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
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...
1 answer
1 answer