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    How do you fill in the table without using k for experiment 3?Name:  Screenshot 2017-01-30 19.27.35.png
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    (Original post by kiiten)
    How do you fill in the table without using k for experiment 3?Name:  Screenshot 2017-01-30 19.27.35.png
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    You are given the rate equation so that you know the dependence of the rate on the concentration of all components.
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    (Original post by charco)
    You are given the rate equation so that you know the dependence of the rate on the concentration of all components.
    But im not sure how to work it out

    (NOTE: ive just written the numbers below and not the x10 part)

    So 4.8 / 1.6 = 3

    10.4 / 3 =3.5 but the rate has to be 10.4

    Maybe B is 6.6 x 3 but it needs to be squared??
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    (Original post by kiiten)
    But im not sure how to work it out

    (NOTE: ive just written the numbers below and not the x10 part)

    So 4.8 / 1.6 = 3

    10.4 / 3 =3.5 but the rate has to be 10.4

    Maybe B is 6.6 x 3 but it needs to be squared??
    Rate = k[A][ B2

    If [A] doubles so does the rate
    if [ B] doubles the rate increases by a factor of 4
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    (Original post by charco)
    Rate = k[A][ B2

    If [A] doubles so does the rate
    if [ B] doubles the rate increases by a factor of 4
    Can you explain how you get B please.

    I know that A is x3 so the rate must be x3. But the rate must be the same so maybe B is divided by sqrt 3?

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    Bump :cry:
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    (Original post by kiiten)
    Bump :cry:
    Rate = k[A][B ]2

    From experiments 1 & 3

    [B ] doubles so you would expect that the rate would increase by a factor of 4.

    BUT

    The rate actually decreases by a factor of 2

    So the rate is actually 1/8 of what you would expect if [A] did not change.

    Hence the decrease is due to a change in [A]

    ... as the rate is first order wrt [A] then the decrease in rate must be proportional to the decrease in [A]

    Hence, the concentration of [A] in experiment 3 is 1/8 of the concentration of [A] in experiment 1
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    (Original post by kiiten)
    How do you fill in the table without using k for experiment 3?Name:  Screenshot 2017-01-30 19.27.35.png
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    This type of question really doesn't need to be overcomplicated, it's literally just a matter of plugging numbers into a calculator.

    So, it's not a matter of not using k, you need to find k by rearranging the equation and then using the data from experiment 1. I won't do the calculation for you, I won't insult your intelligence by rearranging an equation for k and plugging numbers into a calculator haha.

    Once you have k, the rest is a piece of cake, it's again just a matter of rearranging the equation to find the value that you need.

    If you've explicitly been told not to use k (which isn't mentioned in the question), then you can do what I assume charcot has talked about. In other words, for experiment 3, you'll notice that the rate has halved, and that B has doubled. Because B is to the power 2, when it is doubled, the rate multiplies by 4, but because it is halved, this means that A must be 8 times smaller than the original value of 4.8 x10^-2. Does that make sense? It's kind of like a puzzle. But at the end of the day, just use a calculator to check your answer anyway, unless you have a non-calculator exam, you'll never be asked to do a question involving tables like that without a calculator.

    If none of that makes sense then let me know and I'll try and explain it better
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    Yh infact charcot explained it a lot better than I did. Let me try and explain it a bit better with simpler numbers. So, let's say that 6 x 9^2 = 486 right. Now let's double the 9 and halve the 486, so we have 6n x 18^2 = 243. We need to find 'n' now to make the equation work, so 18^2 = 324, 243/324 = 0.75. So this means that 6n = 0.75, thus n = 0.75/6 = 1/8. So that just proves why you need to divide A by 8 when B doubles and the rate halves. Do you see a bit better now?
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    lol sorry, your name is charco not charcot.. I was thinking of charcot feet in diabetes hahaha, medicine has taken over my life
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    (Original post by charco)
    Rate = k[A][B ]2

    From experiments 1 & 3

    [B ] doubles so you would expect that the rate would increase by a factor of 4.

    BUT

    The rate actually decreases by a factor of 2

    So the rate is actually 1/8 of what you would expect if [A] did not change.

    Hence the decrease is due to a change in [A]

    ... as the rate is first order wrt [A] then the decrease in rate must be proportional to the decrease in [A]

    Hence, the concentration of [A] in experiment 3 is 1/8 of the concentration of [A] in experiment 1
    (Original post by AortaStudyMore)
    Yh infact charcot explained it a lot better than I did. Let me try and explain it a bit better with simpler numbers. So, let's say that 6 x 9^2 = 486 right. Now let's double the 9 and halve the 486, so we have 6n x 18^2 = 243. We need to find 'n' now to make the equation work, so 18^2 = 324, 243/324 = 0.75. So this means that 6n = 0.75, thus n = 0.75/6 = 1/8. So that just proves why you need to divide A by 8 when B doubles and the rate halves. Do you see a bit better now?
    Yes i kind of understand it now. I could have used k to fill in the table and it probably would have been easier but i wanted to know how to do it without. My earlier posts probably dont make sense because i was referring to experiment 4 :3

    But just to check is this right for experiment 4?:

    comparing A for experiment 1 and 4
    4.8 / 1.6 = 3

    rate must be 1/3 but stays the same so the concentration of B is 6.6 x 3

    B is order 2 so its 6.6 x sqrt 3 = 0.114 (11.4x10^-2)

    Note: i havent included the x10 for the concentrations just because its easier to write
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    (Original post by kiiten)
    Yes i kind of understand it now. I could have used k to fill in the table and it probably would have been easier but i wanted to know how to do it without. My earlier posts probably dont make sense because i was referring to experiment 4 :3

    But just to check is this right for experiment 4?:

    comparing A for experiment 1 and 4
    4.8 / 1.6 = 3

    rate must be 1/3 but stays the same so the concentration of B is 6.6 x 3

    B is order 2 so its 6.6 x sqrt 3 = 0.114 (11.4x10^-2)

    Note: i havent included the x10 for the concentrations just because its easier to write
    Looks good to me ...
 
 
 
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