I need the correct way to find the rate order of the reaction of oxalic acid and

Question

I need the correct way to find the rate order of the reaction of oxalic acid and permanganate. This is based on a lab with experimental data. The rates from the trials were computed from the change of molarity divided by an average time. I need to find the X and Y exponent values and then plug it into the rate law formula to solve for K, the rate constant. The lab says that the X and Y exponents can be rounded to the nearest whole integer. I am extremely confused with this :(

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
not sure if that is right....

HELP. I will rate ASAP! thank you

Explanation / Answer

for oxalic acid 6*10^-4 = k[.31-.15]^X = k*.16^X 8*10^-4 = k[.63-.47]^X = k*.16^X So this clearly indicates it is not dependent on Oxalic Acid concentration X = 0 Also we get to know that from the same reactions that it also doesnt depend on concentration of KMnO4 so Y = 0 (Make the same equations) Hence X = 0 and Y = 0 its a 0th order reaction.

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