Time Ratio in First Order Reaction Calculation

❓ Question

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:

$$t_1 \text{ and } t_2$$

respectively.

The ratio:

$$\frac{t_1}{t_2}$$

will be:

1. $\frac43$
2. $\frac32$
3. $\frac23$
4. $\frac34$

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 First order reaction
👉 Integrated rate law
👉 Logarithmic concentration relation.

Main idea:

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

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🔷 Step 1 — Formula for First Order Reaction 💯

For first order reaction:

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

where:

• $[A]_0$ = initial concentration
• $[A]$= concentration after time $t$

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🔷 Step 2 — Find $t_1$

When concentration becomes:

$$\frac{[A]_0}{4}$$

then:

$$t_1= \frac{2.303}{k} \log\frac{[A]_0}{[A]_0/4}$$
$$= \frac{2.303}{k}\log4$$

Now:

$$\log4=2\log2$$

So:

$$t_1= \frac{2(2.303)\log2}{k}$$

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🔷 Step 3 — Find $t_2$

When concentration becomes:

$$\frac{[A]_0}{8}$$

then:

$$t_2= \frac{2.303}{k} \log8$$

Now:

$$\log8=3\log2$$

Thus:

$$t_2= \frac{3(2.303)\log2}{k}$$

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🔷 Step 4 — Find Ratio

$$\frac{t_1}{t_2} = \frac{2}{3}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Log values simplify na karna

❌ First order formula bhool jaana

❌ $\log4$ aur $\log8$ directly compare kar dena

Remember:

$$\boxed{ \log(2^n)=n\log2 }$$

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✅ Final Answer

$$\boxed{ \frac23 }$$

(Option 3)

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