Consider the following mechanism: (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] (a) What is the overall equation? Select the single best answer. ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq) ClO−(aq) + I−(aq) ⇌ IO−(aq) + Cl−(aq) ClO−(aq) + I−(aq) ⇌ IO−(aq) + H2O(l) + Cl−(aq) ClO−(aq) + I−(aq) → IO−(aq) + Cl−(aq) (b) Identify the intermediate(s), if any. Select the single best answer. No intermediates Cl−, OH−, I−, ClO−, IO− HClO, OH−, HIO HClO, OH−, HIO, H2O (c) What are the molecularity and the rate law for each step? Select the single best answers. (1): bimolecular unimolecular termolecular rate = k1([HClO][OH−])/([HClO][OH−]) k1[HClO][OH−] k1[ClO−][H2O] (2): bimolecular unimolecular termolecular rate = k2[HIO][Cl−] k2([HIO][Cl−])/([I−][HClO]) k2[I−][HClO] (3): bimolecular unimolecular termolecular rate = k3[OH−][HIO] k3([H2O][IO−])/([OH−][HIO]) k3[H2O][IO−] (d) Is the mechanism consistent with the actual rate law: rate = k[ClO−][I−]? no yes

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

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.