chemistry
posted by unknown .
consider the reaction 2HgCl2 + C2O42 > 2Cl21 + 2CO2 + Hg2Cl2
The initial rate of this reaction was determined for several concentrations of HgCl2 and C2O42 and the following rate data was obtained.
TRIAL (1)
[HgCl2](M)= .105
[C2O42](M)= .15
RATE= 1.8 x 10^5
TRIAL (2)
[HgCl2](M)= .105
[C2O42](M)= .30
RATE= 7.1 X 10^5
TRIAL (3)
[HgCl2](M)= .052
[C2O42](M)= .30
RATE= 3.5 x 10^5
TRIAL (4)
[HgCl2](M)= .052
[C2O42](M)= .15
RATE= 8.9 x 10^6
TRIAL (5)
[HgCl2](M)= .080
[C2O42](M)= .10
RATE= ?
a)What is the rate law for this reaction?
b)What is the numerical value for the rate constant?
c)What would be the rate for the last set of data?
I am destroying my brain cells trying to figure this problem out so if anyone can help me I will be very greatful thanks!!!
rate = k(HgCl2)^{x} (C2O4)^{y}
You are to determine x, y, and k.
Compare trial 1 and trial 2.
(HgCl2) changes by factor of 0.105/0.105 = 1 (that is, it doesn't change).
(C2O4) changes by a factor of 0.30/0.15 = 2
rate changes by 7.1E5/1.8E5 = 4
Thus, any change in rate must be due to change in C2O4. We chose trial 1 and trial 2 specifically for that reason because (HgCl2) did not change.
change in rate = (change in C2O4)^{y}.
4=(C2O4)^{y}
4 = (2)^{y}
Thus, y must be 2 (or squared).
Now pick one where C2O4 is same but HgCl2 changes. That might be trial 1 and trial 4.
(HgCl2) changes by 0.105/0.052 = 2
(C2O4) doesn't change.
rate changes by ratio of 1.8E5/8.9E6 = 2.
Thus, rate = 2 = (HgCl2)^{x}
2 = (2)^{x} or x must be 1.
Now you know the rate equation is
rate = (HgCl2)(C2O4)^{2}
Take ANY set, 1 is as good as any, and substitute rate, (HgCl2), (C2O4), and calculate k.
Then trial 5 simply becomes sustitution of the HgCl2 and C2O4 given, you know x and y and k, so calculate rate.
Check my work. It's easy to make a mistake trying to type these exponents. If I made a typo I'll correct on a follow up.
Post any work if you have questions and explain in detail what you don't understand.
im sorry there's no work. I don't understand what were doing except for when we divide the changes
You mean I went through that whole thing and you didn't catch a glimpse? That took an hour. But here is what we did in summary. Study this, then go back over my step by step process and see if that helps.
The rate of an equation is governed, in this case, by this rate expression.
rate = k(HgCl2)^x(C2O4)^y.
The rate can be measured. Those have been tabulated in the table given in the problem. x and y must be determined experimentally. They have no relationship to the coefficients in the balanced equation you gave in the problem. The concentrations are given in the table, too. Here is what we do. We look for two trials where the concentration of one of the reagents is the same. In trial 1 and trial 2, the (HgCl2) stays the same BUT the (C2O4) changes by a factor of 2. So any change in the rate MUST be due to the C2O4. How much did that change. It changed by a factor of 4 (it went from 1.8E5 to 7.1E5). Therefore, if the C2O4 went up by 2 and the rate went up by 4, then the exponent y must be 2.
I did the same thing for trial 1 and trial 4 because the C2O4 stayed the same (look at the table to see) but (HgCl2) changed. I went through the same process for (HgCl2) and rate and found that the exponent x was 1. Now we know the rate equation is
rate = k(HgCl2)^1(C2O4)^2
So we pick trial 1, and substitute 1.8E5 for rate, k is what we solve for, (HgCl2) = 0.105 from the table, and (C2O4) = 0.15 from the table. We solve for k. I left that for you. Now we know k, x, and y.
You are asked to solve for the rate in trial 5.
rate = k(HgCl2)(C2O4)^2
rate = substitute your value for k (0.08)(0.10).
Solve for rate. I'll be happy to check your value.
rate for trial 5 = k(.080)(.10) = .008
No.
rate for trial 5 = k(0.08)(0.1)^2
You forgot the squared part.
I found k = 0.00762
I found rate for trial 5 = 6.1 x 10^6 but check my work. Check my arithmetic.
rate constant is 4 because from trial 1 to 2 it was +4 and from 3 to 4 it was +4
rate constant is k and k is evaluated this way.
rate = k(HgCl2)(C2O4)^2
Use trial 1 but any will do.
rate = 1.8E5 from the table.
k is what we solve for.
(HgCl2) = 0.105 in the table.
(C2O4) = 0.15 in the table.
substitute in the equation.
1.8E5=k(0.105)(0.15)^2
solve for k.
k = 0.00762
Respond to this Question
Similar Questions

Chemistry
How do you find rate of production of Cl? 
chemistryplease help
I don't know how to find generic rates and how to find rate of production? 
Chemistry
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? 
Chemistry
A chemist does a reaction rate analysis on the following reaction: 2CO (g) + O2 (g) → 2CO2 (g) She collects the following data: Trial Initial Concentration of CO (M) Initial Concentration of O2 (M) Instantaneous Reaction Rate … 
Chemistry urgent!
When a solution of sodium oxalate (Na2C2O4) is added to a solution of lanthanum (III) chloride, lanthanum oxalate (molar mass 541.86 g/mol) The balanced net ionic equation for the reaction is: 3C2O4^2 (aq) + 2La^3+ (aq) > La2(C2O4)3 … 
chemistry
For the reaction, 2 XO + O2 = 2 X02, some data obtained from measurement of the initial rate of reaction at varying concentrations are given below. run # [XO] [O2] rate, mol Ll sl 1 0.010 0.010 2.5 2 0.010 0.020 5.0 3 0.030 0.020 … 
Chemistry
For the reaction, 2 XO + O2 = 2 X02, some data obtained from measurement of the initial rate of reaction at varying concentrations are given below. run # [XO] [O2] rate, mol Llsl 1 0.010 0.010 2.5 2 0.010 0.020 5.0 3 0.030 0.020 … 
Chemistry
For the reaction, 2 XO + O2 = 2 X02, some data obtained from measurement of the initial rate of reaction at varying concentrations are given below. run # [XO] [O2] rate, mol Llsl 1 0.010 0.010 2.5 2 0.010 0.020 5.0 3 0.030 0.020 … 
Chemistry
For the reaction, 2 XO + O2 = 2 X02, some data obtained from measurement of the initial rate of reaction at varying concentrations are given below. run # [XO] [O2] rate,mol Ll s1 1 0.010 0.010 2.5 2 0.010 0.020 5.0 3 0.030 0.020 … 
science
what substance is being reduced in the following redox reaction?