Question
An organic compound undergoes first order decomposition. The time taken for its decomposition to $\frac{1}{8}\text{th}$ and $\frac{1}{10}\text{th}$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. The value of $\left[ \frac{t_{1/8}}{t_{1/10}} \right]$ is
[Given $\log_{10} 2 = 0.3$]
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
0.9
First Order Decomposition: Key Concepts
For a first-order reaction, the relationship between the rate constant ($k$), time ($t$), initial concentration ($[A]_0$), and concentration at time $t$ ($[A]_t$) is given by the integrated rate law:
$k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right)$This can be rearranged to solve for time:
$t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right)$Time Ratio Calculation
We are given times for specific fractional decompositions:
- $t_{1/8}$: Time for decomposition to $\frac{1}{8}$th of initial concentration. Here, $[A]_t = \frac{1}{8}[A]_0$.
- $t_{1/10}$: Time for decomposition to $\frac{1}{10}$th of initial concentration. Here, $[A]_t = \frac{1}{10}[A]_0$.
Using the integrated rate law:
- For $t_{1/8}$: $t_{1/8} = \frac{1}{k} \ln\left(\frac{[A]_0}{\frac{1}{8}[A]_0}\right) = \frac{1}{k} \ln(8)$
- For $t_{1/10}$: $t_{1/10} = \frac{1}{k} \ln\left(\frac{[A]_0}{\frac{1}{10}[A]_0}\right) = \frac{1}{k} \ln(10)$
Determining the Ratio $\left[ \frac{t_{1/8}}{t_{1/10}} \right]$
To find the required ratio, we divide the expression for $t_{1/8}$ by the expression for $t_{1/10}$:
$\frac{t_{1/8}}{t_{1/10}} = \frac{\frac{1}{k} \ln(8)}{\frac{1}{k} \ln(10)} = \frac{\ln(8)}{\ln(10)}$We can simplify $\ln(8)$ using logarithm properties:
$\ln(8) = \ln(2^3) = 3 \ln(2)$So the ratio becomes:
$\frac{t_{1/8}}{t_{1/10}} = \frac{3 \ln(2)}{\ln(10)}$Using the change of base formula for logarithms, $\frac{\ln(a)}{\ln(b)} = \log_b(a)$. Therefore:
$\frac{t_{1/8}}{t_{1/10}} = \frac{3 \log_{10}(2)}{\log_{10}(10)}$Given that $\log_{10}(10) = 1$ and $\log_{10}(2) = 0.3$:
$\frac{t_{1/8}}{t_{1/10}} = \frac{3 \times 0.3}{1} = 0.9$Was this answer helpful?
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