The dimerization of NO2 has a rate constant at 25 degrees Celcius of 2.45x10^-2 L/mol min. What will be the concentration of NO2 after 120 seconds, given a starting concentration of NO2 of 11.5 mol/L?

7.4 mol/L

Explanation:

To solve problems involving the determination of the concentration of a reactant after a time given the rate constant for the reaction, one needs to know the rate law.

Given the units of the rate constant one can learn about the rate law of the reaction.

One can deduce the order by remembering that the rate has units of concentration per time = molarity / t = M/t.

Now M is equal to mol/L and one is working in units of time in minutes, as is the case in this question where k = 2.45 x 10⁻² L/mol min, and knowing:

rate = k[A]^n

You need to ask yourself what are the units I need to multiply with the L/mol min given in the question to get mol/L.

mol/Lmin = [A]^n (L/molmin) ⇒ if n = 2

mol/Lmin = mol²/L² x L/mol·min Therefore, the rate order is 2 and

r = k[A]²

In general the units of k given a general order n is 1 / (M ^n-1 s)

For a second order reaction, the integrated rate law is:

1/[A]t = 1 / [A]₀ + kt

So for t = 2 min, we have:

1 / [A]t = 1 / 11.5 mol/L + 2.45 x 10⁻² L/mol min (2 min)

= 0.0870 L/mol + 0.0490 L/mol = 0.136 L/mol

⇒ [A]t = 1 / 0.136 mol/L = 7.35 mol/L