### Units of k

Posted:

**Wed Feb 28, 2018 9:15 pm**Can someone go over how we determine the units of k for different orders of reactions? What impacts do the units make on our calculations?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=147&t=28691

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Posted: **Wed Feb 28, 2018 9:15 pm**

Can someone go over how we determine the units of k for different orders of reactions? What impacts do the units make on our calculations?

Posted: **Wed Feb 28, 2018 9:18 pm**

Basically you take the rate law (rate=k[a] for first order) and the units of rate are always mol.L-1.s-1 and then the units of concentration are always mol.L-1 and then you plug those in to find units of k, which would be s-1.

Posted: **Wed Feb 28, 2018 9:19 pm**

The units of k can change the units for rate when you are using it in a calculation

Posted: **Thu Mar 01, 2018 8:41 am**

Rate = k[A]^n, where n is the exponent that represents order of the reaction. Concentration is always going to have units of M/L, which is the equivalent to mol.L^{-1}.s^{-1}. Be sure to apply the exponent to the units of the concentration as appropriate. You want to isolate k to be alone on one side of the equal sign to find its units. Since the units of rate are always mol.L^{-1}.s^{-1}, you can divide the units of the rate by the units of [A]^n.

Posted: **Thu Mar 01, 2018 10:04 am**

Also the time units do not always have to be in seconds. Any time unit works, just make sure that you units are consistent throughout the calculation.

Posted: **Thu Mar 01, 2018 3:41 pm**

the units themselves have no impact on the numerical answer, unless they involve a conversion factor (for example mmol to mol)

Posted: **Fri Mar 02, 2018 9:20 am**

Since Rate= mol•L^-1•S^-1

AND [A]= mol•L-1

First order rate = K1*[A]

And calculate the unit of K1

AND [A]= mol•L-1

First order rate = K1*[A]

And calculate the unit of K1

Posted: **Fri Mar 02, 2018 11:07 am**

Unit of K is in a format so that the rate becomes mol/L S

Posted: **Fri Mar 02, 2018 11:11 am**

Units of k start out at mol/(L*s) at the zero order and every time you go up one order you can multiply by L/mol to get the new units of k.

Posted: **Fri Mar 02, 2018 1:04 pm**

I think the units of k is the same units as concentration.

Posted: **Fri Mar 02, 2018 1:14 pm**

For a 1st order reaction, k units = s^-1

2nd order reaction, k units = L/(mol*s)

and increasing the powers of L and mol as you go up in reaction order so that rate will be in unit of mol/(L*s)

2nd order reaction, k units = L/(mol*s)

and increasing the powers of L and mol as you go up in reaction order so that rate will be in unit of mol/(L*s)

Posted: **Fri Mar 02, 2018 2:15 pm**

I know this question has been answered many times, but the simplest response is just to do dimensional analysis and multiply everything out. The units of k will be whatever makes the equation work out, so they differ according to the order of the reaction.

Posted: **Fri Mar 02, 2018 2:45 pm**

The unit of the rate of reaction is mol•L^-1•S^-1, so you just have to manipulate the unit of K so that when put into the equation will give you mol•L^-1•S^-1.

Posted: **Fri Mar 02, 2018 2:56 pm**

An easy way of rememberin gthis is to to take (M^1-x)/s where s is the order of the reaction.

For example, for a 1st order reaction 1-1=0 and M^0=1 so the unit for k will be 1/s for 1st order rxn.

For example, for a 1st order reaction 1-1=0 and M^0=1 so the unit for k will be 1/s for 1st order rxn.