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:
Answer & explanation
Correct answer: option 3
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:
\[ \frac{4.96 \times 10^{-3}\, \text{mol}^{-1}\text{s}^{-1}}{2.48 \times 10^{-3}\, \text{mol}^{-1}\text{s}^{-1}} = \left( \frac{2.4\, \text{M}}{0.6\, \text{M}} \right)^n \]
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: \(0.5\).