Lindemann Theory

Question 1

What is kobs?

Answer 1

This represents the observed or measured rate constant

Question 2

What does Lindemann theory explain?

Answer 2

The pressure dependence of gas phase reactions

Question 3

Why does this reaction not fit with understanding of the time?
CH3N2CH3(A) -> C2H6(P) + N2
The rate found is d[P]/dt= kobs[A]

Answer 3

This shows the decomposition of A as first order kinetics but the rate increased with temperature
This implies the molecule must overcome an energy barrier but first order kinetics appeared to preclude collision activation

Question 4

What reaction mechanism did Lindermann suggest for:
A -> P + M
(m is a gas)

Answer 4

A + M -> A* + M (k1/k-1)
A* -> P (k2)

A* is a collisionally excited species
M is a molecule that provides energy by colliding with A but does not take part in the reaction

Question 5

What is the rate of formation of P in the lindermann mehanism?

Answer 5

d[P]/ dt= k2[A]*

Question 6

How can we evaluate this ?

d[P]/ dt= k2[A]*

Answer 6

Find an expression for [A]
d[A]/dt= k1[A][M]- k-1[A][M] - k2[A]
Then apply the steady state approximation to find [A]- find the concentration of intermediates
d[A]/dt= k1[A][M]- k-1[A][M] - k2[A]=0
Therefore [A*] =k1[A][M]/ k-1[M]+k2

Question 7

How can you find an expression for kobs?

Answer 7

d[P]/ dt= k2[A]*
[A*] =k1[A][M]/ k-1[M]+k2

So you can say:
d[p]/ dt= k2 x k1[A][M]/k-1[M]+k2
The experiment gave d[P]/dt=kobs[A]
So this would suggest
kobs[A]= k2k1[A][M]/k-1[M]+k2
So kobs= k2k1[M]/k-1[M]+k2

Question 8

Outline how overall you get K obs from start to beginning

Answer 8

1) work out different equation with respect to P
2) find [A*] using steady state
3) put this back into differential p equation
4) compare this with experiment observation to find kobs

Question 9

What happens at high pressures?

Answer 9

At high pressures [M] will be high and so less A* will be made which will decrease the rate of reaction 2
K-1[M]» k2
You can use this to simplify the rate equation
Rate becomes independent of M

Question 10

How do you know if the reaction is dependent on pressure?

Answer 10

If the [M] term is involved in the rate equation

Question 11

What happens at low pressures?

Answer 11

At low pressure A* is unlikely to collide and deactivate and so it is more likely to go on and form P
K-1[M]&laquo_space;k2
Use this to simplify the rate equation
M controls kobs

Question 12

What is k infinity?

Answer 12

When [M] becomes sufficiently large, the resulting kobs reaches a maximum limiting value
It becomes independent of M and so increasing pressure more will not increase K
This is referred to as k inifinity