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Table on initial rates help

dippers

ImageUploadedByStudent Room1379153209.982127.jpg

I'm not too sure how to complete this table about initial rates

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Reply 1

ChemEng

Original post by dippers

I'm not too sure how to complete this table about initial rates

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rearrange rate equation to get in terms of k then using the first row substitute values into the rate equation to find k

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now for the second row u need to find rate and you now know k and you have both P and Q so substitute everything into the rate equation

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for 3rd row rearrange formula to get in terms of Q and substitute everything else in.

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and again for last row rearrange to get in terms of P

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Reply 2

dippers

Thank you so much! That was really helpful! I don't suppose you know how to go about this question?

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Reply 3

ChemEng

Original post by dippers

Thank you so much! That was really helpful! I don't suppose you know how to go about this question?

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which question?

Reply 4

dippers

Sorry I must of forgotten to add it. ImageUploadedByStudent Room1379239373.130796.jpg

someone told me the answers but I'm not sure how they got it. Someone said you could do the same method that you did for the other question but I wasn't sure because it didn't involve h+.

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Reply 5

ChemEng

Original post by dippers

Sorry I must of forgotten to add it. ImageUploadedByStudent Room1379239373.130796.jpg

in this question H+ does not exist in the rate equation. This means H+ is zero order. So you can ignore H+ for parts (a)i and (a)ii.
The rate equation also tells us that H2O2 and I- are first order. This means if for example the concentration of H2O2 doubles as it does from experiments 1 to 2 the initial rate also doubles so the initial rate for experiment 2 will be twice the initial rate of experiment 1 which is 2.3x10-6 multiplied by 2 which gives 4.6x10-6.

Now if you look at experiment 3 you can see that the concentration of H2O2 and I- do not change compared to experiment 1 therefore the initial rate is the same as experiment 1 = 2.6x10-6

For experiment 4 concentration of H2O2 increases by x5 (compared to expt 1) and I- halves. so here we simply multiply the initial rate from experiment 1 by 5 and then divide by 2 and you get 5.75x10-6.

for (a)ii the answer you have written down in incorrect.
to find the rate constant k you have to use the rate equation
you can use values from any experiment but if you are not sure always use the values that are already given so in this case use values from experiment 1 as the initial rate and all the concentration are already given.
Now simply substitute concentrations of H2O2 and I- and the initial rate into the rate equation. Rearrange for K.

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(edited 10 years ago)

Reply 6

dippers

ImageUploadedByStudent Room1379424606.587586.jpg

could you please help me with cii. Thanks! I'm not sure how to go about it.

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Reply 7

ChemEng

Original post by dippers

could you please help me with cii. Thanks! I'm not sure how to go about it.

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This is quite difficult to explain but so far you've got the right idea. From the rate equation you understood you need 2NO₂ on the LHS of the second step and from the overall equation you found you need a CO on the LHS of the second step.

Now one of the products we need is NO so we can make this on the first step.
So for your first step you can have:

NO₂ + NO₂

\longrightarrow

NO + NO₃

Now as NO₃ does not exist in the overall equation we have to use it in the second step to get rid of it. And also in the second step we need to produce CO₂.

Your second step can be:

NO₃ +CO

\longrightarrow

CO₂ + NO₂

To check if these steps are correct you can cancel everything that appears both on the left and right hand side and if you do you will see that you get the overall equation so these steps are correct.

You can do it other ways such as producing N₂O₄ in the first line but as long as you get the overall equation when you cancel you will get the marks.