Question

The half life of first order reaction is \(850\) s. The initial concentration of the reactant is \(0.06\) mol \(\text{dm}^{-3}\). What concentration would remain after \(1200\) s ?

• A. \(0.023\) mol \(\text{dm}^{-3}\)
• B. \(0.035\) mol \(\text{dm}^{-3}\)
• C. \(5.25\) mol \(\text{dm}^{-3}\)
• D. \(6.25\) mol \(\text{dm}^{-3}\)

Hint:

Find k from k = 0.693 / t half, then use the integrated first order equation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a first-order reaction, \(k = \dfrac{0.693}{t_{1/2}}\) and \([A] = [A]_0 e^{-kt}\).

Step 2: Find the rate constant:
\(k = \dfrac{0.693}{850} = 8.15 \times 10^{-4}\) s\(^{-1}\).

Step 3: Use the integrated equation:
\[ \log\dfrac{[A]_0}{[A]} = \dfrac{kt}{2.303} = \dfrac{8.15 \times 10^{-4} \times 1200}{2.303} = 0.4247 \]
\[ \dfrac{[A]_0}{[A]} = 10^{0.4247} = 2.66 \]

Step 4: Find the concentration:
\([A] = \dfrac{0.06}{2.66} = 0.0226 \approx 0.023\) mol dm\(^{-3}\).

Step 5: Why the other options are wrong.
0.035 mol dm\(^{-3}\) would need less than one half-life to pass. 5.25 and 6.25 are larger than the initial 0.06, which cannot be since the concentration only falls.

Final Answer:
The remaining concentration is about 0.023 mol per dm3. \[ \boxed{\text{(A) }0.023\ \text{mol dm}^{-3}} \]