Answer:
1. The overall equation is ClO-(aq)+I-(aq) → Cl-(aq)+IO-(aq)
2. The intermediates include: HClO(aq), OH-(aq) and HIO(aq)
3. The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]
4. No,
The rate depends on [OH-], so it's not consistent with the actual rate law
Explanation:
1
Given
(1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]
(2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]
(3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]
Add up the three equations
ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]
I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]
OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]
Remove all common terms {H2O(l) + I-(aq) +HClO(aq) +OH-(aq) +HIO(aq) => HClO(aq) + OH-(aq) + HIO(aq)
+H2O(l)}
We're left with
ClO-(aq) + I-(aq) => Cl-(aq) + IO-(aq)
2.
There are intermediates generated but they are not visible in the overall equation.
The intermediates include: HClO(aq), OH-(aq) and HIO(aq)
3.
The three steps are bimolecular.
The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]
4. Let K represents equilibrium constant
At step 1,
K1 = [HClO][OH-]/[ClO-]
Simplify;
K1 [ClO-]= [HClO][OH-]
K1[ClO-]/[OH-] = [HClO]
Determine the rate at step 2
= k2[I-][HClO]
= K1k2[I-][ClO-]/[OH-]
= k[ClO-][I-]/[OH-]
The answer is no