Question

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?

Answer 1

Answer:

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