Question

Part A:

A certain first-order reaction (A→products) has a rate constant of 6.30×10⁻³ s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?

Part B:

A certain second-order reaction (B→products) has a rate constant of 1.30×10⁻³ M−1⋅s−1 at 27 ∘C and an initial half-life of 264 s . What is the concentration of the reactant B after one half-life?

Answer 1

K   = 2.303/t log[A]0/[A]

K   = 6.30*10^-3 s^-1

t    = 2.303/(6.30*10^-3)log(100/6.25)

      = 2.303*1.204/(6.30*10^-3)

        = 440s

          = 7.34minutes

part-B

n = 2 for second order

t1/2   = 1/K[B]0

[B]0   = 1/t1/2*K

          = 1/(264*1.3*10^-3) = 1/0.3432   = 2.9137M

After one half life the concentration

[B]   = 2.9137/2    = 1.45685M