Question

In a first order reaction 20 millimole of reactant is lowered to 10 millimole in \(0.3010\) minute. Find rate constant of the reaction?

• A. \(3.010 \text{minute}^{-1}\)
• B. \(0.602 \text{minute}^{-1}\)
• C. \(2.303 \text{minute}^{-1}\)
• D. \(0.301 \text{minute}^{-1}\)

Hint:

Reactant halves in 0.3010 min, so this is the half life.

Answer 1

The Correct Option is C

Step 1: Use the integrated law:
For first order, $k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}$.

Step 2: Substitute:
$k = \frac{2.303}{0.3010}\log\frac{20}{10} = \frac{2.303}{0.3010} \times 0.3010 = 2.303$ min$^{-1}$.
The time $0.3010$ equals $\log 2$, so it cancels exactly.

Final Answer:
$k = 2.303$ min$^{-1}$, option (C). \[ \boxed{2.303\ \text{min}^{-1}} \]