Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m

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