Orders Of Reaction

Hi,
If you have a second order reaction but which consists of two first order reactants so rate=k[A] will a concentration time graph of either of these reactions display a constant half life or will they both show one of a typical second order reaction: rate=k[A]^2.
Thanks.

If you keep one component concentration constant then the other will show first order kinetics.

Thanks Charco i understand. So would a method of keeping the concentration on one constant by having it in excess?

Secondly if you did not keep either constant neither would display a constant half life would they?

And thirdly, if you had a first order reaction like rate=k[A] but you also had zero order reactants which were not in excess (therefore concentration not constant) would the concentration time graph of these be in theory identical to the one of [A].

My understanding of the above question is that it is possible to have a a zero order reactant (not in excess) in a reaction as long as it is not involved in the rate determining step. Could you just confirm this please thanks.

You seem to be getting the idea ok ...

Do you know if what i said was true because i don't lol (third point especially)

If it is true, i came across a question which had the equation:
HCOOH + BR2 --> goes to some products
It said HCOOH was in excess and gave a graph of [Br2] against time, this graph was a first order reaction graph.

My point is that if they did not specify that HCOOH was in excess, while the conc of bromine against time graph was first order would it be true that there would not be enough evidence to suggest the reaction was first order WITH RESPECT TO bromine as it could be HCOOH if it was not specified to be in excess...

Believe me, if i could pay you i would.

If a reagent is in large excess then effectively it's concentration does not change as the reaction proceeds. The reaction then shows pseudo-first order kinetics providing that the order wrt the other component is first.

Rate = k[A][ B]

if [ B] is constant then

Rate = k'[A]

Don't forget that the 'order' per se has to be with respect to some component. Yes, you can talk about the 'overall' reaction order, but you actually measure each order individually.

Your graphs must be of one of the components of the reaction or one of the products changing with time.

thanks, i understand that the graph must be of one of the reactants or products and everything you say makes sense. but if in the previous example if they did not say HCOOH was in excess then surely it would deplete at the rate of the rate determining step and so give a first order graph. By this i mean you would know that the reaction is first order - with respect to a reactant - but not which one?? So in other words you could not answer the question they gave. talking about conc time graphs btw.

or am i misunderstanding

But they DID tell you that it was in excess, presumably for that very reason.

I'd like it if I could just get the answer tbh. I have a chem mock in a day so...

The "answer" will not help you revise for your mock...

... but understanding the method will!