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The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

Trial      [A]      [B]         Rate
            (M)     (M)          (M/s)
1         0.50    0.010      3.0×10−3
2         0.50    0.020       6.0×10−3
3         1.00 0  .010       1.2×10−2


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s

The initial rate for the formation of 'c' at 25[tex]\rm ^\circ C[/tex] has been calculated to be 2.25 [tex]\rm \times\;10^-^2[/tex] M/s.

The rate of formation of the product has been dependent on the concentration of reactants.

The given reaction has been:

A + 2B [tex]\rightarrow[/tex] C

The rate of reaction can be given as:

Rate = k [tex]\rm [A]^a\;[B]^b[/tex]

By the experimental data:

[A]          [B]            Rate

0.5 M     0.01 M     3 [tex]\rm \times\;10^-^3[/tex] M/s

0.5 M     0.02 M    6 [tex]\rm \times\;10^-^3[/tex] M/s

1 M         0.0.1 M     1.2 [tex]\rm \times\;10^-^2[/tex] M/s

For the value of [A] = 0.5 M,

The possible rate can be: 3 [tex]\rm \times\;10^-^3[/tex] M/s or 6 [tex]\rm \times\;10^-^3[/tex] M/s

The ratio of two possible rates has been directly proportional to the concentration of b.

[tex]\rm \dfrac{Rate\;2}{Rate\;1}\;=\;\dfrac{[B2]^b}{[B1]^b}[/tex]

[tex]\rm \dfrac{6.0\;\times\;10^-^3}{3.0\;\times\;10^-^3}\;=\;(\dfrac{0.02}{0.01})^b[/tex]

2 = [tex]\rm 2^b[/tex]

b = 1

Since rate 3 and rate 1 have the same concentration of [B], the possible value for a can be calculated with their ratio.

[tex]\rm \dfrac{Rate\;3}{Rate\;1}\;=\;\dfrac{[A3]^a}{[A1]^a}[/tex]

[tex]\rm \dfrac{1.2\;\times\;10^-^2}{3.0\;\times\;10^-^3}\;=\;(\dfrac{1}{0.5})^a[/tex]

4 = [tex]\rm 2^a[/tex]

a = 2

The value of rate constant k can be calculated by substituting the values for the rate 1.

Rate =  k [tex]\rm [A]^a\;[B]^b[/tex]

3 [tex]\rm \times\;10^-^3[/tex] M/s = k [tex]\rm [0.5]^2\;[0.01]^1[/tex] M

3 [tex]\rm \times\;10^-^3[/tex] = k (0.25) (0.01)

3 [tex]\rm \times\;10^-^3[/tex] = 0.0025 k

k = 1.2 [tex]\rm M^-^2\;s^-^1[/tex]

To calculate the initial rate for the formation of c at [A] = 0.5 M, and [B] = 0.075 M.

Rate = 1.2 [tex]\rm [0.5]^2\;[0.075]^1[/tex]

Rate = 0.0225 M/s

Rate = 2.25 [tex]\rm \times\;10^-^2[/tex] M/s

The initial rate for the formation of 'c' at 25[tex]\rm ^\circ C[/tex] has been calculated to be 2.25 [tex]\rm \times\;10^-^2[/tex] M/s.

For more information about the rate of reaction, refer to the link:

https://brainly.com/question/25159894

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