Asked by Jvaline
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...
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...
Answers
Answered by
Jake
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
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
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