CHEMISTRY INITIAL RATES! Help!

Scroll to see replies

I got the final rate as 1x10^-5 hopefully thats right. Seems we have some good teachers here so you wont be needing me :smile:

Reply 21

greentron6

Original post by Poooky

Try and explain it back to me :smile:

(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 :biggrin:

I can't even explain it to you! I'm actually terrible! I don't think there is another way for you to explain this, I'm actually just bad! Sorry I've spent so much of your time! I guess I'll just have to revise the topic again!

Thank you for your help! Any suggestions for succeeding in A2 Chem?

Reply 22

username2455183

Original post by chhhhelsie

I got the final rate as 1x10^-5 hopefully thats right. Seems we have some good teachers here so you wont be needing me :smile:

Correct!

Original post by greentron6

Hmm what part is specifically troubling you? Is it changing two variables at once?

Just do loads and loads of past paper questions and check the answers and go through any questions you can't do- basically what you're doing now! :biggrin:

Reply 23

greentron6

Original post by Poooky

Correct!

Hmm what part is specifically troubling you? Is it changing two variables at once?

Just do loads and loads of past paper questions and check the answers and go through any questions you can't do- basically what you're doing now! :biggrin:

I'm stuck with the comparison you made between A and B. How did one halve and the other double?

Reply 24

username2455183

Original post by greentron6

I'm stuck with the comparison you made between A and B. How did one halve and the other double?

That's with respect to their initial concentrations- so compare the concentrations of A and B in exp 2 with 4 in my drawing, or 2 and 3 in your question. How are they changing?

Reply 25

greentron6

Original post by Poooky

That's with respect to their initial concentrations- so compare the concentrations of A and B in exp 2 with 4 in my drawing, or 2 and 3 in your question. How are they changing?

B doubles in conc., and A halves in conc. and the initial rate for both stay the same.

Reply 26

username2455183

Original post by greentron6

B doubles in conc., and A halves in conc. and the initial rate for both stay the same.

Ignore A for a second, imagine you did an experiment, keeping A and C the same and you doubled B. You know B is first order. What do you expect the rate to be?

Reply 27

greentron6

Original post by Poooky

Ignore A for a second, imagine you did an experiment, keeping A and C the same and you doubled B. You know B is first order. What do you expect the rate to be?

The rate will also double.

Reply 28

username2455183

Original post by greentron6

The rate will also double.

Yup. So now do another experiment, this time keeping B and C the same as your first experiment. You half A. You're TOLD that the rate has now halved. So B,C the same. A halves. Rate halves. What is the order of A?

Reply 29

greentron6

Original post by Poooky

First order.

Reply 30

username2455183

Original post by greentron6

First order.

Correct. So in all, you have done 2 experiments and there was an initial experiment (for comparison)

MY ROWS ARE MESSED UP! Will make a proper table to make it easier!

Initial: A B C Rate
1st exp A Double B C Double Rate

Now compare 1st exp, and 2nd exp:
1st exp A Double B C Double Rate
2nd exp Half A Double B C Half double rate =1/2 *2=1*Rate= Rate

Initial: A B C Rate
2nd exp Half A Double B C Half double rate =1/2 *2=1*Rate= Rate

Therefore, comparing initial and exp, same rate! A halves, B doubled, both first order

Any clearer? :redface: