The Student Room Group
Reply 1
If your dividing powers you take them away, if your multiplying them you add them.

E.g.
k=moldm3s1/(moldm3)2[br][br]k=moldm3s1/mol2dm6[br][br]k=mol1dm9s1k = moldm^-3s^-1/(moldm^-3)^2[br][br]k = moldm^-3s^-1/mol^2dm^-6[br][br]k = mol^-1dm^9s^-1

Hope that in anyway helps :biggrin:

(BTW I suck at LaTeX)
you have the rate equation:

rate = k [A]^x ^y

rearrange to get

rate / ([A]^x ^y) = k


units of rate are mol dm-3 s-1
units of concentration are mol dm-3

so you end up with mol dm-3 on top and bottom of the equation, you can cancel these (only once remember)

which will leave you with something like

s-1 / (mol dm-3 mol dm -3) = s-1 / (mol2 dm-6)


when you "move" the mol2 dm-6 to the top, the powers become negative.

leaving you with (in this example)

mol-2 dm6 s-1


it's hard to explain over the net - hope this helps :smile:
For Example

rate = k[A]
^ Rearrange to get k
= k = rate/[A]

Concentration units = moldm^-3
Rate = moldm^-3s^-1

So now substitute units into rearranged equation k = rate/[A]
k = (moldm^-3s^-1) / (moldm^-3)(moldm^-3)
Which cancels down to k = (s^-1) / (moldm^-3)
SO units for k = mol^-1dm^3s^-1
Reply 4
For a zero order reaction, the unit of K is the unit of rate, i.e. mol dm-3 s-1

For a 1st order reaction, take the unit for a zero order reaction and subtract the unit of concentration i.e. (mol dm-3 s-1) - (mol dm-3) = s-1

For a 2nd order reaction, subtract another i.e. (s-1) - (mol dm-3) = mol-1 dm+3 s-1

etc.