## Molecularity

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Samantha Jimenez 4k
Posts: 29
Joined: Fri Sep 28, 2018 12:21 am

### Molecularity

In my notes, I have that the molecularity is based on the number of species in an elementary step but would it be different if the order of those individual species was something other than 1?

VivianaHF2L
Posts: 29
Joined: Fri Sep 28, 2018 12:20 am

### Re: Molecularity

The molecularity would be the same as the overall order.

Samantha Jimenez 4k
Posts: 29
Joined: Fri Sep 28, 2018 12:21 am

### Re: Molecularity

could you clarify what you mean by overall order? so can we avoid the elementary steps and just look at the overall order if it is given?

VivianaHF2L
Posts: 29
Joined: Fri Sep 28, 2018 12:20 am

### Re: Molecularity

You would still have to find the overall order but they are based on the orders taken from each elementary step.

Samantha Jimenez 4k
Posts: 29
Joined: Fri Sep 28, 2018 12:21 am

### Re: Molecularity

Could you give an example as to how we would use the orders from the elementary steps to get the molecularity?

VivianaHF2L
Posts: 29
Joined: Fri Sep 28, 2018 12:20 am

### Re: Molecularity

For example on problem 49 on edition 6, each step has it's own molecularity and order. Step 1 is just taking the reactants HBr and NO2 and multiplying them. Each one has a coefficient of 1 and so individually they have an order of one. However, when added k[HBr][NO2] the order becomes two. This influences the molecularity which is bi-molecular.