## Units of k [ENDORSED]

$\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$

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### Units of k

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?

Alyssa Parry Disc 1H
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### Re: Units of k

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.

SantanaRodriguezDis1G
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### Re: Units of k

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

Abigail Urbina 1K
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### Re: Units of k

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.

Sarah_Stay_1D
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### Re: Units of k

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.

Ishita Monga 1B
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### Re: Units of k

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

Yeyang Zu 2J
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### Re: Units of k

Since Rate= mol•L^-1•S^-1
AND [A]= mol•L-1

First order rate = K1*[A]

And calculate the unit of K1

Josh Moy 1H
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### Re: Units of k

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

David Minasyan 1C
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### Re: Units of k

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.

Grace Han 2K
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### Re: Units of k

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

Emily Mei 1B
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### Re: Units of k

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)

Liz White 1K
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### Re: Units of k

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.

Yu Chong 2H
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### Re: Units of k

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.

Sophie 1I
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### Re: Units of k  [ENDORSED]

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.