CHEMISTRY INITIAL RATES! Help!

You know that both reactants are first order, so to get their combined effect just substitute how the concentration has changed (in this case, tripling and doubling) into the rate equation

Is the answer 2.4 x 10^-3?

Yes, it is!

Now I have another dilemma...

There are 3 reactant concentrations this time.. I tried to compare, however I got them wrong. A and B are first order and C is second? How did this work? I don't get it...

(P.S. I'm an AS student, go it's pretty hard getting my head around A2 Chem...)

Use experiment 1 and 2 ==> You get that B is first order (both B conc and rate decrease by 1/4)
Use 2 and 3 ==> C stays the same. B doubles, therefore rate must double. A halves, what has happened to the rate? It is the same as before, so A must also be first order.
x+y+z is 4, therefore the order of C must be 2

Yes, it is!

Now I have another dilemma...

There are 3 reactant concentrations this time.. I tried to compare, however I got them wrong. A and B are first order and C is second? How did this work? I don't get it...

(P.S. I'm an AS student, go it's pretty hard getting my head around A2 Chem...)

Hey!
Thanks for the reply!
I don't get the 2 and 3 experiment. Could you explain it in more detail please?

Sure-

C stay the same, so we expect no change in rate from C.

B doubles, and because we've deduced that B is first order from exp 1 and 2, we expect the rate to also double.

We need to figure out the effect of A on the rate. Compare the rate of 2,3- it's the same. From before, using ONLY changes in C and B, we expect the rate of 3 to be double of 2 (4*10^5). Now we see that A has halved, and compare the expected rate of 3 to the actual rate- is has also halved. So we can deduce A is also first order

Lol don't worry bro i got u k? I struggled with this and yes it was difficult but super easy once you understandCompare experiments 1 and 2 conc. of A and C doesn't change however Conc of B does change, it has gone smaller by 4 times and the rate has also gone down by 4 times thus B is 1st order right?compare experiments 2 and 3 we know that if conc of B is 4x smaller then the trate is 4 times less right? so if the conc of B is 2x more then the rate is 2x more yea? So we know the rate when conc of B is linear so if the initial rate is 2x10^-5 and conc B doubles the rate becomes 4x10^-5 however it seems that when we make conc of A 2x smaller the rate also gets 2x smaller so A is a first order reaction.Look at the top where it says x+y+z=4 so if we know that X and Y are 1 and 1 then 2+z=4 right? so z=2 so C is a second order reaction ok?

Yeah...
I think I'm a lost cause...... A2 Chem is trying to end before I've even started...

Is there any other way you can explain or a video that can explain this?

Yeah...
I think I'm a lost cause...... A2 Chem is trying to end before I've even started...

Is there any other way you can explain or a video that can explain this?

I'll try drawing it out on something- seeing it visually might help

That would be heavily appreciated....

First do you understand how the order of A is 1?

No. That's the part that threw me off course. However, I understand why B is first order, because when the concentration times by 4, so does the rate. Therefore, 4^{1 = 4, which indicates the first order.}
And i know you'll get C due to the equation x+y+z=4.

Shooooot I meant B!!

I've broken down exp 2 and 3 in your question- start from 2 in my crudely drawn picture, understand how to go to 3 from 2, and then go from 3 to 4.
The way I've one it is that the total effect from going from 2 to 4 is the same as in YOUR example above ie one halves, the other doubles.

Did that help?

I think I kinda understand.....

Thank you for spending your time and effort trying to explain this to me! I really appreciate it!