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Chemical Kinetics homework

  1. Sep 3, 2006 #1
    Hello everyone,

    I have been working on several exercises for the past several days, and there are some I just can't seem to do. I would appreciate some help! =)

    1) We have the following reaction: C2H5Br (alc) + OH(-) (alc) --> C2H5OH (l) + Br(-)
    The (-) indicates it's an ion.
    The reaction is 1st order in the reactants. We have a [ ] for each of the reactants, and the rate of disappearance of C2H5Br.
    -We need to find the value of the rate constant. Do we use the formula r=k[A] and get k? Why did they give us the rate of disappearance though?
    - If we were to double the ethyl alcohol, what would happen to the disappearance rate? I have to say, I have no idea how to do that... Does it just mean that since the concentration is less than, the rate of disappearance is less than.. as well?

    2) Here, I am sure I missed something important, but I just can find clues anywhere, not in the book, not in the lecture, nowhere...
    We have a reaction (combustion of ethylene). If we decrease the concentration of 1 of the 2 reactant at a rate = (some number), what are the rates of change in the concentration of the products?
    I don't understand how I can do that if we don't know the r of the other reactant.

    3) One I tried OVER and OVER and it's not giving a reasonable answer.
    If you want to help with just one question, please start with this one =)

    We have a 1st order rate reaction, and the rate constant for the decomposition of reactant A. We start with x amount of A in moles, in a volume of 2.0L (I don't know what to do with that 2.0 L. Do I just ignore it? Do I divide the x amount of A by 2 to get an amout per 1L?).
    We need to know the amount of A in moles after 2.5 minutes.
    I know that the formula is:
    ln [A]t = -kt + ln [A]o
    We have [A] at t=0. We have t. We have k. It should all be very simple. But I keep getting answer that are way to high. So, I probably should do something with that 2L. I tried keeping it. I tried dividing by 2. I tried multiplying the t to get seconds, and simplify.
    Any ideas?
    For more clarity, here's the whole thing:

    N2O5 --> 2NOw + O2. K= 6.82 x 10^-3 s^-1. [N2O5]= 0.0250 mol/ 2L. How many moles remain after 2.5 minutes?

    4) Finally, (we had lots of exercises. I could do many, but these, I couldn't, for some reason): We have several temperatures and their corresponding k (in M^-1s^-1).
    We need to find the activation energy and A. My question is, how can I do this if I don't have R (for the Arrhenius equation)?! I plug in everything I can, but this unknown R prevents me from getting an answer. Or do we just need an answer in the form of R(answer)? I don't think so though...

    THANKS A LOT if you can help me. Or even if you try =)

  2. jcsd
  3. Sep 3, 2006 #2
    If you need me to show some more work to be sure I have worked on the problems, feel free to ask. I thought I wouldn't add more to an already long post =)
  4. Sep 3, 2006 #3


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    Well, what number would you use for 'r'? To make sure you understand this, ask yourself what 'r' represents.

    Concentration of what? "Less than" what?

    Please post the exact question - reduces room for misinterpretation.

    After you've done this, explain, step by step, what you would do, starting from writing out and balancing the equation.

    The equation is correct, so you must have made an arithmetic error. If you show us the math you did, we can point out the error.

    Do you know what the different symbols in the Arrhenius Equation represent? You can't "use an equation" unless you understand it. What does 'R' represent?

    Henceforth please post the original question/problem (to the best that is possible) exactly as it appears in your text/homework set.
    Last edited: Sep 3, 2006
  5. Sep 4, 2006 #4

    Thank you for your help AND super fast answer.

    Here's the equation for number 3:

    UPDATE: I forgot to take the ln of the other side. Fixed now...
    Would that be correct?

    ln[A] = ln[A]o - kt
    ln[A] = ln[0.0125] - 0.0063 x 150
    ln[A] = -5.407
    [A] = e-5.407
    [A] = 0.00449 mol

    Moles of A = 0.00449 x 2 = 8.98 x 10^-3 M/2L or 0.00898

    BUT then they ask how much time "the quantity" would take to drop to 0.010 mol. Well, it's already below. Do they mean the first quantity we had? Or is my answer wrong?
    Last edited: Sep 4, 2006
  6. Sep 4, 2006 #5
    So for 1), we can use the rate of disappearance of one reactant to determine k for the whole reaction, because it's first order?

    What I did was: (r and the 2 reactants' values were given):

    r = k [C2H5Br][OH-]
    1.7 x 10^(-7) M/s = kx 0.0477M x 0.100M
    k = (1.7 x 10^-7 M/s) / (4.77 x 10^-4) M2 (squared) = 3.56 x 10^-4 M-1s-1.

    Is this correct?

    Also, in #1, I meant that if we double the amount of ethyl alcohol the reactants are diluted in, then the disappearance rate of one of the reactants would be twice as slow, since the rate of the reaction depends on the concentration (I am saying twice as slow, since it is a first rate reaction).
    Last edited: Sep 4, 2006
  7. Sep 4, 2006 #6
    Now, for #2.
    We have the following equation:
    C2H4 (g) + 3O2 (g) --> 2CO2 (g) + 2H2O
    The equation is balanced.
    We know that C2H4 is decreasing at a rate of 0.37 M/s.
    We need to find the rate of change in the concentration of the products.

    We do not have the rate for 3O2. Is it just the same?

    What I would with the given info is:

    Rate = - ∆[C2H4]/∆t = 1/2 ∆[CO2] /∆t = 1/2 ∆[H2O]/∆t
    Rate= 0.37 M/s

    1/2 ∆[CO2] /∆t = 0.37 M of C2H4 /s x (2 moles of CO2 / 1 mole of C2H4)
    r for CO2 = 0.74 M of CO2/s

    And since we have the same amout of M for H2O, the answer for be the same here too.
    I am not sure of what I just did though, I would appreciate someone's opinion on this work.

  8. Sep 4, 2006 #7
    I had an idea but wasn't sure. Is R always a constant? (R= 8.314 J/M-K)?
    If so, then I think I can go one with the equation. I just wasn't sure R was always the same value no matter what...
    Complete equation: k= Ae^(-Ea/RT)

    Thank you so much =)
  9. Sep 4, 2006 #8
    The question:
    We have 5 different temperature and their corresponding k for a certain equation we know nothing about. We have to find Ea and A.

    Here's my work:

    We need find Ea and A. We are given temperatures and their corresponding k.
    To find Ea:
    I found the values of 1/T and ln k.
    Then, I found the slope of the associated curve, and got -16.
    Such a round number almost looks suspicious, after all these decimals.
    Anyway, so then I took the formula slope= - Ea / R and plugged the values in. I got Ea = 133.024 J/M.

    On to A:
    I used k = Ae^(-Ea/RT)
    I got 2.9 x 10^-2 M s^-1.

    I hope that's right...

  10. Sep 5, 2006 #9
    Hello, it's me again. I wanted to say thanks, and don't bother any further because everything is due today!

  11. Sep 5, 2006 #10


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    Should've seen this earlier...:frown:
    Yup, that's right.

    EDIT: Oh, guess this is too late now. Sorry!
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