2.5
3
f(x) = 0.051x + 0.678R² = 1.000
Second Order Reaction
Reciprocal of [A] v time
1/[A]
Linear (1/[A])
time (s)
1/[A
] (M
¯¹)
K = 0.05
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 35 400
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)
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 35 400
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)
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 35 400
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)