Question

Two reactants A and B decompose independently by first order kinetics. Their half-lives are: $t_{1/2}(A) = 10 min$ $t_{1/2}(B) = 20 min$ Initially, $[A]_0 = 4[B]_0$. After how many minutes will the concentration of A and B become equal?

• A. 10 mins
• B. 20 mins
• C. 30 mins
• D. 40 mins

Answer 1

Answer: (D)

40 mins

Answer 2

For a first-order reaction, the concentration at time $t$ is given by $[X]_t = [X]_0 (\frac{1}{2})^{t/t_{1/2}}$. We are given $t_{1/2}(A) = 10$ min, $t_{1/2}(B) = 20$ min, and $[A]_0 = 4[B]_0$. We need to find $t$ when $[A]_t = [B]_t$.

Setting up the equation: $[A]_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}(A)}} = [B]_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}(B)}}$

Substitute the given values: $4[B]_0 \left(\frac{1}{2}\right)^{\frac{t}{10}} = [B]_0 \left(\frac{1}{2}\right)^{\frac{t}{20}}$

Cancel $[B]_0$ and rearrange: $4 \left(\frac{1}{2}\right)^{\frac{t}{10}} = \left(\frac{1}{2}\right)^{\frac{t}{20}}$ $2^2 \cdot 2^{-\frac{t}{10}} = 2^{-\frac{t}{20}}$ $2^{2 - \frac{t}{10}} = 2^{-\frac{t}{20}}$

Equate the exponents: $2 - \frac{t}{10} = -\frac{t}{20}$ $2 = \frac{t}{10} - \frac{t}{20}$ $2 = \frac{2t - t}{20}$ $2 = \frac{t}{20}$ $t = 40$ min.