$\text{Na}_2\text{S}_2\text{O}_3$ reacts with dilute $\text{HCl}$ and produces a pale yellow precipitate. When $1\ \text{cm}^3$ of $0.1\ \text{mol dm}^{-3}\ \text{HCl}$ is added to $10\ \text{cm}^3$ of $0.02\ \text{mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3$, the precipitate appears slowly. If the test is done again using $1\ \text{cm}^3$ of $0.1\ \text{mol dm}^{-3}\ \text{HCl}$ and $10\ \text{cm}^3$ of $0.05\ \text{mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3$, the precipitate appears more rapidly. What accounts for this?
- AThe activation energy of the reaction is lower when $0.05\ \text{mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3$ is used.
- BThe reaction proceeds by a different pathway when $0.05\ \text{mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3$ is used.
- CThe collisions between reactant particles are more violent when $0.05\ \text{mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3$ is used.
- DThe reactant particles collide more frequently when $0.05\ \text{mol dm}^{-3}\ \text{Na}_2\text{S}_2\text{O}_3$ is used.