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### Determining Order

Posted: Mon Mar 09, 2020 11:21 pm
If we are not given a graph, how do you determine the order of a reaction? Do we have to write out all the reaction mechanism steps?

### Re: Determining Order

Posted: Tue Mar 10, 2020 1:47 am
It depends on how the problem is given to you. If there is the chart, with different concentrations and rates you can determine the order with respect to a reactant. Or if the rate constant is given in units you can determine the overall reaction order.

### Re: Determining Order

Posted: Tue Mar 10, 2020 10:36 am
The rate constant and elementary rxns can determine rate order.

### Re: Determining Order

Posted: Wed Mar 11, 2020 3:32 pm
You can also look at units of the rate constant if it is given and be able to tell what the order is.

### Re: Determining Order

Posted: Wed Mar 11, 2020 4:24 pm
Martina wrote:You can also look at units of the rate constant if it is given and be able to tell what the order is.

Yes! for zero order reactions the units are M/s. First order is s^-1, and second order is 1/M.s.

### Re: Determining Order

Posted: Wed Mar 11, 2020 4:33 pm
Maika Ngoie 1B wrote:
Martina wrote:You can also look at units of the rate constant if it is given and be able to tell what the order is.

Yes! for zero order reactions the units are M/s. First order is s^-1, and second order is 1/M.s.

The way that I remember this is that rate is in the units M/s, then when you solve for k you have to divide by the concentration of the reactant. 0 order is [A]^0 which is 1, so you do not divide by anything and K will still have the units of M/s. 1st order has [A]^1 with units of M, so when you divide the units cancel and you are left with 1/s. 2nd order has [A]^2 with units of M^2. so when you divide by M^2 the result is 1/mS

### Re: Determining Order

Posted: Wed Mar 11, 2020 4:46 pm
Does the larger the order mean that the more collisions will occur?

### Re: Determining Order

Posted: Fri Mar 13, 2020 5:29 pm
You can look at units of rate constant for the orders of each reactant and add them together.