sorry i tired to make it a chart bbut it did not work the 3 numbers in order for each concenraion is as follows
[ClO2]0
0.500
0.100
0.100
and then
[oh-]
o.100
0.100
0.050
Initial rate
5.75* 10^-2
2.30* 10^-1
1.15* 10^-1
i need help with this pre lab question:|
plse help me!
teh following data were obained for the reaction
2ClO2(aq)+ 2OH^-(aq) -> ClO3^-(aq) + ClO2^-(aq) +H2O (l)
Where rate = delta[ClO2]/ delta t
(mol/L) (mol/L) (mol/L*s)
[ClO2]0 [OH-]0 Initial rate
0.0500 0.100 5.75*10^-2
0.100 0.100 2.30*10^-1
0.100 0.050 1.15*10^-1
a) determine the rate law and the value of the rate constant
b) what would be the initial rate for an eperiment with [ClO2]0= 0.175 mol/L and the [OH-]0=0.0844 mol/L
3 answers
Here is how you do part a. That should get you started.
rate = k*(ClO2)^x*(OH^-)^y so substitute run 1 and run 2 into this.
r1 = 5.75 x 10^-2=k*(0.050)^x*(0.100)^y
r2 = 2.30 x 10^-1=k*(0.100)^x*(0.100)^y
Now divide equation for r1 by equation for r2 and solve for x, the only unknown in the equation. I obtain x = 2
rate = k*(ClO2)^x*(OH^-)^y so substitute run 1 and run 2 into this.
r1 = 5.75 x 10^-2=k*(0.050)^x*(0.100)^y
r2 = 2.30 x 10^-1=k*(0.100)^x*(0.100)^y
Now divide equation for r1 by equation for r2 and solve for x, the only unknown in the equation. I obtain x = 2
I can read the chart you posted. If determining x, you do the same kind of thing to determine y 9but make sure the value for x cancels; i.e., you want to use run 2 and run 3. After determining x and y, then you can take any of the runs, substitute the values for concn in those runs along with the values for x and y and determine k, the rate law constant. For part b, simply plug in the values in the part b problem, plug in k just determined in part a, and solve for the rate.
2 answers