Question

For a first order reaction, half life is \(5\) hour. What time is required to reduce \(10\) g of reactant to \(2.5\) g ?

• A. \(3\) hour
• B. \(4\) hour
• C. \(5\) hour
• D. \(10\) hour

Hint:

Going from 10 g to 2.5 g takes two half lives.

Answer 1

The Correct Option is D

Step 1: Using the integrated law
$t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$, with $k = 0.693/5 = 0.1386\ \text{h}^{-1}$.

Step 2: Ratio
$[A]_0/[A] = 10/2.5 = 4$ and $\log 4 = 0.602$.

Step 3: Compute
$t = \dfrac{2.303 \times 0.602}{0.1386} = 10.0$ h.

Step 4: Conclusion
The time is 10 hours, which agrees with the two half life count.

Step 5: Check with the other options
A time of 5 h gives only $10 \to 5$ g, and 3 h or 4 h would leave more than 5 g, so none of these can reach 2.5 g. A time of 10 h gives exactly two halvings, so 2.5 g remains, and so the answer is 10 h.

Final Answer:
The reaction needs 10 hours. This is option (D). \[ \boxed{\text{(D) }10\ \text{h}} \]