In a first order decomposition reaction, the time taken for the decomposition of reactant to one-fourth and one-eighth of its initial concentration are t₁ and t₂(s), respectively. The ratio t₁/t₂ will be:

In a first order decomposition reaction, the time taken for the decomposition of reactant to one-fourth and one-eighth of its initial concentration are t₁ and t₂(s), respectively. The ratio t₁/t₂ will be:

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📷 Question Image:

Short Text Solution:

For a first-order reaction:

$$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$$

• For $t_1$:

$$\frac{[R]_0}{[R]} = \frac{1}{1/4} = 4$$
$$t_1 = \frac{2.303}{k} \log 4$$

• For $t_2$:

$$\frac{[R]_0}{[R]} = \frac{1}{1/8} = 8$$
$$t_2 = \frac{2.303}{k} \log 8$$

So,

$$\frac{t_1}{t_2} = \frac{\log 4}{\log 8} = \frac{2 \log 2}{3 \log 2} = \frac{2}{3}$$

✅ Final Answer: $\mathbf{\frac{2}{3}}$

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📷 Solution Image:

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Conclusion – Video Solution: