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  k[OH-][ClO2]2 = d[ClO2-]/dt
Yes No  d[ClO3-]/dt = k[OH-][ClO2]2
Yes No  k[OH-]2[ClO2] = -d[ClO2]/dt
Yes No  -d[ClO2]/dt = k[ClO2]2[OH-]
Yes No  k[OH-]2[ClO2]2 = -d[ClO2]/dt

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

(Units required.)

What would be the initial rate (-d[ClO2]/dt) for an experiment with [ClO2]0 = 0.140 mol/L and [OH-]0= 0.0660 mol/L?

(Units required.)

Explanation / Answer

First of all we shoul remember that d[ClO2]/dt) will be negative because it is the rate of comsuption.

Know with the data of the table we can find the X and Y for the reaction

rate= k [ClO2]X[OH]Y

For X we use the data where [OH] mantains constant the

ratio of rate = 4.68/1.17 = 4

ratio of [ClO2] = 0.340/0.170 = 2

then 4 = 2X   thus X= 2

For Y we do the same but with [ClO2] constant

ratio of rate = 4.68/2.34 = 2

ratio of [OH] = 0.150/ 0.075 = 2

2=2Y

Thus Y= 1

we have that the rate law will be

rate = [ClO2]2[OH]

Then the answer is

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

Now we can use any data from the table (same line) to calculate k

4.68 mol/ Ls = k (0.340 mol/L )2( 0.150 mol/L)

4.68 mol/ Ls = k (0.116 mol2/L2 )( 0.150 mol/L)

4.68 mol/ Ls = k (0.0174 mol3/L3 )

4.68 mol/ Ls/ (0.0174 mol3/L3 ) = k

269.9 L2/smol2 = k

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

-d[ClO2]/dt = 269.9 L2/smol2[0.140 mol/L]2[0.0660 mol/L]

-d[ClO2]/dt = 0.35 mol /sL

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