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Chemistry

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3) The following set of data was obtained by the method of initial rates for the reaction:
2 HgCl2(aq) + C2O42-(aq) 􀀏 2 Cl-(aq) + 2 CO2(g) + Hg2Cl2(s)
What is the rate law for the reaction?
[HgCl2], M [C2O42-], M Rate, M/s
0.10 0.10 1.3 × 10-7
0.10 0.20 5.2 × 10-7
0.20 0.20 1.0 × 10-6
A) Rate = k[HgCl2][C2O42-]-1 B) Rate = k[HgCl2][C2O42-]2
C) Rate = k[HgCl2]2[C2O42-] D) Rate = k[HgCl2][C2O42-]-2

  • Chemistry -

    I'll do one and leave the other for you.
    Label these rates as 1, 2, and 3.
    Then they follow this scheme.
    rate = k*[Hg2Cl2]x[C2O4^-2]y
    You want to determine x and y, the exponents.
    Look for rates where concn is the same for one of them and different for the other. 1 is ok for this; Hg2Cl2 is the same 0.1 M while oxalate changes from 0.1 to 0.2. Here is what I do.
    Write rate for #2 and divide it by rate for #1.
    rate 2 ...k[Hg2Cl2]^x[C2O4]^y
    ------ = -----------------------
    rate 1....k[Hg2Cl2]^x[C2O4]^y

    5.2E-7...k[0.1]^x[0.2]^y
    ------- = -------------------------
    1.3E-7...k[0.1]^x[0.1]^y

    ks cancel, 0.1^x cancels and we are left with
    4.00 = 2^y
    So 2 to the y power is 4.00 which means y must be 2; therefore, the reaction is second order with respect to oxalate.
    (Sometimes the power is not obvious; if that is the problem, you do it this way.
    4.00 = 2^y
    log 4.00 = y*log 2
    0.602 = y*0.301
    y = 0.602/0.301 = 2.00 and I think you will need to do that to find x power for Hg2Cl2.

  • Chemistry -

    A)

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