pre-lab questin..

part a was to determine the rate law from the given table...
finally i was able to calculate ..rate law which was

k[ClO2]2[OH-]
since i got m =2 n=1 ..overall order was 3 ...

the equation was ..

2ClO2(aq) + 2OH-(aq)---ClO3-(aq) + ClO2-(aq) + H2O(l)

----------------------for the part (b) ... it says..

what would be the initial rate for an experiment with [ClO2]0 = 0.175 mol/L and [OH-]0 = 0.0844 mol/L?

i don't know how to figure out part B any clue .. how to approch for dis problem ..thnks
plz answer ,.and explan in detail .. thnks in advance...

First, You would set up the rate law equation.

r=k[OH-]1[CLO2]2

Next, plug in the numbers you have already been given and have figured out

r= 230 L2/M2xs [0.175]2 [.0844]

Then you should get 5.94x10e-1 as your answer

To find the initial rate for the given experiment, we will need to use the rate law equation and the initial concentrations of the reactants.

First, let's review the rate law equation you have determined:
rate = k[ClO2]^2[OH-]

In this equation, [ClO2] and [OH-] represent the concentrations of ClO2 and OH- ions, respectively, and k is the rate constant. The exponent values (2 and 1) in the rate law equation come from the stoichiometric coefficients of ClO2 and OH- in the balanced chemical equation.

To find the initial rate, we need to substitute the initial concentrations of [ClO2]0 = 0.175 mol/L and [OH-]0 = 0.0844 mol/L into the rate law equation.

rate = k[ClO2]^2[OH-]
rate = k(0.175)^2(0.0844)

Now, you need to know the value of the rate constant, k, in order to calculate the initial rate. The rate constant can be determined experimentally, usually by conducting several experiments with different initial concentrations and measuring the corresponding rates.

If you have been given the value of the rate constant, you can substitute it into the equation and solve for the initial rate. If the rate constant is not provided, then you would need to perform experiments to determine its value.