a) NO₃ (g) + NO (g) → 2 NO₂ (g) 2
b) NO₃ (g) + CO (g) → NO₂ (g) + CO₂ (g) 2
c) NH₃ (g) + HCl (g) → NH₄Cl (s) 2

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)

a) What is the overall equation?
\(\displaystyle ClO^-(aq)\;+\;H_2O\;(l)\;⇄\; \require{cancel}\cancel{HClO\;(aq)}\;+\;\cancel{OH^-\;(aq)}\)
\(\displaystyle I^-\;(aq)\;+\;\cancel{HClO (aq)}\;→\; \cancel{HIO\;(aq)}\;+\;Cl^-\;(aq)\)
\(\displaystyle \cancel{OH^-(aq)}\;+\;\cancel{HIO\;(aq)}\;→\;H_2O\;(l)\;+\;IO^-\;(aq)\)
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\(\displaystyle ClO^-\;(aq)\;+\;I^-\;(aq)\;→\;Cl^-\;(aq)\;+IO^-\;(aq)\)

b) Indicate if there are intermediates and identify them

The intermediates are HClO, HIO, and OH^–

c) What are the molecularity and rate law for each step?

Step 1 molecularity is 2. Rate = k[ClO^–][H₂O]
Step 2 molecularity is 2. Rate = k[I^–][HClO]
Step 3 molecularity is 2. Rate = [OH^–][HIO]

d) The experimental rate law was determined to be rate = k[ClO^–][I^–]. Is the above mechanism a plausible mechanism? Explain.

Write the rate law from the slow step.

Rate = k[I^–][HClO]

An intermediate cannot be in the rate law. Set the rates of the forward and reverse reactions in step one equal. Do not include water in the rates.

\(\displaystyle k[ClO^-]\;=k[HClO][OH^-]\)

Solve for the concentration of the intermediate [HClO].

\(\displaystyle [HClO]\;=\;\frac{k_1}{k_{_1}}\frac{[ClO^-]}{[OH^-]}\)
 
Plug into the rate law written from the slow step.

\(\displaystyle rate\;=\;\frac{k\;k_1}{k_{_1}}\frac{[ClO^-][I^-]}{OH^-}\)
 
\(\displaystyle rate\;=\;k\frac{[ClO^-][I^-]}{[OH^-]}\)
 
This is not a plausible mechanism. The rate law determined from the mechanism does not agree with the experimental rate law. In addition, there is an intermediate in the rate law.

\(\displaystyle O_3\;+\;I^-\;→\;O_2\;+\;IO^-\;\;\;(slow)\)
\(\displaystyle IO^-\;+\;H_2O\;→\;HOI\;+\;HO^-\;\;\;(fast)\)
\(\displaystyle HOI\;+\;I^-\;→\;I_2\;+\;HO^-\;\;\;(fast)\)
 
a) Write the rate law.

rate = k[O₃][I^–]

b) Write the overall reaction.

\(\displaystyle O_3\;+\;I^-\;→\;O_2\;+\;\require{cancel}\cancel{IO^-}\;\;\;(slow)\)
\(\displaystyle \require{cancel}\cancel{IO^-}\;+\;H_2O\;→\;\require{cancel}\cancel{HOI}\;+\;HO^-\;\;\;(fast)\)
\(\displaystyle \require{cancel}\cancel{HOI}\;+\;I^-\;→\;I_2\;+\;HO^-\;\;\;(fast)\)
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\(\displaystyle O_3\;+\;2\;I^-\;+\;H_2O\;→\;O_2\;+\;2\;HO\;+\;I_2\)

c) Identify any intermediates

IO^– and HOI

\(\displaystyle Cl\;(g)\;+\;O_3\;(g)\;⇄\;ClO\;(g)\;+\;O_2\;(g)\;\;\;(fast\;equilibrium)\)
\(\displaystyle ClO\;(g)\;+\;O(g)\;→\;Cl\;(g)\;+\;O_2(g)\;\;\;(slow)\)

a) Write the overall equation.

\(\displaystyle \require{cancel} \cancel{Cl\;(g)}\;+\;O_3\;(g)\;⇄\;\cancel{ClO\;(g)}\;+\;O_2\;(g)\;\;\;(fast\;equilibrium)\)
\(\displaystyle \require{cancel}\cancel{ClO\;(g)}\;+\;O(g)\;→\;\cancel{Cl\;(g)}\;+\;O_2(g)\;\;\;(slow)\)
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\(\displaystyle O_3\;(g)\;+\;O\;(g)\; →\; 2\;O_2\;(g)\)

b) Write the rate law

Write the rate law from the slow step.
rate = k[ClO][O]

The rate law cannot have an intermediate. The first equation is a fast, equilibrium step and the rate of the forward reaction is equal to the rate of the reverse reaction.

\(\displaystyle k_1[Cl][O_3]\;=\;k_{_1}[ClO][O_2]\)

Solve for the concentration of [ClO]

\(\displaystyle [ClO]\;=\;\frac{k_1}{k_1}\frac{[Cl][O_3]}{[O_2]}\)

Plug this into the rate law written for the slow step.

\(\displaystyle \frac{kk_1}{k_1}\frac{[Cl][O_3][O]}{[O_2]}\)
 
\(\displaystyle rate\;=\;\frac{k[Cl][O_3][O]}{[O_2]}\)

c) Identify intermediates and catalysts, if any.

ClO is an intermediate. Cl is a catalyst.