Question

For the $1^{\text{st}}$ order decomposition reaction $\text{A} \to \text{P}$. The value of $\frac{t_{1/8}}{t_{1/10}} \times 10$ will be :-
$t_{1/8} =$ time at which concentration of A become 1/8 of initial concentration.
$t_{1/10} =$ time at which concentration of A becomes 1/10 of initial concentration.

• A. 3
• B. 6
• C. 9
• D. 0.9

Hint:

The time ratio depends only on the logarithm of the concentration reduction factors: $t_1/t_2 = \ln(x_1)/\ln(x_2)$.

Answer 1

The Correct Option is C

For a first-order reaction, $kt = \ln \frac{[A]_0}{[A]_t}$.
For $t_{1/8}$, $[A]_t = \frac{[A]_0}{8}$.
$k t_{1/8} = \ln \left( \frac{[A]_0}{[A]_0/8} \right) = \ln 8$.
For $t_{1/10}$, $[A]_t = \frac{[A]_0}{10}$.
$k t_{1/10} = \ln \left( \frac{[A]_0}{[A]_0/10} \right) = \ln 10$.
Ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \log_{10} 8$.
Given $\log 2 \approx 0.3$, $\log 8 = 3 \log 2 = 0.9$.
So ratio is approximately $0.9$.
Value required: $\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$.