CHM 2046C Module Nine Sample Exam Name:__Answers___

Part G: Reaction Mechanisms 20 points

1) The decomposition of ozone in the upper atmosphere to dioxygen occurs by a two-step mechanism. The first step is a fast reversible step and the second is a slow reaction between an oxygen atom and an ozone molecule:

Step 1: O₃ (g) ß à O₂ (g) + O (g) Fast. Reversible, reaction

Step 2: O₃ (g) + O (g) à 2 O₂ (g) slow

a. Which is the rate determining step?

Step #2

b. Write the rate expression for the rate-determining step.

Rate₂ = k₂[O₃][ O]

c. Write the rate equation for the overall reaction.

From fast step#1:

		[O2] [ O]
K1	=	------------
		[O3]

Algebraic Rearrange:

		K1[O3]
[ O]	=	------------
		[O2]

Substituting into Rate₂ Expression:

		K1[O3]
Rate2	=	k2[O3] X ------------
		[O2]

Rearranging:

		[O3] [O3]
Rateo	=	k2 K1 X ------------
		[O2]

Since a constant times a constant is a constant: k_o = k₂ K₁

		[O3]^2
Rateo	=	ko ----------
		[O2]

d. Which specie(s) is (are) reaction intermediates?

O (g)

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2) The conversion of cis-2-butene to trans-2-butene is catalyzed by iodine I₂.

The mechanism is believed to involve the dissociation of the iodine molecules to two iodine atoms, a process that quickly reaches an equilibrium, followed by the addition of iodine (atom) to one of the carbon atoms of the butene. This allows the rotation around the carbon-carbon bond. (carbon=carbon can not rotate). The iodine then dissociates from the intermediate to facilitate the conversion of another butene molecule.

CH₃ CH₃ CH₃ H

\ / I₂ \ /

C = C à C = C

/ \ / \

H H H CH₃

cis-2-butene trans-2-butene

a. Write the equations for the various steps of this mechanism.

Step#1 I₂ ↔ I∙ + I∙ (quickly means it is a fast equilibrium)

CH₃ CH₃ CH₃ H

\ / \ /

Step#2 I∙ + C = C à I— C — C∙ Slow

/ \ / \

H H H CH₃

CH₃ CH₃ CH₃ H

\ / \ /

Step#3 I— C — C∙ ßà I— C — C∙ Fast

/ \ / \

H H H CH₃

CH₃ H CH₃ H

\ / \ /

Step#4 I— C — C∙ à C = C + I∙ Fast

/ \ / \

H CH₃ H CH₃

b. Derive the rate equation.

Step #2 is the rate determining step.

Rate₂ = k₂ [I∙] [cis-C₄H₈]

From Step#1:

		[I∙] [ I∙]
K1	=	------------
		[I2]

Algebraic Rearrange:

[I∙]² = K₁[I₂] [I∙] = (K₁[I₂])^½

Rate_o = k₂ (K₁[I₂])^½[cis-C₄H₈]

A constant times a constant is a constant

k_o = k₂ K₁^½

therefore:

Rate_o = k_o [I₂]^½[cis-C₄H₈]

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c) Consider the following reaction in basic solution:

I^1- + ClO ^1- à Cl ^1- + IO ^1-

The rate equation is: Rate = k [I^1-][ ClO ^1-] / [OH^1-]

Suggest a mechanism:

The rate equation indicates that the reaction is not a simple single step bimolecular reaction between the iodide ion and the hypochlorite ion. There are four principle species initially in the solution:

I^{1- ;} ClO^1- ; OH^1- ; and water HOH.

Therefore with a denominator in the rate expression, then at least the first step must be an equilibrium or an equilibrium for a rate determining step.

Assuming the first step is a bimolecular reaction, which two species might interact? All are negative charged except water so perhaps any one of the following reactions with water might be Step#1:

Step#1

I^1- + HOH à HI + OH^1- not likely since HI is a strong acid and would not be molecular

Or: ClO ^1- + HOH à HClO + OH^1- more likely, a hydrolysis reaction see Chapter 18

Reaction with iodide is the only reasonable first step, since hydroxide reacting with water produces the same product as reactant and could not be the first step.

Step#2 The Intermediate from Step#1, HClO, would react with one of the three negative ions:

The hypochlorous acid might then collide and react with an iodide ion:

HClO + I^1- à HlO + Cl^1-

(HClO reacting with hydroxide is the reverse of step#1 and the HClO reacting with hypochorite ion produces the same product as reactant. Therefore Iodide is the best choice.)

Step#3

The hypoiodous acid is in equilibrium with its anion, the hypoiodite ion:

HlO + OH^1- ß à IO^1- + HOH

Since a base is present, the equilibrium shifts to the right as a fast step. Steps #1 & #3 are fast steps, therefore Step#2 must be the rate determining step and produces a rate expression that is consistent with the known rate expression given.

Rate₂ = k₂ [I^1-] [HClO]

From Step#1:

		[HClO] [OH^1-]
K1	=	-------------------
		[ClO ^1-] [HOH]

Algebraic Rearrange:

		K1[HOH] [ClO ^1-]
[ HClO]	=	-----------------------
		[OH^1]

The concentration of water is 55.5 M and can be considered a constant in the above expression and a constant times a constant is another constant:

k₃ = K₁[HOH] = K₁[55.5]

therefore:

		k3 [ClO ^1-]
[ HClO]	=	--------------
		[OH^1]

Substituting in the Rate expression of Step#2:

		k2 [I^1-] k3 [ClO ^1-]
Rate2	=	-----------------------
		[OH^1]

And a constant times a constant is another constant:

k_o = k₂∙k₃

Now the overall observed rate expression is derived from the above mechanism.

		k0 [I^1-][ClO ^1-]
Rate0	=	-----------------------
		[OH^1]