Question

Saved Help Save & E The decomposition of hydrogen peroxide is a first-order process with a rate constant of 106 x 10- 3 min-1 How long will it take for the concentration of H202 to drop from 0.0200 M to 0.0120 M? 481 min 7.55 min <1 min

Answer 1

Q1. correct option is : 481 min

Q2. correct option is : The partial pressure of hydrogen will increase.

Explanation

Q1. The integrated first order rate law is given as

ln([A] / [A]o) = -(k) * (t)

where [A]o = initial reactant concentration = 0.0200 M

[A] = final reactant concentration = 0.0120 M

k = rate constant = 1.06 x 10-3 min-1

t = time passed

Substituting the values,

ln(0.0120 M / 0.0200 M) = -(1.06 x 10-3 min-1) * (t)

ln(0.6) = -(1.06 x 10-3 min-1) * (t)

-0.5108 = -(1.06 x 10-3 min-1) * (t)

t = (-0.5108) / (-1.06 x 10-3 min-1)

t = 481.9 min