Question

A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _________ min. (Round off to the Nearest Integer).

Hint:

Write [A] and [B] as exponentials in t and set [A]/[B] = 16.

Answer 1

Correct Answer: 108

Rate constants: \(k_A = \ln 2/54\), \(k_B = \ln 2/18\).
\([A] = [A]_0 e^{-k_A t}\) and \([B] = [B]_0 e^{-k_B t}\) with \([A]_0 = [B]_0\).
\(\dfrac{[A]}{[B]} = e^{(k_B - k_A)t} = 16\).
\(k_B - k_A = \ln 2\left(\dfrac{1}{18} - \dfrac{1}{54}\right) = \ln 2 \times \dfrac{2}{54} = \dfrac{\ln 2}{27}\).
So \(\dfrac{t}{27}\ln 2 = \ln 16 = 4\ln 2\), giving \(t = 108\) min.
Answer: 108