Question

A reacts to form $P.A$ plot of the reciprocal of the concentration of $A$ vs time is a straight line. When the initial concentration of $A$ is $1.0 \times 10^{-2} M$, its half-life is found to be $20\, \min$. When initial concentration of $A$ is $3.0 \times 10^{-3} M$ the half-life will be

• A. 20 min
• B. 40 min
• C. 56 min
• D. 67 min

Answer 1

Answer: (D)

Since, the plot of the reciprocal of the concentration of $A$ versus time is a straight line. It indicates that the given reaction is of second order.
For a second order reaction,
$t_{1 / 2}=\frac{1}{k a} 20(\min )$
$=\frac{1}{k \times 1.0 \times 10^{-2} M}$ ... (i)
$t_{1 / 2}=\frac{1}{k \times 3.0 \times 10^{-3} M}$ ... (ii)
$\frac{20}{t_{1 / 2}}=\frac{1 \times 3.0 \times 10^{-3}}{1.0 \times 10^{-2} \times 1}$
$=66.66 \approx 67\, \min$