When solid potassium halides are treated with concentrated sulfuric acid, the equations below apply: Reaction 1: $2\text{KCl} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{HCl}$ Reaction 2: $2\text{KBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{SO}_2 + \text{Br}_2 + 2\text{H}_2\text{O}$ Reaction 3: $8\text{KI} + 5\text{H}_2\text{SO}_4 \rightarrow 4\text{K}_2\text{SO}_4 + \text{H}_2\text{S} + 4\text{I}_2 + 4\text{H}_2\text{O}$ In each reaction, what is the greatest change in the oxidation number of sulfur?
- Areaction 1: $0$, reaction 2: $0$, reaction 3: $4$
- Breaction 1: $0$, reaction 2: $2$, reaction 3: $4$
- Creaction 1: $0$, reaction 2: $2$, reaction 3: $8$
- Dreaction 1: $0$, reaction 2: $4$, reaction 3: $8$