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:
<span>Trial [A] [B] Rate </span><span>
<span> (M)
(M) (M/s)
</span>
<span>1 0.50
0.010 3.0×10−3 </span>
<span>2 0.50 0.020 6.0×10−3 </span>
<span>3 1.00
0 .010 1.2×10−2</span></span>
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
= 0.531 mol O₂