Question

The reaction A(aq) ---> B(aq) is a first order reaction with respect to A(aq). The concentration of A(aq) is reduced to 22.3 % of its initial value in 1.53 minutes. What is this reaction's half-life (in s)?

Answer 1

Solution:-

Let, the initial concentration of A = 100%

then, final concentration of A = 22.3%

Now, for first-order reaction,

k = (2.303 / t) log₁₀ ([Initial concentration] / [Final concentration])

Where,

k = Rate constant

t = time (= 1.53 min. = 1.53 × 60 s = 91.8 s)

k = (2.303 / 91.8) log₁₀ (100 / 22.3)

k = 1.635 × 10^-2  s^-1

Now,

Half life (t½) = (0.693) / k = (0.693) / (1.635 × 10^-2) = 42.39 s

Hence, half-life of the reaction is = 42.39 s

--------------------------------------------------------------