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Equilibrium problem, anyone?

  1. Jun 15, 2011 #1

    1. The problem statement, all variables and given/known data

    When equal volumes of A and B are combined in a 3.5L flask, their initial concentrations were each 1.75mol/L. Once equilibrium is reached, the equilibrium concentration of c, is [c] = 0.65mol/L . determine the Kc for this reaction

    2. Relevant equations

    Usually, there is an initial value given, then the equilibrium value given, then you subttract those and plug them into the rest of the table. However... I dont know what to do inorder to get the answer when jus the eq val is given of c , and no data for d is given (assuming you need to find x first)

    3. The attempt at a solution

    I created the ice table

    2A : 1.75-2x
    B : 1.75 - x
    c: 0.65
    d: x

    Attempt to solution was a failure. The answer is 0.022

    Help please!
  2. jcsd
  3. Jun 16, 2011 #2


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    Staff: Mentor

    What is the reaction equation?
  4. Jun 17, 2011 #3
    Sorry, the reaction equation is 2A + B ----> 3C + D

    I still dont know how to do this problem.

    Edit: "2 A (g) + B (g)-----> 3 C (g) + D (g) ΔH0 = -315.9kj
    Last edited: Jun 17, 2011
  5. Jun 17, 2011 #4


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    Staff: Mentor

    Use stoichiometry to calculate equilibrium concentrations of all substances. This will be not much different from creating ICE table.

    Can you express C in terms of x?
  6. Jun 17, 2011 #5
    Okay... well, heres what I did. Since C is 0.65. And change in A and b is -x .

    For A : i did I created the ice table

    2A : 1.75-2(0.65) 0.45
    B : 1.75 - (0.65) 1.1
    c: 0.65
    d: 0.0619

    n = c/v , 0.65/3.5L 1/3 * that mol value = 0.0619 for d
    answer still doesnt come out as 0.022.
  7. Jun 17, 2011 #6
    Ive tried it using several methods now... used stoic of 0.65 to calculate all values, then subtracted by x (ended up giving neg values). Used stoic of 0.65 to calculate D , giving 0.216, then used x to subtract the other two values (A and B) . Used 0.65 as X in d and used x to calculate A and B.... nothing is working
  8. Jun 17, 2011 #7


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    Staff: Mentor

    Solving the problem is not about plugging numbers randomly in hope something will work.

    If you have produced x of D, can you tell - just by looking at stoichiometric coefficients - how much C was produced? Was it x? 2x? 3x? Something else?
  9. Jun 17, 2011 #8
    Yes.... Like i said..

    If X is 0.65 for c. Then d will be 0.65/3 Which will be 0.216.

    Change for a : -2x. Change for b : -x. change for c : +3x . Change for d = x

    I dont understand what youre trying to tell me

    Apparently you neglect any coefficiant in change when you have x. So i did 1.7-0.62 , giving me 1.1 for both a and b. then plugged those values in and got to 0.04456 Which is still wrong. However. 0.04456 / 2 = 0.022. I dont know why you do divided by 2 though.
    Last edited: Jun 17, 2011
  10. Jun 17, 2011 #9
    I understand it a bit more. Its saying that 3c= 0.65. for 3 moles of C. making D 0.216.

    And A , since your neglecting the change value "2" coefficiant when subtraction. you need to use stoic : 2/3 * 0.65 to get the change avalue for that , which is 0.4333.

    then doing 1.7-0.433 to get the equilib value for A. then for b you get 1.7- 0.216 and get that value.

    I reinput the numbers and got 0.0245 . Which is a bit off...

    figured maybe its a rounding error, made the eq. values whole numbers. got to 0.0233
    Last edited: Jun 17, 2011
  11. Jun 17, 2011 #10
    does anyone know....?
  12. Jun 17, 2011 #11


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    Staff: Mentor

    Exact value seems to be 0.022384... Perhaps you used rounded down numbers in your calculations. Unfortunately your explanation of what you did is chaotic to me, I see correct elements, but I am not sure everything is right.
  13. Jun 17, 2011 #12
    Okay. So this is what I did.

    2A + b ---------> 3C + D.

    Since 3C is 0.65 mol/l at equilibrium , per 3 moles, I used stoic calculations fo first figure out D at equilibrium.

    Ie) 1/3 * 0.65 = D . Which gave me the value for D.

    Then, I used stoic calculations to figure out the equilibrium value of A.

    Ie) 2/3 * 0.65 = Mol/L equilibrium of A.
    . Since i found the equilibrium value of A, I then subtracted 1.7-0.43 which gave me 1.27.

    For B I used 1/3 * 0.65 which gave me 0.216 mol/l at equilibrium of b.

    So, I did 1.7 - 0.216 = 1.484 Mol/l of B .

    Thus giving me a final keq quation of [0.65]^3[0.216] / [1.27]^2[1.484] = 0.245
  14. Jun 17, 2011 #13


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    Staff: Mentor

    Why 1.7 and not 1.75? Why 0.43 and not the exact value?

    Again, why 1.7? And why 0.216? If anything, 0.21(6) rounds up to 0.217, but you shouldn't use neither of these values.

    Use guard digits in your calculations, don't round down intermediate results.
  15. Jun 20, 2011 #14
    1. The problem statement, all variables and given/known data

    A mixture of 9.22 moles of A, 1011 moles of B, and 27.83 moles of C is placed in a 1 litre container. constant temp. At equilibrium the number of moles of B is 18.32 Calculate the eq constant for this reaction

    2. Relevant equations

    A + 2B <-> 3C

    3. The attempt at a solution

    I need help.. on how to solve this.. heres what i did:

    Ice table:

    A + 2B <-> 3C

    9.22 10.11 27.8
    +x +2x -3x ( for some reason the change is backwords)

    Since equilib moles of B is 18.32, I did 18.32 - 10.11 , giving me 8.21 moles of 2B at equilibrium. This means that A will be 8.21mol divided by 2 , giving me 4.105. 9.22 + 4.105= 13.32 moles of A at equlibrium.

    For C i used stoic, 3/2 * 8.21 mol = 12.315. at equilibrium. So, 27.83 -3(12.315) Gives me a negetive value, So now im stuck. Wth am i doing wrong?
    Help please!
  16. Jun 20, 2011 #15
    "For C i used stoic, 3/2 * 8.21 mol = 12.315. at equilibrium. So, 27.83 -3(12.315)"
    Huh why are you multiplying with 3 again?? you have already done that at the 3/2 part.

    27.83 -12.315 = 15,51 Slow down man :D
  17. Jun 20, 2011 #16
    What? No i havent. When did i multiply with 3? Its said that value of equilibrium for 2 moles is 8.21. , ALl i did was 1/2 * 8.21 to give me the value of A. Then went back to the value it is for 2 moles (8.21) and did 3/2 * 8.21 , giving me 12.315. When did i already multiply by 3? oh, yes. So.. i see what youre saying. Why am i supposed to neglect the change on that value? Is it because the stoic ratio GIVES me that *3 value? so i dont need to do "change again" since i already calculated it?
    Last edited: Jun 20, 2011
  18. Jun 20, 2011 #17
    You know that you gain 8.21 moles of b. According to the reaction equation you gain 1/2 moles of A and lose 3/2 moles C for every mole of B you gain. Now you multiply 3/2 * 8.21 to know how many moles of C you lost. That numbers you arrived at was 12.315
    Since you lost that amount, you subtract it from how many moles of C you had at the start, which was 27.83

    27.83-12.315 = 15.51

    When you did your final calculation you said "27.83 -3*(12.315)" and that's the error. 12.315 is the amount of moles of c lost, not 3*12.315

    Edit: You ADD/SUB the change, but the change wasn't 3*12.315 but 12.315!!
    Last edited: Jun 20, 2011
  19. Jun 20, 2011 #18
    I some what get it... not really though, does that mean I never really bother subt/adding the change? only lookign at if its positive or negetive (from what you determine) .

    For example: if i say C is instead +3x, the only difference is i add 27.83 + 12.315?

    If A was -x I would just subtract by that single value? I dont need to do +2(4.12) etc, since i already have that +2 right?
  20. Jun 20, 2011 #19
    What does it mean if the reaction moves to the right or left? I dont understand. Heres a sample question.

    C7H14 ----> C7H8 +3H2

    Using only the info given. which of the following changes would increase the molar concentration of C7H8

    a) increase pressure at constant temp, b) inc temperature , c ) decrease concentration. d) add a catalyst

    Apparently the answer is a, but i chose C. I know this relationship. : decreasing the reactant conc will make the reaction move to the left, , does that mean the conc of nh3 will decrease?
  21. Jun 20, 2011 #20


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    Staff: Mentor

    Decrease concentration of what? If you decrease concentration, concentration decreases - that's quite logical.

    Please try to be less chaotic, it is impossible to help you not knowing what is the question, what is the information given and what you mean.

    First: you were already told twice to properly capitalize formulas. Second: assuming you mean NH3, there is no ammonia in the reaction. That probably means you meant something else, but I am not going to guess what.
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