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

To solve this problem, we need to understand the concept of a first-order decomposition reaction and the specific terms given, such as \( t_{1/8} \) and \( t_{1/10} \).

A first-order decomposition reaction is given by:

\(\ln \left(\frac{[A]_0}{[A]}\right) = kt\)

where \( [A]_0 \) is the initial concentration of A and \( [A] \) is the concentration at time \( t \). The rate constant is \( k \).

For first-order reactions, the time taken for the concentration to decrease to a fraction of its initial value is given by:

\(t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]}\right)\)

Given:

• \( t_{1/8} \) is the time at which the concentration of A becomes \( \frac{1}{8} \) of the initial concentration.
• \( t_{1/10} \) is the time at which the concentration of A becomes \( \frac{1}{10} \) of the initial concentration.

The expressions for \( t_{1/8} \) and \( t_{1/10} \) are:

• \(t_{1/8} = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_0 / 8}\right) = \frac{1}{k} \ln(8)\)
• \(t_{1/10} = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_0 / 10}\right) = \frac{1}{k} \ln(10)\)

Our task is to calculate \( \frac{t_{1/8}}{t_{1/10}} \times 10 \):

\(\frac{t_{1/8}}{t_{1/10}} = \frac{\ln(8)}{\ln(10)}\)

Calculating:

\(\ln 8 = \ln(2^3) = 3 \ln 2\)

\(\ln 10 = \ln(10) \text{ is approximately } 2.302\)

Thus:

\(\frac{t_{1/8}}{t_{1/10}} = \frac{3 \ln 2}{\ln 10} \approx \frac{3 \times 0.693}{2.302} \approx \frac{2.079}{2.302} \approx 0.903\)

Finally, multiplying by 10:

\(\frac{t_{1/8}}{t_{1/10}} \times 10 \approx 0.903 \times 10 = 9.03\)

The closest option is:

9