For the simple decomposition reaction below, rate = k[AB]2 and k = 0.30 L/mols.
AB(g) A(g) + B(g)
If the initial concentration of AB is 1.20 M, what is [AB] after 10.0 s?
Is that rate = k[AB]^2? And I assume the units are L/mol*s. If so then it is a second order reaction and
(1/A) - (1/Ao) = kt
Substitute and solve for A
To find the concentration of AB after 10.0 seconds, we need to use the rate equation and integrate it with respect to time.
The rate equation for the given decomposition reaction is rate = k[AB]^2, where k is the rate constant.
Integrating the rate equation gives us:
∫d[AB]/[AB]^2 = k*dt
Integrating both sides of the equation will give us the relationship between the initial concentration of AB and its concentration at a given time t.
∫d[AB]/[AB]^2 = ∫k*dt
To solve the integral, we can use partial fraction decomposition:
∫d[AB]/[AB]^2 = ∫(A/[AB] + B/[AB]^2)d[AB]
Where A and B are constants that we need to determine.
Integrating both sides gives us:
-1/[AB] = Aln([AB]) - B/[AB] + C
Where C is the constant of integration.
Rearranging the equation gives us:
ln([AB]) = -1/(A - B[AB]) + C
We can now substitute the initial concentration of AB, [AB]0, and the time t into the equation to find [AB] at a specific time:
ln([AB]) = -1/(A - B[AB]) + C
ln([AB]0) = -1/(A - B[AB]0) + C
Now we can solve for A, B, and C using the given information and then use the equation to find [AB] at 10.0 seconds.