Question

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

Hint:

For first-order reactions, the unit of $k$ is always time$^{-1}$ (e.g., s$^{-1}$, min$^{-1}$). Concentration units cancel inside the logarithmic term.

Answer 1

Concept: For a first-order reaction, the rate constant is given by: \[ k = \frac{2.303}{t} \log \left( \frac{[R]_0}{[R]} \right) \] where $[R]_0$ is initial concentration, $[R]$ is concentration after time $t$.
Step 1: Calculate concentration ratio. \[ \frac{[R]_0}{[R]} = \frac{0.6}{0.2} = 3 \] 
Step 2: Substitute values into formula. Given $t = 5$ min: \[ k = \frac{2.303}{5} \log(3) \] 
Step 3: Calculate numerical value. Given $\log 3 = 0.48$: \[ k = \frac{2.303}{5} \times 0.48 \] \[ k = \frac{1.10544}{5} = 0.221088\ \text{min}^{-1} \] 
Step 4: Final answer. The rate constant is approximately: \[ \mathbf{0.221\ min^{-1}} \]