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.

If products A and B produce product C, a general equation for the formation of C is of the kind rate = k*[A]^a * [B]^b

[ ] is used for the concentration of each compound.

When k, m and n are unknown, chemists run lab trials to calculate them..

You have these data from 3 trials

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

Trials 1 and 2 are run at constant [A] which permits to calculate the exponent b, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^a * [B1]^b

rate 2 = 6.0*10^-3 = k [A2]^a * [B2]^b

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

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

6.0 / 3.0  =( 0.020 / 0.010)^b => 2 = 2^b => b = 1

 
In the same way trials 1 and 3, which were run at constan [B], are used to calculate a

rate 3 / rate 1 = 12 / 3.0   = (1.0)^a / (0.5)^a => 4 = 2^a => 2^2 = 2^a => 2 = a

Now use any of the data to find k

With the second trial: rate = 6*10^-3 m/s = k (0.5)^2 * (0.02) =>

k = 6.0*10^-3 M/s / (0.05 M^3) = 0.12 M^-2 s^-1

With the calcualted values of k, a and b you 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

4.5 * 10^-3 M/s

Ithink it’s one of those things that you put the object on