Question

1) A first-order reaction has a half-life of 15.5 s . How long does it take for the concentration of the reactant in the reaction to fall to one-eighth of its initial value?

2) The rate constant for a certain reaction is k = 4.30×10−3 s−1 . If the initial reactant concentration was 0.350 M, what will the concentration be after 8.00 minutes?

Answer 1

1) For first order recation,

half life t_1/2 = 0.693 /k where k is rate constant

k = 0.693/ t_{1/2 --- Eq (1)}

k = 1/t ln { [A]_o/[A]_t} -----Eq (2)

From Eqs (1) and (2),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t} ------Eq (3)

Given that

half life t_1/2 = 15.5 s

time t = ?

Initial amount = [A]_o

Final amount [A]_t = [A]_o/8

Substitute all the values in Eq (3),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t}

0.693/ 15.5 = (1/t) ln {[A]_o/ [A]_o/8}

t = 15.5 x ln {8} / 0.693

= 46.5 s

t = 46.5 s

Therefore, after 46.5 seconds the concentration of the reactant in the reaction to fall to one-eighth of its initial value.

2)

Given that k = 4.30×10⁻³ s⁻¹

It is rate constant of first order reaction because units of first order reaction rate constant = s^-1

For first order recation,

k = 1/t ln { [A]_o/[A]_t} -----Eq (2)

Given that

time t = 8 min = 480 s

Initial concentration [A]_o = 0.35 M

Final concentration [A]_t = ? M

k = (1/t) ln {[A]_o/ [A]_t}

4.30×10⁻³ s⁻¹ = (1/480 s ) ln {0.35 M/ [A]_t}

[A]_t = (0.35 M)  x e ^{-[4.30×10−3 s−1 x 480 s]}

   = 0.044 M

Final concentration = 0.044 M

Therefore, 0.044 M is the concentration of reactant after 8 min.