What will be the initial rate of a reaction if its constant is 10^–3 min^–1 and the concentration of the reactant is 0.2 mol dm³?

0.0002 mol dm^–3 min^–1

The correct answer is option 3. 0.0002 mol dm^–3 min^–1

To calculate the initial rate of a reaction given the rate constant and the initial concentration of the reactant, we use the rate law equation that corresponds to the reaction order. For a first-order reaction, the rate law is typically expressed as:

\( \text{Rate} = k \cdot [\text{Reactant}] \)

where:

\( k \) is the rate constant,

\( [\text{Reactant}] \) is the concentration of the reactant.

Given:

\( k = 10^{-3} \) min\(^{-1}\) (rate constant)

\( [\text{Reactant}] = 0.2 \) mol dm\(^{-3}\) (initial concentration of the reactant)

Calculation:

Substitute the given values into the rate law equation:

\(\text{Rate} = (10^{-3} \text{ min}^{-1}) \cdot (0.2 \text{ mol dm}^{-3}) \)

\(\text{Rate} = 10^{-3} \cdot 0.2 \)

\( \text{Rate} = 0.0002 \text{ mol dm}^{-3} \text{ min}^{-1} \)

Conclusion:

Therefore, the initial rate of the reaction, given the rate constant of \( 10^{-3} \) min\(^{-1}\) and the initial concentration of the reactant of \( 0.2 \) mol dm\(^{-3}\), is: 0.0002 mol dm\(^{-3}\) min\(^{-1}\).