JEE Advanced · Chemistry · 18. Chemical Kinetics
An organic compound undergoes first-order decomposition. The time taken for its decomposition to \(1 /\) 8 and \(1 / 10\) of its initial concentration are \(t_{1 / 8}\) and \(t_{1 / 10}\) respectively. What is the value of \(\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10\) ? \(\left(\log _{10} 2=0.3\right)\)
- A 9
- B 12
- C 45
- D 3
Answer & Solution
Correct Answer
(A) 9
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
t_{1 / 8}=\frac{2.303 \log 8}{k}=\frac{2.303 \times 3 \log 2}{k} \\
t_{1 / 10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \\
{\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10=\frac{\left(\frac{2.303 \times 3 \log 2}{k}\right)}{\left(\frac{2.303}{k}\right)} \times 10=9}
\end{array}\)
t_{1 / 8}=\frac{2.303 \log 8}{k}=\frac{2.303 \times 3 \log 2}{k} \\
t_{1 / 10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \\
{\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10=\frac{\left(\frac{2.303 \times 3 \log 2}{k}\right)}{\left(\frac{2.303}{k}\right)} \times 10=9}
\end{array}\)
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