Question

Half life of a first order reaction is \(900\) second. If initial concentration of reactant is \(0.08\,\text{mol dm}^{-3}\) find concentration that remains after \(35\) minute ?

• A. \(0.159\,\text{mol dm}^{-3}\)
• B. \(0.0159\,\text{mol dm}^{-3}\)
• C. \(1.05\,\text{mol dm}^{-3}\)
• D. \(0.759\,\text{mol dm}^{-3}\)

Hint:

Convert 35 minutes to seconds, count the half lives, then divide the initial concentration by 2 to that power.

Answer 1

The Correct Option is B

Step 1: Understand the concept
For a first order reaction, the concentration halves every half life. After \(n\) half lives, \([A] = [A]_0/2^{n}\) with \(n = t/t_{1/2}\).

Step 2: Convert time
\(t = 35 \text{ min} = 35 \times 60 = 2100\) s. Half life is 900 s, so \(n = 2100/900 = 2.333\).

Step 3: Calculate the rate constant route
\(k = 0.693/900 = 7.7 \times 10^{-4}\ \text{s}^{-1}\). Then \(\log \dfrac{[A]_0}{[A]} = \dfrac{kt}{2.303} = \dfrac{7.7 \times 10^{-4} \times 2100}{2.303} = 0.7021\), so \([A]_0/[A] = 5.04\).

Step 4: Result
\[ [A] = \frac{0.08}{5.04} = 0.0159\ \text{mol dm}^{-3} \]
The other options are larger than the initial concentration or are the wrong order of magnitude, which cannot be so since concentration only falls.

Final Answer:
The remaining concentration is 0.0159 mol per dm^3. This is option (B). \[ \boxed{\text{(B) }0.0159\ \text{mol dm}^{-3}} \]