I am having a difficult time trying to find the correct way to find the rate ord

Question

I am having a difficult time trying to find the correct way to find the rate order of the reaction of oxalic acid and permanganate.

my date looks something like this:
Oxalic Acid
trial 1 rate: 6*10^-4; initial concentration=0.31 M final =0.15 M
trial 2 rate: 8 *10^-4 initial concentration=0.63 M final= 0.47 M
Permanganate
Trial 1 rate=3.7*10^-5; initial concentration 0.01 M final 0.01M
Trial 2 rate= 4.9*10^-5 initial concentration= 0.02 M final = 0.02M

The ratios I have come up with in order to solve look like this:

k(H2C2O4)x1/kH2C2O4x2 and K(MnO4-Y1/KMnO4-Y2

X and Y are the exponents I need to solve for and then I Must find the value of K, the rate constant. I am so confused! :((

HELP. I will rate ASAP! thank you

Explanation / Answer

Let oxalic acid be compound X and permanganate by compound Y
then

[X] represents conc of X and [Y] represents conc of Y at any instant t

Assuming elementary rate law, we can write rate of reaction to be

rate, r = k[X]a[Y]b

Let us first find rate w.r.t X

for this keep the conc of Y constant

so when, [X] = 0.31M , rate = 6e-4

so, 6e-4 = k[0.31]a[Y]b

similarly when [X] = 0.63M, rate = 8e-4

again we can write,

8e-4 = k[0.63]a[Y]b

take ratio then you will get

(8/6) = (0.63/0.31)^a

solve to get a = 0.406

similarly find rate w.r.t Y by keeping conc of X constant

follow same procedure

when [Y] = 0.01M, rate = 3.7e-5

and

when [Y] = 0.02M, rate = 4.9e-5

take ratio

(4.9/3.7) = (0.02/0.01)^b

solve to get b = 0.406

so rate = k[X]0.406[Y]0.406

to find K, the data here is insufficient as you need the value of rate and [X] and [Y]

what you have provided here doesn't give the value of rate when distinct amount on X and Y are used. that is we don't have the value of rate and conc of X and Y to calculate k.

Provide that information, and I will be able to help you.

or else,

you can work it yourself

you have rate law as rate = k[X]0.406[Y]0.406

if you have values for [X], [Y] and rate for a particular trial just substitute and find k

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