HALOGENATION OF PROPANONE

Hint

To obtain the steady state concentration:

Step	Procedure	Example: 0 = + aX - bX + c
1	Move all terms containing the species of interest to the left hand side of the equation. (Remember sign changes.)	- aX + bX = + c
2	Rewrite the left hand side of the equation, bringing the species of interest to the front of a bracket (i.e. [Pe]( . . . )).	X ( - a + b ) = + c
3	Divide both sides of the equation by the term in the bracket.	X ( - a + b ) / ( - a + b ) = + c / ( - a + b )
4	Cancel terms (where appropriate).	X = + c / ( - a + b )

Steady State Concentration of the enol form of Propanone {[P_e]_ss}

Applying the Steady State Approximation to the overall rate of production of the enol form of proponone, gives

0 = + k₁[H⁺][P_k] - k_-1[P_e][H⁺] - k₂[P_e][I₂]

On rearrangement, the steady state concentration of the enol form of propanone is found to be

[P_e]_ss = + k₁[H⁺][P_k] / (k_-1[H⁺] + k₂[I₂])

[P_e]_ss = - k₁[H⁺][P_k] / (k_-1[P_e][H⁺] - k₂[P_e][I₂])

[P_e]_ss = + k₁[H⁺][P_k] / (k_-1[H⁺] - k₂[I₂])

[P_e]_ss = - k₁[H⁺][P_k] / (k_-1[H⁺] + k₂[I₂])