In a first-order reaction, the concentration of the reactant decreases from 800 mol/dm3 to 50 mol/dm3 in 2 × 104 sec. The rate constant of reaction in sec.–1 is:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. 1.386 × 10–4
To calculate the rate constant of a first-order reaction, we can use the following equation:
\[ k = - \frac{1}{t} \ln \left(\frac{A_f}{A_i}\right) \]
Where:
\( k \) is the rate constant,
\( t \) is the time taken for the concentration to change (in this case, \( 2 \times 10^4 \) sec),
\( A_f \) is the final concentration of the reactant (50 mol/dm\(^3\)),
\( A_i \) is the initial concentration of the reactant (800 mol/dm\(^3\)).
Plugging in the values:
\[ k = - \frac{1}{2 \times 10^4 \text{ sec}} \ln \left(\frac{50 \text{ mol/dm}^3}{800 \text{ mol/dm}^3}\right) \]
Simplifying the fraction:
\[ k = - \frac{1}{2 \times 10^4 \text{ sec}} \ln(0.0625) \]
Calculating the natural logarithm:
\[ k \approx - \frac{1}{2 \times 10^4 \text{ sec}} (-2.7726) \]
Simplifying further:
\[ k \approx 1.386 \times 10^{-4} \text{ sec}^{-1} \]
Therefore, the rate constant of the reaction in sec\(^{-1}\) is approximate \(1.386 \times 10^{-4}\).
Hence, the correct answer is option (3) \(1.386 \times 10^{-4}\).