Question

Pre-Equilibria Conditions in Reaction Mechanisms 5Br (aq) + BrO3- (aq) + 6H+(aq) + 3Br2(0) + 3H20(1) The above reaction is expected to obey the mechanism: BrO3(aq) + H+ (aq) HBrO3(aq) Fast equilibrium HBrO3(aq) + H+ (aq) H2BrO3+ (aq) Fast equilibrium H2BrO3(aq) + Br"(aq) *34 (Br-BrO2)(aq) + H20(1) Slow (Br-BrO2)(aq) + 4H+(aq) + 4Br"(aq) *4, products Fast Choose, from the list below, correct expressions for the overall rate law which are completely consistent with the above mechanism. d[Br-Broz]/dt = K[Br"][Bro3"][+]2 -d[H+]/dt = k[+]4 -d[Br"]/dt = k[H+]?[Br][BrO3-] -d[+]/dt = K[H+]?[BrO3-][Br] -d[BrO 3 )/dt = k[Br"][BrO3"][+]2 -d[H+]/dt = k[Br"]²[Bro3"][H+]
hi! please show all work step by step so I can follow along and understand! thank you!

Answer 1

The slow step in the given elementary steps of the mechanism would mean that the intermediate, H₂BrO₃⁺(aq), would accumulate and would also be able to turn the previous two elementary steps back towards the reactants' side. So, the overall rate law will be formed from the first three elementary steps of the given mechanism as the fourth step is a fast step and it would not affect the rate.

The overall rate must depend on the slow elementary step. So, writing the overall rate from the molecularity of the third (slow) step:

Rate = k₃. [H₂BrO₃⁺].[Br^-]

but [H₂BrO₃⁺] and k₃ should not figure in the overall rate as these quantities are intermediates or concerned with them. Finding the overall rate in terms of the reactants figuring in the overall equation would give the rate law required. Refer to the pictures attached below:

- ---- Overall rate of the reaction is based on slow step. Rate = kg. [H, B40]. [br] for step II, Rate of forward reaction - Rate of → ke [H Baby] [**] = {,[My breathin → [My Blo+] = Rz. [H Brez]. [**] reverse reaction - u. 1

For step I, (similarly as above) k. [Broz] [ 14+] = [H Bez] => [HBxbx] , [Bros - ] [**] Putting the above value in ( we get, [H, Beost] = k ki (828,-] [n+]? Putting the above value of [H, Br in i to get the overall rate, Rate = kg. kaik, [Bredz ] [H + 1² [BE] k, k, - d[br -] = k [be] [Beba] [4t 1² dt where k=k. k.kz 2

Note: The rate for the overall reaction is expressed as -d[Br^-]/dt because Br^-(aq) is the reactant figuring in the third slow step.