5. The rate constant for the reaction 3A  4B is 6.00 x 10-3 L•mol-1min-1. How long will it take the concentration of A to drop from 0.75 M to 0.25 M?

it will take 444.44 min to reach 0.25 M from 0.75 M

Explanation:

since the reaction constant is k = 6.00*10⁻³ L*mol⁻¹min⁻¹. For a reaction rate of the form

-dCa/dt=k*Caⁿ

doing a dimensional analysis

[dCa/dt] = mol / (L*min) = [k]*[Caⁿ] = L / (mol*min) * (mol/L) ⁿ

then only n=2 can comply with the dimensional analysis, therefore we get a the second order reaction. Thus

-dCa/dt=k*Ca²

-dCa/Ca² = k*dt

-∫dCa/Ca² = k*∫dt

(1/Ca₂ - 1/Ca₁) = k*Δt

Δt = 1/k * (1/Ca₂ - 1/Ca₁)

replacing values

Δt = 1/k * (1/Ca₂ - 1/Ca₁) = 1 / (6.00*10⁻³ L*mol⁻¹min⁻¹) * (1/0.25 M - 1/0.75 M) = 444.44 min