Units of k
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Units of k
What are the units of k in zero-, first-, and second- order reactions? How are these units derived?
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Re: Units of k
Zero-order: M/s
First order: s^-1
Second order: L/(mol*s)
These units are derived from the rate law -- we know the rate of reaction must be M/s, so the units of k change to accommodate that.
First order: s^-1
Second order: L/(mol*s)
These units are derived from the rate law -- we know the rate of reaction must be M/s, so the units of k change to accommodate that.
Last edited by Sirajbir Sodhi 2K on Mon Mar 05, 2018 10:52 am, edited 1 time in total.
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Re: Units of k
If you are having trouble figuring out the units of k, write out the entire rate law with the units. You know that the rate must be in mol/Ls, so the units of k must cancel out the units of concentration of the reactants until you get mol/Ls.
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Re: Units of k
Units of the rate are almost always mol/L*s and units of concentration are almost alway mol/L. Therefore the units of k can be determined based on the reaction order and the already determined units.
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Re: Units of k
The easiest way to determine the units of the rate constant k is to use the following "formula":
M^(1-X)/S
where x is the order of the reaction. Examples:
Zero Order: M(^1-0)/S ----> units: M/S
First Order: M^(1-1)/S ----> units: 1/S and so on
M^(1-X)/S
where x is the order of the reaction. Examples:
Zero Order: M(^1-0)/S ----> units: M/S
First Order: M^(1-1)/S ----> units: 1/S and so on
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Re: Units of k
You can also solve for the units of k for each order! Since rate always begins with mol/(L.s)
Zero order(mol/L.s) = k[A mol/L]^0
mol/(L.s) = k
First order (mol/L.s) = k[A mol/L]
/s = k[A]
Second order (mol/L.s) = k[A mol/L]^2
(mol/(L.s)) = k[A]^2 (mol^2/L^2)
L/(mol.s) = k[A]^2
Zero order(mol/L.s) = k[A mol/L]^0
mol/(L.s) = k
First order (mol/L.s) = k[A mol/L]
/s = k[A]
Second order (mol/L.s) = k[A mol/L]^2
(mol/(L.s)) = k[A]^2 (mol^2/L^2)
L/(mol.s) = k[A]^2
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Re: Units of k
0 order: M/s or mol/L.s
1st order: 1/s
2nd order: 1/M.s or L/mol.s
3rd order: 1/M^2.s or L^2/mol^2.s
etc.
1st order: 1/s
2nd order: 1/M.s or L/mol.s
3rd order: 1/M^2.s or L^2/mol^2.s
etc.
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Re: Units of k
Will we have any questions pertaining to any reaction larger than the 3rd order on the final?
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Re: Units of k
I highly doubt that we would be asked to find anything higher than the 3rd order. But we may be asked questions where the overall order is greater than 3.
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Re: Units of k
There might be questions that ask to derive 3+ order equations but I doubt we will be asked to do much with them at all.
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Re: Units of k
I would remember that zero order would be M/Sec and then divide M each time you go down an order. Here's what I mean:
Zero order would be M(molarity)/second
so divide by M to get units for first order. Notice how M cancels out.
1st order units: s^-1
Divide M again to get 2nd order units
s^-1*M^-1
And continue dividing Molarity each time you go down an order. Hope that helps!
Zero order would be M(molarity)/second
so divide by M to get units for first order. Notice how M cancels out.
1st order units: s^-1
Divide M again to get 2nd order units
s^-1*M^-1
And continue dividing Molarity each time you go down an order. Hope that helps!
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