# Rate of Reaction

Watch
Announcements
#1
Hi, I'm not sure if I'm just being silly here or something but..

For the reaction 2A+B --> C+D

We say, increasing the concentration of A/B increases the number of successful collisions per unit time hence rate of reaction increases.

Similarly, increasing the pressure increases the number of molecules per unit volume, hence there are more successful collisions per unit time hence rate of reaction increases.

However, my issue lies when we have a reaction like the following:

A-->C+D
(decomposition of A)

How can we still use the argument there are more collisions per unit time by increasing concentration, when no collisions need to occur.

If no collisions need to occur, why does increasing the concentration/pressure effect the rate of reaction at all?

Thanks for your help, this is bugging me.
0
6 years ago
#2
(Original post by Hilton184)
Hi, I'm not sure if I'm just being silly here or something but..

For the reaction 2A+B --> C+D

We say, increasing the concentration of A/B increases the number of successful collisions per unit time hence rate of reaction increases.

Similarly, increasing the pressure increases the number of molecules per unit volume, hence there are more successful collisions per unit time hence rate of reaction increases.

However, my issue lies when we have a reaction like the following:

A-->C+D
(decomposition of A)

How can we still use the argument there are more collisions per unit time by increasing concentration, when no collisions need to occur.

If no collisions need to occur, why does increasing the concentration/pressure effect the rate of reaction at all?

Thanks for your help, this is bugging me.
Energy is needed for the decomposition i.e. to break apart 'A'. This energy comes from collisions.
0
6 years ago
#3
(Original post by Hilton184)
Hi, I'm not sure if I'm just being silly here or something but..

For the reaction 2A+B --> C+D

We say, increasing the concentration of A/B increases the number of successful collisions per unit time hence rate of reaction increases.

Similarly, increasing the pressure increases the number of molecules per unit volume, hence there are more successful collisions per unit time hence rate of reaction increases.

However, my issue lies when we have a reaction like the following:

A-->C+D
(decomposition of A)

How can we still use the argument there are more collisions per unit time by increasing concentration, when no collisions need to occur.

If no collisions need to occur, why does increasing the concentration/pressure effect the rate of reaction at all?

Thanks for your help, this is bugging me.
It is still possible for A to collide with A. And the Maxwell-Boltzman distribution still applies.
0
#4
(Original post by Protoxylic)
It is still possible for A to collide with A. And the Maxwell-Boltzman distribution still applies.
So for a reaction to occur, a molecule of A must still collide with another molecule of A for a reaction to occur- A wont just decompose on its own providing it had enough energy to do so?

Posted from TSR Mobile
0
6 years ago
#5
(Original post by Hilton184)
So for a reaction to occur, a molecule of A must still collide with another molecule of A for a reaction to occur- A wont just decompose on its own providing it had enough energy to do so?

8
Posted from TSR Mobile
I'm saying it is possible for two A molecules to collide and for that to provide energy, other ways of providing the AE are possible.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (394)
56.61%
I don't have everything I need (302)
43.39%