Question

A certain substance, initially at 0.10 M in solution, decomposes by second-order kinetics. If the rate constant for this process is 0.40 L/mol • min, how much time is required for the concentration to reach 0.020 M?

Answer 1

Answer:-

Given:-

Initial concentration of substance [A0] = 0.10 M = 0.10 mol / L

Final concentration of substance [A] = 0.020 M = 0.020 mol / L

rate constant (k2) =  0.40 L/mol.min

time (t) = ?

According to the second-order kinetics the rate of reaction depends upon the concentration of two reactants or squre of the single reactant present in the reaction. So according to the formula.

1 /  [A] = 1 / [A0] + k2 t

where

[A] = Final concentration of substance

[A0] = Initial concentration of substance

k2 = rate constant of second order reaction

t = time

therefore

1 /  Final concentration of substance [A] = 1 / Initial concentration of substance [A0] + k2 time (t)

1 /  0.020 mol / L = 1 / 0.10 mol / L   + 0.40 L/mol.min time (t)

0.40 L/mol.min time (t) = 1 /  0.020 mol / L - 1 / 0.10 mol / L

0.40 L/mol.min time (t) = 0.10 - 0.020 /  0.020 0.10 mol / L

0.40 L/mol.min time (t) = 0.080 /  0.002 mol / L

0.40 L/mol.min time (t) = 40 L / mol

time (t) = 40 L / mol / 0.40 L/mol.min

time (t) = 100 minutes (i.e the answer)