k units  [ENDORSED]

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Phi Phi Do 2E
Posts: 27
Joined: Fri Sep 29, 2017 7:05 am

k units

Postby Phi Phi Do 2E » Sun Mar 04, 2018 10:42 pm

Why is the constant K in different units depending on the rate order?

Jessica Patzlaff 1A
Posts: 28
Joined: Fri Sep 29, 2017 7:05 am

Re: k units

Postby Jessica Patzlaff 1A » Sun Mar 04, 2018 10:49 pm

There is a different equation for each rate order, and thus a different k unit to cancel out to get the final units.

Sophie Krylova 2J
Posts: 59
Joined: Fri Sep 29, 2017 7:06 am

Re: k units

Postby Sophie Krylova 2J » Sun Mar 04, 2018 10:51 pm

From the general formula R=k[X]^n you can see that the rate units are fixed and so are the concentration units. Therefore, k units will change depending on what power the concentration is raised to.

Jaewoo Jo 2L
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am

Re: k units

Postby Jaewoo Jo 2L » Sun Mar 04, 2018 10:57 pm

You need to find the K with the right units so that it cancels with the rate order equation.

604807557
Posts: 51
Joined: Wed Nov 16, 2016 3:02 am

Re: k units

Postby 604807557 » Sun Mar 04, 2018 10:58 pm

It’s different units depending on the rate order because each order has different powers the concentration is raised to so using the rate law you get different units for k in each rate order.

Wenjie Dong 2E
Posts: 53
Joined: Fri Jun 23, 2017 11:40 am

Re: k units

Postby Wenjie Dong 2E » Sun Mar 04, 2018 11:03 pm

It's because the unit of the reaction rate is always the same.

Jessica_Singh_1J
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

Re: k units

Postby Jessica_Singh_1J » Mon Mar 05, 2018 1:25 pm

As a reminder:

the units for k in a 0 order reaction are: mol/L*s , for a 1st order reaction are: 1/s, for a 2nd order reaction are: L/mol*s.

Ryan Sydney Beyer 2B
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Re: k units  [ENDORSED]

Postby Ryan Sydney Beyer 2B » Mon Mar 05, 2018 1:51 pm

The units for k depend on what order the reaction is on. This is because the concentrations - which will be to the zero, first, or second order for this. So mol / L will be raised to some exponent. We know that the rate of the reaction is going to have units of mol / L * s, so we just have to make sure that the units of k made the right side of the equation having these units.


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