k units [ENDORSED]
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Re: k units
There is a different equation for each rate order, and thus a different k unit to cancel out to get the final units.
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Re: k units
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.
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Re: k units
You need to find the K with the right units so that it cancels with the rate order equation.
Re: k units
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.
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Re: k units
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.
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.
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Re: k units [ENDORSED]
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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