Calculating rate constant for reaction between potassium permanganate and oxalic acid

I'm writing a lab report investigating the effects of temperature on the rate constant in the reaction between potassium permanganate and oxalic acid. For my practical I've measured the time of colour change (purple to colourless).

I've confirmed that the rate expression is: $$\mathrm{rate}=k[\ce{KMnO4}][\ce{H2C2O4}]$$

However, I am unsure about how to calculate the rate constant for the given reaction. I have rate as: $$\mathrm{rate}=\frac{[\ce{MnO4^-}]}{t}$$ where the $$\ce{MnO4^-}$$ ion is disappearing. However, calculating this and using the rate expression only gives me an average value. I understand that I would use the integrated rate laws if the reaction was second-order with one reactant, but here there are two reactants so I am unsure what to do.

Or am I just overthinking this?

• Note that measuring the rate constant of this reaction is very tricky as the reaction is autocatalyzed by the forming Mn^2+ ions. Jan 27 at 11:18

If a reaction is second order in $$\ce{A}$$ and $$\ce{B}$$, with $$\ce{[A]_0 ≠ [B]_0}$$, and with a rate law given by $$\pu{d[A]/dt = d[B]/dt = k[A][B]}$$, the integrated law is : $$\ce{\frac{1}{[B]_0 - [A]_0}ln{ {\frac{[A]_0 [B]}{[A][B]_0}}}} = kt$$ As it is pretty long to establish this law, I recommend that you open a book on chemical kinetics like A. A. Frost and R. G. Pearson, Kinetics and Mechanism, John Wiley & Sons, New York, 4th ed. 1965, p. 16.