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: Use half-lives:
Number of half-lives elapsed: $n = \dfrac{1200}{850} = 1.412$.

Step 2: Halve repeatedly:
$[A] = [A]_0 \times \left(\tfrac{1}{2}\right)^{n} = 0.06 \times 2^{-1.412}$.

Step 3: Evaluate:
$2^{1.412} = 2 \times 2^{0.412} = 2 \times 1.33 = 2.66$. So $[A] = 0.06/2.66 = 0.0226$ mol dm$^{-3}$.

Step 4: Check:
After 1 half-life the value would be 0.03, after 2 it would be 0.015. 0.023 lies between them, which matches option A.

Final Answer:
After 1.41 half-lives, the concentration drops to about 0.023 mol per dm3. \[ \boxed{\text{(A) }0.023\ \text{mol dm}^{-3}} \]