Two isomers (A andB) of a given compound dimerize as follows:
Both processes are known to be second order in reactant, and k₁ is known to be 0.25 L/mol • s at 25° C, where

In a particular experiment, A and B were placed in separate containers at 25° C, where [A]₀ = 1.0 * 10^-² M and [B]₀ = 2.5 *10^-² M. It was found that [A] = 3[B] after the reactions progressed for 3.0 min.
Calculate the half-life for the reaction involving A.

Question

Two isomers (A andB) of a given compound dimerize as follows:

Both processes are known to be second order in reactant, and k₁ is known to be 0.25 L/mol • s at 25° C, where

In a particular experiment, A and B were placed in separate containers at 25° C, where [A]₀ = 1.0 * 10^-² M and [B]₀ = 2.5 *10^-² M. It was found that [A] = 3[B] after the reactions progressed for 3.0 min.
Calculate the half-life for the reaction involving A.

Quizplus · Accepted Answer

The ratio of [A] to [B] at the beginning of the reaction is 1:3. This means that there is more B available to react with A. Since both reactions are second order, the rate of the overall reaction will be determined by the rate of the reaction involving A.

Using the given rate law, we can set up a ratio of the rate constants for the two reactions:

k[A]^2 = 0.25 L/mol • s

k[B]^2 = 0.25 L/mol • s

k[A]^2/k[B]^2 = [B]^2/[A]^2

(1/k[B])^2/(1/k[A])^2 = 3^2

k[B]/k[A] = 1/9

This means that the rate of reaction involving B is 9 times slower than the rate of reaction involving A.

We can use the given concentration ratios to determine how much of each reactant has reacted at the 3-minute mark:

[A] = [A]0 - 2x

[B] = [B]0 - x

where x is the amount reacted.

At the 3-minute mark, [A] = 3[B], so:

[A]0 - 2x = 3([B]0 - x)

Solving for x, we get:

x = [A]0/5

This means that at the 3-minute mark, 2/5 of each A molecule has reacted.

The half-life of a second-order reaction is given by:

t1/2 = 1/(k[A][A]0)

At the 3-minute mark, [A] = [A]0/5. Plugging in the given values, we get:

t1/2 = 1/(0.25 L/mol • s * (1/5 M)^2)

t1/2 = 4 seconds

Converting to the given units:

t1/2 = 4.0 x 10^-12 s

t1/2 = 4.0 picoseconds

So the answer is choice A, with a half-life of 4.0 x 10^-12 s or 4.0 picoseconds.

Two Isomers (A AndB) of a Given Compound Dimerize as Follows