Question

a. A certain first-order reaction (A→products) has a rate constant of 3.00×10−3 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?

b. A certain second-order reaction (B→products) has a rate constant of 1.85×10⁻³M−1⋅s−1 at 27 ∘C and an initial half-life of 300 s . What is the concentration of the reactant B after one half-life?

Answer 1

a.)

k = 3.00×10⁻³ s⁻¹

T = 45 + 273 = 308 K

[A_o] = 100

[A] = 6.25

t = (1/k) ln [A_o]/[A]

t = (1/3.00×10⁻³) ln (100/6.25)

t = (1/3.00×10⁻³) ln 16

t = (1/3.00×10⁻³) x 2.77

t = 924 sec or 15.4 min

924 sec or 15.4 min is time taken for the concentration of the reactant, [A], to drop to 6.25% of the original concentration.

b.)

t_1/2 = 1/k[A]_o

t_1/2 = 300 s

k = 1.85×10⁻³ M⁻¹⋅s⁻¹

T = 27 + 273 = 303 K

[A]_o = 1/kt_1/2

[A]_o = 1/(1.85×10⁻³ M⁻¹⋅s⁻¹ ) x (300 s)

[A]_o = 1.80 M

The concentration of the reactant B after one half-life is

1/[A]_t = kt + 1/[A]_o

[A]_t = ?

[A]_o = 1.80 M

t = 300 s

1/[A]_t = 1.85×10⁻³ M⁻¹⋅s⁻¹ x 300 + 1/1.80

1/[A]_t = 0.555+ 0.555

1/[A]_t = 1.11

[A]_t = 1/1.11 = 0.90M

The concentration of the reactant B after one half-life is 0.90 M