first order reaction graph kinetics
DESCRIPTION
Graphs and equations for a first order chemical reaction, showing the rate constant and slope of the natural log graphTRANSCRIPT
Rate Law: First-Order Reaction
A → BRate Law:
Rate = k x [A]B appears at same rate that A
disappears
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,0.08 molarReduction in [A]And 0.08 molarIncrease in [B]
Time [A] [B]
0 1.50 0.00
1 1.43 0.08
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
time (s)
Co
nce
ntr
atio
ns
(M)
In second second, rate isSomewhatLower because[A] is lower
Time [A] [B]
0 1.50 0.00
1 1.43 0.08
2 1.35 0.15
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
time (s)
Co
nce
ntr
atio
ns
[M]
After 40 seconds it is easy to see the logarithmic decay and increase of the reactant and product.
0 5 10 15 20 25 30 35 40 450
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Concentration v time
A rearranging to B
time (s)
Co
nce
ntr
atio
ns
(M)
Time [A] [B]
0 1.50 0.00
5 1.16 0.34
10 0.90 0.60
15 0.69 0.81
20 0.54 0.96
25 0.42 1.08
30 0.32 1.18
35 0.25 1.25
40 0.19 1.31
Over the same time period, plotting the natural log of the concentration v time gives a straight line with a negative slope.
0 5 10 15 20 25 30 35 40 45
-2
-1.5
-1
-0.5
0
0.5
1
Ln(Concentration) vs time
A rearranging to B
time (s)
Ln
[A]
Time Ln [A]
0 0.405
5 0.148
10 -0.105
15 -0.371
20 -0.616
25 -0.868
30 -1.139
35 -1.386
40 -1.661
You may recall that we set K for the rate law as 0.050 M/s. Let's check to see if the negative of the slope is that.
0 5 10 15 20 25 30 35 40 45
-2
-1.5
-1
-0.5
0
0.5
1
Ln(Concentration) vs time
A rearranging to B
time (s)
Ln
[A]
Time Ln [A]
0 0.405
5 0.148
10 -0.105
15 -0.371
20 -0.616
25 -0.868
30 -1.139
35 -1.386
40 -1.661
Setting our intervals to 5 s, we see that the slope of the Ln[A] v time line is, indeed, -0.05. So the slope is the negative of the rate constant
Time Ln [A]
0 0.405
5 0.148
10 -0.105
15 -0.371
20 -0.616
25 -0.868
30 -1.139
35 -1.386
40 -1.661
0 5 10 15 20 25 30 35 40 45
-2
-1.5
-1
-0.5
0
0.5
1
f(x) = − 0.0514450425993624 x + 0.407491336200507R² = 0.999905262507506
Natural Log Concentration v time
First Order Reaction
Ln [A]
Linear (Ln [A])
time (s)
Ln
[A]
This is the effect of changing the rate constant to 0.01 M/s. The reaction is much slower.
0 5 10 15 20 25 30 35 40 450
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Concentration v time
A rearranging to B
time (s)
Co
nce
ntr
atio
ns
(M)
Lower Rate Constant
This is the effect of changing the rate constant to 0.10 M/s. The reaction is much faster.
0 5 10 15 20 25 30 35 40 450
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Concentration v time
A rearranging to B
time (s)
Co
nce
ntr
atio
ns
(M)
Higher Rate Constant