Acid-catalysed conversion: $S + H X^{+} SH X^{+}$ (fast, constant $K$ ), then $SH X^{+} + H X_{2} O P + H X^{+}$ (slow, solvent in large excess). Experiment 1 uses $[S] = 0.10 M$ , $[H X^{+}] = 0.20 M$ ; experiment 2 uses $[S] = 0.30 M$ , $[H X^{+}] = 0.50 M$ . The ratio rate $_{2}$ /rate $_{1}$ is

Answer: B

The slow step gives rate

= k [SH X^{+}]

with

[SH X^{+}] = K [S] [H X^{+}]

, so rate

\propto [S] [H X^{+}]

(water is constant). Ratio

= (0.30/0.10) (0.50/0.20) = 3 \times 2.5 = 7.5