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

For a first-order reaction, the rate constant (\(k\)) is given by the following formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[R]_0}{[R]} \right) \] where:
- \([R]_0\) is the initial concentration,
- \([R]\) is the concentration after time \(t\).

Step 1: Calculate concentration ratio.
The concentration ratio is calculated by dividing the initial concentration by the concentration at time \(t\):
\[ \frac{[R]_0}{[R]} = \frac{0.6}{0.2} = 3 \] 
Step 2: Substitute values into the formula.
Given that the time \(t = 5\) minutes, substitute the values into the formula for \(k\):
\[ k = \frac{2.303}{5} \log(3) \] 
Step 3: Calculate numerical value.
Using the value \(\log 3 = 0.48\), substitute it into the equation:
\[ k = \frac{2.303}{5} \times 0.48 \] \[ k = \frac{1.10544}{5} = 0.221088\ \text{min}^{-1} \] 
Step 4: Final answer.
The rate constant is approximately:
\[ \boxed{0.221\ \text{min}^{-1}} \]