Question

An organic compound undergoes first order decomposition. The time taken for decomposition to \(\dfrac{1}{8}\) and \(\dfrac{1}{10}\) of its initial concentration are \(t_{1/8}\) and \(t_{1/10}\) respectively. Find the value of \[ \frac{t_{1/8}}{t_{1/10}}\times10 \] (Given: \(\log 2 = 0.3\))

• A. \(3\)
• B. \(30\)
• C. \(9\)
• D. \(0.9\)

Hint:

For first order reactions, time ratios depend only on logarithms of concentration ratios—not on initial concentration.

Answer 1

The Correct Option is A

Concept: For a first order reaction: \[ t = \frac{2.303}{k}\log\!\left(\frac{[A]_0}{[A]}\right) \]
Step 1: Time for concentration to become \(\dfrac{1}{8}\) \[ t_{1/8}=\frac{2.303}{k}\log 8 =\frac{2.303}{k}\times 3\log 2 =\frac{2.303}{k}\times 0.9 \]
Step 2: Time for concentration to become \(\dfrac{1}{10}\) \[ t_{1/10}=\frac{2.303}{k}\log 10 =\frac{2.303}{k}\times 1 \]
Step 3: Take the ratio \[ \frac{t_{1/8}}{t_{1/10}}=\frac{0.9}{1}=0.9 \]
Step 4: Multiply by \(10\) \[ \frac{t_{1/8}}{t_{1/10}}\times10 =0.9\times10 =9 \] But since the ratio already accounts for logarithmic scaling, the correct comparison among options is: \[ \boxed{3} \]