Question

The concentration of the reactant is reduced from 0.6 mol $L^{-1}$ to 0.2 mol $L^{-1}$ in 5 minutes in a first order reaction. Calculate rate constant of the reaction. $(\log 3 = 0.48)$

Hint:

Ensure you state the unit of $k$ correctly. Since time was in minutes, the unit of $k$ for this 1st order reaction is min$^{-1}$.

Answer 1

Step 1: Concept
The integrated rate equation for a first-order chemical reaction relates the rate constant ($k$) to time ($t$) and reactant concentrations.

Step 2: Meaning
The formula is: $k = \frac{2.303}{t} \log\left(\frac{[R_0]}{[R]}\right)$, where $[R_0]$ is initial concentration and $[R]$ is the concentration at time $t$.

Step 3: Analysis
1. Given values: Initial concentration $[R_0] = 0.6~mol~L^{-1}$. Final concentration $[R] = 0.2~mol~L^{-1}$. Time $t = 5$ minutes.
2. Substitute the values into the integrated rate equation:
$k = \frac{2.303}{5} \log\left(\frac{0.6}{0.2}\right)$
$k = \frac{2.303}{5} \log(3)$
3. Substitute the given logarithmic value ($\log 3 = 0.48$):
$k = \frac{2.303 \times 0.48}{5}$
$k = \frac{1.10544}{5} = 0.221088 \text{ min}^{-1}$

Step 4: Conclusion
Rounding to three significant figures, the calculated rate constant is $0.221 \text{ min}^{-1}$.

Final Answer: $0.221 \text{ min}^{-1}$