Practicing Success
Rate of a reaction changes from \(2.48 × 10^{-3}\, \ mol^{-1}s^{-1}\) to \(4.96 × 10^{-3}\, \ mol^{-1}s^{-1}\) when concentration of reactant, is changed from \(0.6\, \ M\) to \(2.4\, \ M\) respectively, the order of reaction is: |
2 Zero 0.5 3 |
0.5 |
The correct answer is option 3. 0.3. To determine the order of the reaction, we can use the rate equation: \[ \text{Rate} = k \times [\text{Reactant}]^n \] where: \( \text{Rate} \) is the rate of the reaction, \( k \) is the rate constant, \( [\text{Reactant}] \) is the concentration of the reactant, and \( n \) is the order of the reaction. Given that the rate of the reaction changes from \(2.48 \times 10^{-3}\, \text{mol}^{-1}\text{s}^{-1}\) to \(4.96 \times 10^{-3}\, \text{mol}^{-1}\text{s}^{-1}\) when the concentration of the reactant changes from \(0.6\, \text{M}\) to \(2.4\, \text{M}\), we can set up a ratio: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[\text{Reactant}]_2}{[\text{Reactant}]_1} \right)^n \] Given: Solving for \( n \): \[ 2 = (4)^n \] \[ \log 2 = n \times \log 4 \] \[ n = \frac{\log 2}{\log 4} \] \[ n = \frac{0.3010}{0.6021} \] \[ n \approx 0.5 \] So, the order of the reaction is approximately \(0.5\). Therefore, the correct answer is option 3: \(0.5\). |