Question

The half-life for the first order decomposition of A is 200.0 sec. How much time must elapse for the concentration of A to decrease to (a) one-half: (b) one-sixteenth: (c) one-ninth of its initial concentration?

Answer 1

For a first-order reaction,
t_1/2 = 0.693/ K
K = rate constant

we know t_1/2 = 200 sec
therefore; K = 0.693/t_1/2 = 0.693/200 = 3.465 x 10^-3 sec^-1

(a) one-half;
t_{50% left} = t_1/2 = 200 sec
this is half life time = 200 secs

(b) one-sixteenth (1/16)
t_{6.25% left} = t_1/16 = 4 x t_1/2 = 4 x 200 sec = 800 sec

(c) one-ninth (1/9)

[A] = [A]₀ x e^-kt
ln[A]/[A]₀ = -KT

therefore;

ln(1/9) = - K x t_1/9
t_{1/9 left} = - ln(1/9)/ 3.465 x 10^-3 sec^-1 = 634 sec