1. Zero, First & Second Rate orders
Chapter 14 Part III

2. Integrated Rate Law aA  Products Rate= -∆[A]/∆t= k[A]m
For any reaction:
Every reaction has a rate law of this form.
We will develop the integrated rate law for
n=1: first order
n=2: second order
n=0 : zero order
aA  Products
Rate= -∆[A]/∆t= k[A]m

3. First Order Rate Laws 2N2O5(soln) -> 4NO2(soln) + O2(g)
Rate =(- ∆[N2O5]/∆t) = k[N2O5]
As it is first order, this means that if the [N2O5] doubles the production of also NO2 & O2 doubles.
By integrating the formula (calculus)
ln [N2O5] = -kt + [N2O5]t=0
This is still the integrated rate law with [Reactant] as a function of time

4. 1st order integrated rate law: Rate as function of time
For any reaction aA-> products
Rate =(- ∆[A]/∆t) = k[A]
ln [A] = -kt + [A]t=0
The equation depends on time. If [A] initial is known and k is known, the [A] may be calculated at t.

5. ln [A] = -kt + [A]t=0 The equation is in the form y = mx +b.
A plot of x (time) versus y (ln[A]) will yield a straight line.
Therefore another way to determine rate order, if the plot of ln[A] versus time gives a straight line, then it is first order. If it is not a straight line, it is not first order.
This rate law may also be expressed:
ln ([A]t=0/[A])=kt

6. 2N2O5(g) -> 4NO2(g) + O2(g)
The decomposition of this gas was studied at a constant volume to collect these results.
Verify the rate order and find the rate constant for rate =(- ∆[N2O5]/∆t)

7. First Order

8. Calculate k Slope = ∆y/∆x = ∆(ln[N2O5])/∆t Using first and last points
Slope = (-2.303)/(400s-0s)
= /400s
k = x 10-3 s-1

9. Half life of a first order reaction
The time required to reach one half the original concentration.
Designated by t 1/2
We can see the half life from the data is 100 sec.
Note is always takes 100 seconds.

10. Butadiene forms its dimer 2C4H6 (g)  C8H12 (g)
Second Order Rate Laws
Butadiene forms its dimer
2C4H6 (g)  C8H12 (g)

11. aA-> products If the reaction is first order: ln([A]0/[A])=kt
By definition, when t= t 1/2 then,
[A] = [A]0/2
ln([A]0/[A]0/2) = kt 1/2
ln(2)=kt 1/2
t 1/2= 0.693/k

12. Second Order Rate Laws Butadiene forms its dimer
2C4H6 (g) - > C8H12 (g)

13. 2nd Order Data [C4H6] (mol/L) Time (sec) 0.01000 0.00625 1000 0.00476
1000
1800
2800
3600
4400
5200
6200

15. Time (sec)
[C4H6] (mol/L)
ln[C4H6]
1/[C4H6]
100.0
1000
160.0
1800
210.1
2800
270.3
3600
319.5
4400
370.4
5200
414.9
6200
480.8

16. 2nd order reaction using integrated law for first order

17. 2nd order reaction and 2nd order integrated rate law