A reaction is said to be diffusion controlled when it occurs as fast as the reactants diffuse through the solution. The following is an example of a diffusion controlled reaction.
The rate constant for this reaction is 1.4x1011 M-1s-1 at 25oC and the reaction obeys the following rate law.
rate = k(H3O+)(OH-)
Calculate the rate of reaction in a neutral solution (pH=7.00).
What's the problem with substituting 10^-7 for hydronium and hydroxide ions along with k and solving?
So would it be
rate = (10^-7)*(10^-7)*(1.4X10^11)
That's what I would do.
To calculate the rate of reaction in a neutral solution, we need to determine the concentrations of H3O+ and OH- ions.
In a neutral solution, the concentration of H3O+ ions (also known as hydronium ions) is equal to the concentration of OH- ions. Therefore, in the rate law equation, both concentrations can be considered as the same value.
Since the concentration of each ion is equal in a neutral solution, we can denote their concentration as [H3O+] = [OH-] = x, where x is the concentration in mol/L.
Substituting this into the rate law equation, we get:
rate = k(x)(x)
The rate constant (k) is given as 1.4x10^11 M-1s-1. We can plug in this value into the equation:
rate = (1.4x10^11 M-1s-1)(x)(x)
We can simplify this equation to:
rate = 1.4x10^11 x^2 M-2s-1
Now, we need to determine the value of x, which represents the concentration of H3O+ (or OH-) ions in a neutral solution.
In a neutral solution, the pH is 7, which corresponds to a H3O+ concentration of 10^-7 M (using the pH scale where pH = -log[H3O+]). Therefore, x = [H3O+] = [OH-] = 10^-7 M.
Plugging this value into the rate equation:
rate = 1.4x10^11 (10^-7)^2 M-2s-1
Simplifying further:
rate = 1.4x10^11 (10^-14) M-2s-1
Finally, we can convert the rate units to a more commonly used notation. Since M-2s-1 can be written as L2mol-2s-1, we have:
rate = 1.4x10^11 x 10^-14 L2mol-2s-1
rate = 1.4x10^-3 L2mol-2s-1
Therefore, the rate of reaction in a neutral solution (pH = 7.00) is 1.4x10^-3 L2mol-2s-1.