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1. CHEMICAL KINETICS
FIRST ORDFER RATE LAW

2. First order rate law
For a reaction: A = B + C
Rate = k [A]
-da/dt = k [A]
The rate law can be integrated to give the concentration of a component following a given length of time
interval.
Separating the variables we obtain:
dA/[A] =K dt
Integrating the equation
∫ dA/[A] = -K dt
ln[A] = -kt + C
We can evaluate the constant C by substituting into the equation the so called boundary conditions:
At time t = 0 , [A] = [A0]
ln[A0] = -k x 0 + C

3. ln[A0] = C
Substituting for the value of C back in our equation we obtain:
ln[At] = -kt + ln[A0]
Or ln[At ] – ln [A0] = -kt
ln {[At]/[A0]} = -kt
[At]/[A0] = e-kt
[At] = [A0] e-kt
[AT] is the concentration of species A at any time t, A0 is the initial concentration (t = 0) of A
The time at which the concentration falls to one half its initial value is called the HALF-LIFE of the reaction.
For a first order reaction, the half -life can be worked out as follows;
When [At]/[A0] = ½ = e-k t1/2
Taking logarithms on both sides, ln(1/2) = -k t1/2 where t1/2 is the half-life.
-0.693 = -k t1/2
t1/2 = 0.693/ k
The half-life for a first order reaction is independent of the concentration, it only depends on the rate
constant k.

4. We should also note that a plot of logarithm of [At]/[A0] against time t, gives a straight line whose slope
is the value of the rate constant k. But a plot of [At] against time is shown in the figure below.

5. The decomposition of dinitrogen pentoxide is first order. The rate constant is 4.80 X 10-4 s-1
N2O5(g) = 4 NO2(g) + O2(g)
Some of the experimental data is as follows;
[N2O5] M Time ( s)
0.0165 0
0.0083 1500
0.0041 3000
0.0020 4500

6. Decomposition of nitrogen pentoxide against time. The red line is the tangent to the
initial part of the curve to give the initial slope.

7. Example: The decomposition of nitrogen pentoxide to nitrogen dioxide and oxygen is first order with a rate
constant of 4.80 x 10-4 at 45 0C
(a) If the initial concentration is 1.65 x 10-2 moles litre-1 , what is the concentration after 825 s ?
(b) How long would it take for the concentration of N2O5 to decrease to 1.00 x 10-2 moles litre-1 from its initial
value given in (a)?
(a) [A t]/[A0] = e-kt
[At=825 = [A0] e-4.80x10(-4) x825
[A825] = 1.65 X 10-2 X e-0.396
= 1.65 x 10-2 x 0.673
= 0.0111 moles litre-1
(b) ln(1.00 x 10-2/ 1.65 x 10-2) = -kt
-0.501 = -4.80 x 10-4 x t
t = 1043 s
t = 17.4 mins

8. The second order rate law:
Rate = k [A]2
-d[A]/dt = k [A]2
Separation of variables as before leads to
-d[A]/ [A ]2 = k dt
1/ [A] = k t + C
We evaluate the constant C as before: at t = 0; [A] = [A0]
1/[A0] = k x 0 + C
1/[A0] = C
Substituting in our differentiated equation
1/[A] = k t + 1/ [A0]
We could also find the expression for the half-life for a second order reaction as follows
When [A] = ½ [A0]

9. So we have from our previous equation:
1/{(1/2)[A0]} - 1/[A0] = k t1/2
1/[A0] = k t1/2
t1/2 = 1/ k [A0]
The half-life (t1/2) for a second order reaction depends on the initial concentration A0

10. First order kinetics diagrams
RELATIONSHIP BETWEEN REACTANTS AND
PRODUCTS AS FUNCTIONS OF TIME
FIRST ORDER RATE LAW

11. Second Order
Kinetics
The decomposition of nitrogen dioxide
is an example to illustrate a second
order rate law.