Download Answers to Kinetics Practice Problems

Chapter 12
Rates of Chemical Reactions
Note to teacher: You will notice that there are two different formats for the Sample
Problems in the student textbook. Where appropriate, the Sample Problem contains the
full set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act on
Your Strategy, and Check Your Solution. Where a shorter solution is appropriate, the
Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check
Your Solution step is also included in the shorter Sample Problems.
Solutions for Practice Problems
Student Textbook page 476
1. Problem
The following reaction is exothermic.
2ClO(g) → Cl2(g) + O2(g)
Draw and label a potential energy diagram for the reaction. Propose a reasonable
activated complex.
Solution
Since the reaction is exothermic, the products should be lower than the reactants on
the potential energy diagram. The activated complex exists between the reactants and
products, and is of higher energy than the reactants.
possible activated complex
Cl . . . Cl
...
...
Potential Energy (kJ)
O ... O
2ClO(g)
∆H
Cl2(g) + O2(g)
Reaction Progress
A reasonable activated complex might be a species where the bonds between Cl and
O in the reactants are breaking, while new bonds between two Cl atoms and between
two O atoms are forming.
2. Problem
Consider the following reaction.
AB + C → AC + B ∆H = +65 kJ, Ea(rev) = 34 kJ
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Draw and label a potential energy diagram for this reaction. Calculate and label
Ea(fwd). Include a possible structure for the activated complex.
Solution
The reaction is endothermic, since ∆H > 0. The products will be of higher energy
than the reactants.
possible activated complex
A ... C
...
Potential Energy (kJ)
B
AC + B
Ea(fwd) = 99 kJ
∆H = +65 kJ
AB + C
Reaction Progress
For an endothermic reaction,
Ea(fwd) = ∆H + Ea(rev)
= 65 kJ + 34 kJ
= 99 kJ
The activation energy for the forward reaction is 99 kJ.
A possible structure for the activated complex could involve simultaneous bond
breaking in the reactants and the formation of new bonds in the products.
3. Problem
Consider the reaction below.
C + D → CD ∆H = −132 kJ, Ea(fwd) = 61 kJ
Draw and label a potential energy diagram for this reaction. Calculate and label
Ea(rev). Include a possible structure for the activated complex.
Solution
Since this reaction is exothermic, the potential energy diagram will be similar to
Figure 6.12.
possible activated complex
Potential Energy (kJ)
C ... D
C+D
Ea(rev) = 193 kJ
∆H = −132 kJ
CD
Reaction Progress
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For an exothermic reaction,
Ea(rev) = ∆H + Ea(fwd)
= 132 kJ + 61 kJ
= 193 kJ
(Note: Use the positive value of ∆H.)
A possible structure of the activated complex might involve partial bond formation
between C and D.
3. Problem
In the upper atmosphere, oxygen exists in other forms other than O2(g). For example,
it exists as ozone, O3(g), and as single oxygen atoms, O(g). Ozone and atomic oxygen
react to form two molecules of oxygen. For this reaction, the enthalpy change is
−392 kJ and the activation energy is 19 kJ. Draw and label a potential energy
diagram. Include a value for Ea(rev). Propose a structure for the activated complex.
Potential Energy (kJ)
Solution
The reaction is
O3(g) + O(g) → 2O2(g) ∆H = −392 kJ; Ea(fwd) = 19 kJ
Since this reaction is exothermic, the potential energy diagram will be similar to
Figure 6.12.
O3 + O
Ea(rev) = 411 kJ
∆H = −392 kJ
2O2
Reaction Progress
For an exothermic reaction,
Ea(rev) = ∆H + Ea(fwd)
= 392 kJ + 19 kJ
= 411 kJ
A possible structure of the activated complex might involve partial bond breakage in
the O3 molecule coupled with partial bond formation to produce two O2 molecules.
Solutions for Practice Problems
Student Textbook page 478
5. Problem
A chemist proposes the following mechanism. Write the overall balanced equation.
Identify any reaction intermediate.
Step 1 NO2(g) + F2(g) → NO2F(g) + F(g)
Step 2 NO2(g) + F(g) → NO2F(g)
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Solution
Each step in the mechanism is an elementary reaction. Add the two steps together,
and cancel out chemical species that occur on both sides, to get the overall reaction.
Step 1 NO2(g) + F2(g) → NO2F(g) + F(g)
Step 2 NO2(g) + F(g)
→ NO2F(g)
2NO2(g) + F2(g) + F(g)
→ 2NO2F(g) + F(g)
The overall reaction is:
2 NO2(g) + F2(g)
→ 2 NO2F(g)
The reaction intermediate is the molecular species produced in step 1 and used in
step 2. It does not appear in the overall final equation. The reaction intermediate in
this reaction is F(g)
6. Problem
A chemist proposes the following mechanism for a certain reaction. Write the equation for the chemical reaction that this mechanism describes, and identify any reaction intermediate.
Step 1 A + B → C
Step 2 C + A → E + F
Solution
Each step in the mechanism is an elementary reaction. Add the two steps together,
and cancel out chemical species that occur on both sides, to get the overall reaction.
Step 1 A + B
→C
Step 2 C + A
→E+F
2A + B + C
→C+E+F
The overall reaction is:
2A + B
→E+F
The reaction intermediate is the molecular species produced in step 1 and used in
step 2. It does not appear in the overall final equation. The reaction intermediate in
this reaction is C.
7. Problem
A chemist proposes the following mechanism for a certain reaction. Write the equation for the chemical reaction that this mechanism describes, and identify any reaction intermediate.
Step 1 X + Y
→ XY
Step 2 XY + 2Z
→ XYZ2
Step 3 XYZ2
→ XZ + YZ
Solution
Add the three steps together, and cancel out chemical species that occur on both
sides, to get the overall reaction.
Step 1 X + Y
→ XY
Step 2 XY + 2Z
→ XYZ2
Step 3 XYZ2
→ XZ + YZ
X + Y + XY + 2Z + XYZ2
The overall reaction is
X + Y + 2Z
→ XY + XYZ2 + XZ + YZ
→ XZ + YZ
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The reaction intermediate is the molecular species produced in one step and used in
another step. It does not appear in the overall final equation. The reaction intermediates in this reaction are XY and XYZ2.
8. Problem
A chemist proposes the following mechanism for the reaction between 2-bromo-2methypropane, (CH3)3CBr(aq), and liquid water. Write the overall balanced equation.
Identify any reaction intermediate.
Step 1 (CH3)3CBr(aq) → (CH3)3C+(aq) + Br–(aq)
Step 2 (CH3)3C+(aq) + H2O(l) → (CH3)3COH2+(aq)
Step 3 (CH ) COH +
→ H+ + (CH ) COH
3 3
2 (aq)
(aq)
3 3
(aq)
Solution
Each step in the mechanism is an elementary reaction. Add the three steps together,
and cancel out chemical species that occur on both sides, to get the overall reaction.
Step 1 (CH3)3CBr(aq)
→ (CH3)3C+(aq) + Br–(aq)
+
Step 2 (CH3)3C (aq) + H2O(l)
→ (CH3)3COH2+(aq)
Step 3 (CH3)3COH2+(aq)
→ H+(aq) + (CH3)3COH(aq)
(CH3)3CBr(aq) + (CH3)3C+(aq) + H2O(l) + (CH3)3COH2+(aq) →
(CH3)3C+(aq) + Br–(aq) + (CH3)3COH2+(aq) + H+(aq) + (CH3)3COH(aq)
The overall reaction is (CH3)3CBr(aq) + H2O(l) → H+(aq) + Br–(aq) + (CH3)3COH(aq)
The reaction intermediate is the molecular species produced one step and used another step. It does not appear in the overall final equation. The reaction intermediates in
this reaction are (CH3)3C+(aq) and (CH3)3COH2+(aq).
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