Homework Help: Chemistry

Posted by Olivia on Wednesday, October 27, 2010 at 3:19am.

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 - DrBob222, Wednesday, October 27, 2010 at 2:00pm
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 - Anonymous, Wednesday, March 7, 2012 at 8:08pm
A)

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