Rate Law from Initial Rates (B) The above reaction was studied (at a particular

Question

Rate Law from Initial Rates (B)

The above reaction was studied (at a particular temperature) and the following data (for rate = -d[ClO2]/dt) were obtained:

Which of the following expressions for the rate law are consistent with the above experimental data.

Note that the rate constant may be different for each of the choices below (i.e. k in choice A may be not equal to k in choice B etc).

Yes No  -d[ClO2]/dt = k[ClO2]2[OH-]
Yes No  k[OH-]2[ClO2]2 = -d[ClO2]/dt
Yes No  d[ClO2-]/dt = k[OH-]2[ClO2]
Yes No  -d[OH-]/dt = k[ClO2]2[OH-]
Yes No  k[ClO2]2[OH-] = d[ClO2-]/dt

For Rate = -d[ClO2]/dt, calculate the rate constant (k).

What would be the initial rate (-d[ClO2]/dt) for an experiment with [ClO2]0 = 0.205 mol/L and [OH-]0 = 0.0242 mol/L?

Explanation / Answer

Let the rate law be,

rate = k[ClO2]^x.[OH-]^y

k = rate constant

x and y are order with respect to each reactant

From run 1 and 3, [OH-] is constant

rate 1/rate 3 = (2.02/5.06 x 10^-1) = (0.220/0.110)^x

taking log on both sides,

x = 2

From run 1 and 2, [ClO2] is constant

rate 1/rate 2 = (2.02/1.01) = (0.190/0.095)^y

taking log on both sides,

y = 1

So the rate expression for the reaction becomes,

-d[ClO2]/dt = k[ClO2]^2[OH-]

-d[OH-]/dt = k[ClO2]^2[OH-]

d[ClO2-]/dt = k[ClO2]^2[OH-]

Rest would be No.

Rate constant k,

k = 2.02/(0.22)^2.(0.190) = 219.661 L^2.mol-2.s-1

Initial rate when [ClO2]o = 0.205 mol/L and [OH-]o = 0.0242 mol/L

Initial rate = 219.661 x (0.205)^2.(0.0242) = 0.223 mol.L-1.s-1

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