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Anika [276]
3 years ago
15

Consider the following mechanism: (1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast] (2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [s

low] (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
Chemistry
1 answer:
mash [69]3 years ago
8 0

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

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P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

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Firstly we determine the mole fraction of solute.

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1) Increasing the pressure          A) Shift to the left  

2) Removing hydrogen gas        B) Shift to the right    

3) Adding a catalyst                     C) No effect

Explanation:

  • <em>Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>

<em></em>

<u><em>1) Decreasing the pressure:</em></u>

  • When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
  • The reactants side (left) has 4.0 moles of gases and the products side (right) has 2.0 moles of gases.
  • So, decreasing the pressure will shift the reaction to the side with more moles of gas (left side).

<u><em>so, the right match is: A) Shift to the left.</em></u>

<em><u>2) Adding hydrogen gas:</u></em>

  • Adding hydrogen gas will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of hydrogen gas by addition.

<u><em>so, the right match is: B) Shift to the right.</em></u>

<u><em></em></u>

<u><em>3) Adding a catalyst:</em></u>

  • Catalyst increases the rate of the reaction without affecting the equilibrium position.
  • Catalyst increases the rate via lowering the activation energy of the reaction.
  • This can occur via passing the reaction in alternative pathway (changing the mechanism).
  • The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
  • in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
  • with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.

<u><em>so, the right match is: B) No effect.</em></u>

<u><em></em></u>

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