Consider the nitration of benzene using mixed conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ and $\mathrm{H}\mathrm{N}{\mathrm{O}}_3$ . If a large amount of $\mathrm{K}\mathrm{H}\mathrm{S}{\mathrm{O}}_4$ is added to the mixture, the rate of nitration will be:

• A

  Faster
• Correct answer
  B

  Slower
• C

  Unchanged
• D

  Doubled

Solution

$Q=\frac{\text{Product}}{\text{Reactant}}$ as product increase $Q>K$

$Q$ has became equal to $K$ product has to decrease so equilibrium will shift backward direction.

If we add $KHS{O}_4^{-},\text{conc}\text{.}HS{O}_4^{-}$ increases, equilibrium shifts backward.

Hence, the correct answer is option: 'B'.