For a reaction the initial concentration of [KMnO4] was 1.3 and for [H2C2O4] was .7M. The reaction took 5min. What is the value of the rate constant at this temperature

K= -0.286 Mol^-1.S^-1

Explanation:

FIRSTLY, WE WRITE A CHEMICAL REACTION EQUATION FOR THE REACTION BETWEEN OXALIC ACID AND POTASSIUM PERMANGANATE.

KMnO2(aq) + H2C2O4(aq) >  Mn2+(aq) + CO2(g) + H2O(l)

(purple)            (colorless)

THE RATE OF THE REACTION IS MEASURED AND MONITORED BY CHECKING FOR THE DISAPPEARANCE OF THE PURPLE COLOR OF THE POTASSIUM PERMANGANATE.

The rate of the reaction is given by : -Δ[KMnO4]/ΔT = -Δ[ H2C2O4]/ΔT = Δ[ Mn2+]/ΔT  

When all of the purple color is gone, we can say that the reaction is finished and that the final  concentration of the KMnO4 is now zero. Since the initial time for ΔT is zero, the equation for the rate of

disappearance of the potassium permanganate becomes:

Rate = Δ[KMnO4]/ΔT= (0-[initial KMnO4]) ÷ (elapsed time -0) = [Initial KMnO4 ] ÷ elapsed time

Hence, the rate of the chemical reaction is thus;

-1.3/5 = -0.26M/s

Now, to get the rate constant for the reaction, we need to know the rate law and this can be expressed as follows:

For a reaction A+B---> C The general rate law would be: Rate = k[A]^m[B]^n

In this particular case,

Rate =k [KMnO4][ H2C2O4]

Kindly note that m and n in this case is 1 each. We have calculated our rate from above. Hence:

-0.26 = k(1.3)(0.7)

k = -0.26÷(1.3)(0.7)

K= -0.286 Mol^-1.S^-1