units of k
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 50
- Joined: Sat Aug 17, 2019 12:16 am
units of k
I know the units of k is dependent on the order of the rate equation, but how do we use that knowledge to determine the units of k?
-
- Posts: 103
- Joined: Fri Aug 09, 2019 12:17 am
Re: units of k
Since the unit of rate is M/s, if the reaction is zero-order, the unit of k is the same as the unit of rate. If the reaction is first order, then the reactant will have the unit M and the unit for k would be 1/s since the unit of rate is always M/s. If the reaction is second order, the reactant will have the unit M^2, which means that the unit of k has to be 1/M*s since the unit of rate is always M/s.
-
- Posts: 111
- Joined: Wed Sep 18, 2019 12:19 am
Re: units of k
If the reaction is zero-order, K has units of M/s
If the reaction is first-order, K has units of 1/s
If the reaction is second-order, K has units of 1/M.s
If the reaction is first-order, K has units of 1/s
If the reaction is second-order, K has units of 1/M.s
Re: units of k
For first order, the units are just s^-1
For second order, the units are L x mol^-1 x s^-1
For third order, the units are L^2 x mol^-2 x s^-2
The pattern continues in this way.
For second order, the units are L x mol^-1 x s^-1
For third order, the units are L^2 x mol^-2 x s^-2
The pattern continues in this way.
Re: units of k
For the units of k, the zeroth order is M/s, the first order is 1/s, and the second order is 1/M*s. A good way to remember this is to divide by M when going from zeroth to first to second order.
zeroth order: M/s --> divide by M --> first order: 1/s --> divide by M --> second order:1/M*s
zeroth order: M/s --> divide by M --> first order: 1/s --> divide by M --> second order:1/M*s
-
- Posts: 122
- Joined: Fri Sep 28, 2018 12:18 am
Re: units of k
When you are given a problem sometimes they may give you a rate constant to solve for something else like concentration. For example, a given rate constant can be written with the following units: L/mol*s. This is the same as 1/M*s because M=mol/L so if we have L/mol then this is the inverse of mol/L so it will equal 1/M. Then we multiplied 1/M by s to get 1/M*s. This indicates that we will be using rate laws or half-life equations for the second order.
-
- Posts: 112
- Joined: Sat Jul 20, 2019 12:16 am
Re: units of k
I found this video to be very helpful in visualizing why the units are different for zeroth, first, and second-order reactions.
https://www.khanacademy.org/science/che ... constant-k
https://www.khanacademy.org/science/che ... constant-k
-
- Posts: 49
- Joined: Sat Sep 07, 2019 12:18 am
Re: units of k
for zero-order reactions, K has units of M/s
for first-order reactions, K has units of 1/s
for second-order reactions, K has units of 1/M.s
for first-order reactions, K has units of 1/s
for second-order reactions, K has units of 1/M.s
-
- Posts: 46
- Joined: Wed Feb 20, 2019 12:17 am
Re: units of k
Always go based on the order of the reaction:
If the reaction is zero-order, K has units of M/s
If the reaction is first-order, K has units of 1/s
If the reaction is second-order, K has units of 1/M.s
If the reaction is zero-order, K has units of M/s
If the reaction is first-order, K has units of 1/s
If the reaction is second-order, K has units of 1/M.s
Return to “Method of Initial Rates (To Determine n and k)”
Who is online
Users browsing this forum: No registered users and 2 guests