$\text{Na}_2\text{S}_2\text{O}_3$ reacts with dilute $\text{HCl}$, producing a pale yellow precipitate. When $1\,\text{cm}^3$ of $0.1\,\text{mol dm}^{-3}$ $\text{HCl}$ is mixed with $10\,\text{cm}^3$ of $0.02\,\text{mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3$, the precipitate appears slowly. When the experiment is repeated 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 quickly. What is the reason 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.