Iodine and propanone react in line with the equation below: $\text{I}_2(aq) + \text{CH}_3\text{COCH}_3(aq) \rightarrow \text{CH}_3\text{COCH}_2\text{I}(aq) + \text{HI}(aq)$. When the concentration of propanone is raised while the total reaction volume is kept unchanged, the initial rate of the reaction also rises. What might account for this?
- AA greater proportion of collisions are successful at the higher concentration.
- BThe particles are further apart at the higher concentration.
- CThe particles have more energy at the higher concentration.
- DThere are more collisions per second between particles at the higher concentration.