Question

What is the rate law for the reaction below:A + B + C Dif the following data were collected?Exp [A]0 [B]0 [C]0 Initial Rate1 0.4 1.2 0.7 2.32x10-32 1.3 1.2 0.9 7.54x10-33 0.4 4.1 0.8 9.25x10-24 1.3 1.2 0.2 7.54x10-3This is the explanation I was given but I still don't understan it. Can you please show the steps for solving the exponents of the rate laws "with algebra"?:The general form of the rate law has to be:rate = k [A]x [B]y [C]zUsing the experiments given to set up several equations, we get:2.32x10-3 = k(0.4)x(1.2)y(0.7)z7.54x10-3 = k(1.3)x(1.2)y(0.9)z9.25x10-2 = k(0.4)x(4.1)y(0.8)z7.54x10-3 = k(1.3)x(1.2)y(0.2)zWith algebra, we find x=1, y=3, and z=0. Plugging these values in, we get the rate law as:rate = k [A] [B]3 [C]0 = k [A] [B]3We can now plug in any experiment into the rate law above to solve for k. Plugging in experiment 1:2.32x10-3 = k(0.4)(1.2)3k = 3.36x10-3The complete rate law:rate = 3.36x10-3 [A] [B]3

Answer 1

Answer:

rate = k  [A] [B] ³

Explanation:

Set up a table with the data given in the question to study the dependence of the reaction rate on the concentrations of reactants.

[A]             [B]             [C]             Rate           Experiment #

0.4            1.2             0.7          2.32 x 10⁻³        (1)

1.3            1.2             0.9          7.54 x 10⁻³        (2)

0.4             4.1            0.8          9.25 x 10⁻²        (3)

1.3              1.2             0.2         7.54 x 10⁻³        (4)

The rate law will have the form : rate =  k [A] ^x   [B]  ^y    [C]   ^z

Comparing experiments (2) and (4) we have to conclude that  the rate  is zero order with respect to  [C]  since keeping [A]  and [B]  the same and varying [C] did not change the rate (i.e ,no dependence on  [C]) .

Now we know the rate law has the form rate =  k [A] ^x  [B]  ^y

Comparing (1) and (4) we keep [B] constant and increase [A] by a factor of  1.3/.4 = 3.25 and the rate increased by a factor of 0.00754 / 0.00232 =3.25, so we can conclude that the rate law is first order with respect to  [A]

Finally, comparing (1) and (3) while keeping  [A]  constant  increasing [B] by a factor of 4.1/1.2 = 3.416, the rate increased by a factor of 0.0925/0.00232 = 40, it is not entirely clear the dependence with respect to  [B] .

In this case we can always set up the following equation which is obtained by dividing  equation (3) by (1)

(4.1 / 1.2)^x = 0.0925/0.00232

taking natural log to both sides of the equation

x ln 3.4167 = ln 40

x = 3.69/1.23 = 3

So the dependence with respect to   [C] is three.

The rate law is :

rate = k  [A] [B] ³