[soupastupid]

I need some explaining for uni molecular reactions

lets use
[A] --> Product
but in order for A to become product, it has to go through a transition state A*

to get a A*
we need a binary collision between A and A
and even when they do collide, A* will most likely become deactivated
only rarely does A* become product.
Am i correct so far??

my question is
why is it that at high pressures (>1atm) A* is more likely to become deactivated than at low pressures?  Is it because A* is more likely to collide with another A?  Or would that give it more energy (A**)?  If A* did become deactivated by A, wouldn't the A have to be same A it had collided before to become A*? Because a different A would the same energy as the initial two A and therefore, the energy would just be transferred (A* + A --> A + A*).

Because I feel like the reasoning is backwards.
At low pressures, you're less likely to get A* because collisions less frequent but yet the probably of getting product is higher?

Im really confused.
or am i thinking too hard?

thanks
-Andrew

[tamim83]

why is it that at high pressures (>1atm) A* is more likely to become deactivated than at low pressures?  Is it because A* is more likely to collide with another A?  Or would that give it more energy (A**)?  If A* did become deactivated by A, wouldn't the A have to be same A it had collided before to become A*? Because a different A would the same energy as the initial two A and therefore, the energy would just be transferred (A* + A --> A + A*).

So, let's say that the activation reaction for A is:

A + A A* + A

The forward and reverse reactions have rate constants k₁ and k₂ and A* can then decompose with rate constant k₃:

A*   Products

At high pressures, you would have more "deactivating" collisions (k₂ will be large) and less "activated" molecules to decompose into the products.  

Does this help some?