Question

) Consider the reaction A à products. The rate law for this reaction is rate = k[A] where k=7.02 ´ 10^-3 M^-1 s^-1 at a particular temperature.

If the initial [A] = 0.0500 M, what is the initial rate?

What is the value of the half-life for this initial concentration of A?

What will be the half-life if the initial [A] = 0.0250 M

What will be the half-life if the initial [A] = 0.100 M

What will be the [A] after 2.00 minutes when we start with an initial [A] = 0.0500 M

How much time will it take for the [A] to decrease from 0.0500 M to 0.0250 M

How much time will it take for the [A] to decrease from 0.100 M to 0.0750 M

Answer 1

Note: You have mentioned the unit of the rate constant as M^-1s^-1 (which is for second-order reaction), whereas the rate law as rate = k[A] (which is for the first-order reaction).

So, I'm going according to the rate law.

Then the unit of 'k' should be only s^-1 not M^-1s^-1

According to the given data:

The initial rate = 7.02*10^-3 s^-1 * 0.05 M = 3.51*10^-4 Ms^-1

Formula: t_1/2 = 0.693/k = 0.693/7.02*10^-3 s^-1 ~ 99 s

The half-life will be the same whatever may be the value of [A]₀ since the half-life period in the first order reaction is independent of the initial concentration of A, i.e. t_1/2 = 99 s for any value of [A]₀.

If t = 2 min = 2*60 = 120 s and [A] = 0.05 M

Formula: k = (1/t) ln([A]₀/[A])

i.e. [A] = [A]₀ * e^-kt

= 0.05 * e^{-7.02*10^-3 * 120}

= 2.15*10^-2 M

If [A]₀ = 0.05 M, [A] = 0.025 M, then t = t_1/2 = 99 s

If [A]₀ = 0.1 M, [A] = 0.075 M, then t = t_1/4 = (1/k) ln([A]₀/[A]) = (1/7.02*10^-3) * ln(0.1/0.075) ~ 41 s