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CHEMISTRY INITIAL RATES! Help!

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1. Hi, I've completed the first two sections of Part A (they are both 1st order), however I am stuck on the the 3rd section (highlighted in blue). Could someone provide me with an explanation of how to do this?

2. 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
3. (Original post by Poooky)
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
Thank you!
4. Is the answer 2.4 x 10^-3?
5. (Original post by chhhhelsie)
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...)
6. (Original post by greentron6)
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
7. (Original post by Poooky)
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
Hey!
I don't get the 2 and 3 experiment. Could you explain it in more detail please?
8. (Original post by greentron6)
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...)
Lol don't worry bro i got u k? I struggled with this and yes it was difficult but super easy once you understand
Compare 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?
9. (Original post by greentron6)
Hey!
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
10. (Original post by Poooky)
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
(Original post by Big white)
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 me before I've even started...

Is there any other way you can explain or a video that can explain this?
11. (Original post by greentron6)
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
12. (Original post by greentron6)
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?
Choose 2 experiments to compare where only 1 of the conc changes(let's start of with this)
13. (Original post by Poooky)
I'll try drawing it out on something- seeing it visually might help
That would be heavily appreciated....
14. (Original post by greentron6)
That would be heavily appreciated....
First do you understand how the order of A is 1?
15. (Original post by Poooky)
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, 41 = 4, which indicates the first order.
And i know you'll get C due to the equation x+y+z=4.
16. (Original post by greentron6)
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, 41 = 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!!
17. (Original post by Poooky)
Shooooot I meant B!!
Well, yes I understand B!

18. 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?
19. (Original post by Poooky)

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!
20. (Original post by greentron6)
I think I kinda understand.....

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

(Basically because rate equations are just components being multiplied, you can separate them and work out what happens which each bit, without it affecting the other bit, eg 2*3*4, you can separate to (2*3)*4=6*4=24. In the same way, you can separate out the effect of changing two components, and just change one, and then the other)

You're welcome

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