JEE Advanced levelSingle correct optionDifficulty 3 of 3o3039
The solvolysis of (CX6HX5)X2CH−Cl is described by RClkX1kX−1RX++ClX− followed by fast capture of RX+ by solvent with rate constant k2, so that kobs=k1/(1+(k−1/k2)[ClX−]). Without added chloride kobs=8.0×10−4s−1, and with 0.10M added LiCl it is 5.0×10−4s−1 (ignore any other salt effects). The kobs in the presence of 0.30M chloride is
A2.86×10−4s−1
B5.00×10−4s−1
C1.67×10−4s−1
D6.78×10−4s−1
Answer: A
From 8.0×10−4/5.0×10−4=1.6=1+(k−1/k2)(0.10) we get k−1/k2=6.0M−1. At 0.30M: kobs=8.0×10−4/(1+6.0×0.30)=8.0×10−4/2.8=2.86×10−4s−1.