Question

For a first order reaction, (A) \(\rightarrow\) Products, the concentration of A changes from \(0.1\) M to \(0.025\) M in \(40\) minutes. The rate of reaction when the concentration of A is \(0.01\) M is

• A. \(3.47\times 10^{-4}\) M/min
• B. \(3.47\times 10^{-5}\) M/min
• C. \(1.73\times 10^{-4}\) M/min
• D. \(1.73\times 10^{-5}\) M/min

Hint:

Find the rate constant from the 0.1 M to 0.025 M data, then multiply it by 0.01.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a first order reaction, \(k=\dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}\) and rate \(=k[A]\).

Step 2: Find k
\(k=\dfrac{2.303}{40}\log\dfrac{0.1}{0.025}=\dfrac{2.303}{40}\times0.602=0.0347\ \text{min}^{-1}\).

Step 3: Check
0.1 M to 0.025 M is two half-lives, so \(t_{1/2}=20\) min and \(k=\dfrac{0.693}{20}=0.0347\ \text{min}^{-1}\).

Step 4: Rate at 0.01 M
Rate \(=0.0347\times0.01=3.47\times10^{-4}\) M/min.

Final Answer:
The rate at 0.01 M is \(3.47\times10^{-4}\) M/min. \[ \boxed{\text{(A) }3.47\times10^{-4}\ \text{M/min}} \]