Iodine and propanone react as shown in 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 remains constant, the reaction rate also rises. What might explain this?
- AA greater proportion of collisions is 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 between reactant particles per second at the higher concentration.