Final answer:
In SN₂ reactions, stabilizing the transition state leads to a decrease in activation energy and an increase in reaction rate, contrary to the incorrect statement given. Such stabilizing groups help the reaction proceed faster by lowering the energy of the transition state.
Step-by-step explanation:
The incorrect statement regarding SN₂ reaction is that the group which stabilizes the transition state increases the reaction rate. In an SN₂ reaction, the transition state is a point along the reaction pathway where the incoming nucleophile has partially formed a bond with the electrophilic carbon, while the leaving group has partially broken its bond. This reaction mechanism is characterized by a concerted, simultaneous bond-breaking and bond-making process. The correct statement is that any group or atom that can stabilize the transition state by dispersing charge or by other electronic effects will effectively lower the activation energy and increase the reaction rate.
For example, the stabilization of the C=C bond by a conjugated C=O group makes the product more stable and therefore lowers the activation energy of the reaction. Contrary to the incorrect statement provided, if a group actually stabilizes the transition state, it would not increase but rather decrease the energy required to reach that state, helping the reaction to proceed faster.
Now, applying this understanding to the equilibrium principle, it should be noted that while the question provides an equilibrium constant expression (K = [Sn] / [H₂]) for a different system, the concept can be paralleled. A catalyst or stabilizing group that lowers the energy of a transition state effectively acts like a catalyst in an equilibrium system, shifting the equilibrium towards the products without changing the equilibrium constant itself.