Consider two reaction vessels, one containing A and the other containing B with equal concentrations at t = 0 / If both substances decompose by first order kinetics where ka = 4.50e-4s^-1 kb=3.70e-3s^-1 how much time must pass to reach a condition such that [A] = 4.00[B]?

It must pass 7.1 min

Explanation:

For first order kinetics, the concentration of the reactive in function of time can be written as follows:

[A] = [A]₀e^ (-kt)

where:

[A] = concentration of reactant A at time t.

[A]₀ = initial concentration of A

k = kinetic constant

t = time

We want to know at which time the concentration of A is 4 times the concentration of B ([A] = 4.00[B]). These are the equations we have:

[A] = [A]₀e^ (-ka*t)

[B] = [B]₀e^ (-kb*t)

[A] = 4.00[B]

[A]₀ = [B]₀

Replacing [A] for 4[B] and [A]₀ = [B]₀ in the equation [A] = [A]₀e^ (-ka*t), we will get:

4[B] = [B]₀e^ (-ka*t)

if we divide this equation with the equation for [B] ([B] = [B]₀e^ (-kb*t)), we will get:

4 = e^ (-ka*t) / e^ (-kb*t)

Applying ln on both sides:

ln 4 = ln (e^ (-ka*t) / e^ (-kb*t))

applying logarithmic property (log x/y = log x - log y)

ln 4 = ln (e^ (-ka*t)) - ln (e^ (-kb*t))

applying logarithmic property (log xⁿ = n log x)

ln 4 = - ka*t * ln e - (-kb*t) * ln e (ln e = 1)

ln 4 = kb * t - ka * t

ln 4 = (kb - ka) t

t = ln 4 / (3.7 x 10⁻³ s⁻¹ - 4.50 x 10⁻⁴ s⁻¹) = 426. 6 s or 7.1 min