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By the end of this lesson you should be able to:
The rate constant k appears in any rate equation:
rate = k[A]^m[B]^n
It is the proportionality constant linking rate to the concentration terms. At a given temperature, for a given reaction with a given catalyst (or none), k has a single, fixed value. k does not depend on the concentrations of any reactants.
Key properties of k:
This is where students most often lose marks. The units are worked out from the rate equation by inserting the units of each quantity:
rate [mol dm^-3 s^-1] = k [units?] x concentration [mol dm^-3] raised to the overall order
Rearrange:
k = rate / ([concentrations]^overall order)
rate = k
Units of k = units of rate = mol dm^-3 s^-1
rate = k[A]
k = rate / [A] = (mol dm^-3 s^-1) / (mol dm^-3) = s^-1
rate = k[A]^2 (or k[A][B])
k = rate / [A]^2 = (mol dm^-3 s^-1) / (mol dm^-3)^2 = mol^-1 dm^3 s^-1
rate = k[A]^2[B] (or k[A][B]^2, or k[A][B][C])
k = rate / [concentration]^3 = (mol dm^-3 s^-1) / (mol dm^-3)^3 = mol^-2 dm^6 s^-1
| Overall order | Units of k |
|---|---|
| 0 | mol dm^-3 s^-1 |
| 1 | s^-1 |
| 2 | mol^-1 dm^3 s^-1 |
| 3 | mol^-2 dm^6 s^-1 |
You can remember the pattern: each extra order divides by another mol dm^-3, so each extra order subtracts one power of mol dm^-3 from the units.
rate = k[A][B]^2. Experimental data: [A] = 0.25 mol dm^-3, [B] = 0.10 mol dm^-3, rate = 3.0 x 10^-4 mol dm^-3 s^-1.
Overall order = 1 + 2 = 3.
k = rate / ([A][B]^2)
= (3.0 x 10^-4) / (0.25 x 0.010)
= (3.0 x 10^-4) / (2.5 x 10^-3)
= 0.12 mol^-2 dm^6 s^-1
rate = k[A]. When [A] = 0.40, rate = 8.0 x 10^-3 mol dm^-3 s^-1.
k = 8.0 x 10^-3 / 0.40 = 2.0 x 10^-2 s^-1.
rate = k. Rate measured = 5.0 x 10^-4 mol dm^-3 s^-1.
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