Question

If for a first-order reaction, \([A]_0 = 1.0 \text{M}\) and \([A]_t = 0.25 \text{M}\) after 276 s, find the value of the rate constant (k).

• A. \(0.0021 \text{s}^{-1}\)
• B. \(0.0050 \text{s}^{-1}\)
• C. \(0.003 \text{s}^{-1}\)
• D. \(0.006 \text{s}^{-1}\)

Hint:

Use k = (2.303/t) log (A0/At).

Answer 1

The Correct Option is B

Step 1: Write the formula
For first order, \(k = \frac{2.303}{t}\log\frac{[A]_0}{[A]_t}\).

Step 2: Substitute
\(\frac{[A]_0}{[A]_t} = \frac{1.0}{0.25} = 4\), and \(\log 4 = 0.6021\).
\[ k = \frac{2.303 \times 0.6021}{276} = \frac{1.3866}{276} \]
\[ k = 5.02\times 10^{-3}\ \text{s}^{-1} \approx 0.0050\ \text{s}^{-1} \]

Step 3: Check
Concentration drops to a quarter, i.e. two half-lives, so \(t_{1/2} = 138\) s and \(k = 0.693/138 = 0.0050\).

Step 4: Other options
\(0.0021\), \(0.003\) and \(0.006\) would give the wrong time for a fall to a quarter.

Final Answer:
The rate constant is 0.0050 per second. \[ \boxed{\text{(B)}\ 0.0050\ \text{s}^{-1}} \]