For the reaction H_2O_2 + 2H^+ + 2l^- rightarrow l_2 + 2H_2O in acidic aqueous s
ID: 923469 • Letter: F
Question
For the reaction H_2O_2 + 2H^+ + 2l^- rightarrow l_2 + 2H_2O in acidic aqueous solution, a proposed mechanism is H^+ + I^- doubleheadarrow k_-1 k_1 HI HI + H_2O_2 rightarrow k_2 H_2O + HOI HOI + I^- rightarrow k_3 I_2 + OH^- OH^- + H^+ rightarrow k_4 H_2O Verify that this mechanism adds to the correct overall reaction. Using the rate determining step approximation and assuming that the first step is in rapid equilibrium, the second step is slow, and the last two steps are fast detemiine the overall rate law predicted by this mechanism.Explanation / Answer
to prove that the mechanism adds up to cotrrect overall reaction,
add step (1),(2),(3) and (4)
H+ I - + HI + H2O2 + HOI + I- + OH- + H+ -----> HI + H2O + HOI + I2 + OH- + H2O
Rearranging, it becomes
H2O2 + 2H+ + 2I- ---> I2 + 2H2O
So, yes mechanism adds up to correct overall reaction
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Rate depends on the slowest step
rate= K2 [HI][H2O2]
but HI is product of 1st step
use:
K-1/K1= [HI]/[H+][I-]
[HI] =(K-1/K1) * [H+][I-]
Substitute this in rate expression
rate= K2 [HI][H2O2]
rate= K2 [H2O2] *(K-1/K1) * [H+][I-]
rate= K2*K-1/K1 * [H2O2] [H+][I-]
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