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

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

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

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.

To determine the rate law and the value of the rate constant, we need to use the method of initial rates. This method involves comparing the changes in reactant concentrations to the changes in the initial rate of the reaction. In this case, there are two reactants, ClO2 and OH-, so we need to determine how changes in their concentrations affect the rate.

a) To determine the rate law, we can start by comparing the initial rates for the different sets of concentrations provided:

[ClO2]0 = 0.0500 M, [OH-]0 = 0.100 M, Initial rate = 5.75 * 10^-2 M/s
[ClO2]0 = 0.100 M, [OH-]0 = 0.100 M, Initial rate = 2.30 * 10^-1 M/s
[ClO2]0 = 0.100 M, [OH-]0 = 0.050 M, Initial rate = 1.15 * 10^-1 M/s

By comparing the initial rates, we can determine the relationship between the concentrations of ClO2 and OH-.

Let's start by comparing the first and second set of concentrations:

(0.100/0.0500) ^x = (2.30 * 10^-1) / (5.75 * 10^-2)
2^x = 4
x = 2

Now, let's compare the second and third set of concentrations:

(0.100/0.100) ^y = (1.15 * 10^-1) / (2.30 * 10^-1)
1^y = 0.5
y = 0

Therefore, the rate law for this reaction is:

Rate = k[ClO2]^2[OH-]^0
= k[ClO2]^2

Since [OH-] does not affect the rate of the reaction, its exponent is 0.

Now, to determine the value of the rate constant (k), we can choose any set of initial concentrations and substitute them into the rate law equation. Let's use the first set of concentrations:

5.75 * 10^-2 M/s = k(0.0500 M)^2
k = (5.75 * 10^-2 M/s) / (0.0500 M)^2
k ≈ 2.3 M^-2s^-1

Therefore, the rate law for this reaction is: Rate = k[ClO2]^2, and the value of the rate constant (k) is approximately 2.3 M^-2s^-1.

b) To find the initial rate for an experiment with [ClO2]0 = 0.175 mol/L and [OH-]0 = 0.0844 mol/L, we can use the rate law equation:

Rate = k[ClO2]^2

Substituting the given concentrations:

Rate = (2.3 M^-2s^-1)(0.175 M)^2
Rate ≈ 0.0688 M/s

Therefore, the initial rate for the given experiment would be approximately 0.0688 M/s.