Chemistry 114
Third Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. Mechanisms
Below is a proposed mechanism for a reaction.
A+B6C+D
Fast equilibrium
C+E6F
Slow
F6G
Fast
A. (2 points) What is the overall stoichiometry for this reaction?
A+B+C+E+F 6 C+D+F+G
A+B+E6D+G
B. (2 points) Which step is the rate determining step?
Slow step, step 2
C. (6 points) If this mechanism is correct, what should the rate equation for this reaction
be? (Make sure you include the fast equilibrium)
rate = k2[C][E], but C is intermediate... Use step 1 equilibrium k+1[A][B]=k-1[C][D];
[C]=k+1/k-1[A][B]/[D] Substitute in to first equation
rate = k2 k+1/k-1[A][B][E]/[D]
D. (2 points) The following mechanism uses both an intermediate species and a catalyst:
Step 1.
A+B6C+D
Slow
Step 2.
C+E6A+F
Fast
Which of the species; A, B, C, D, or E is the intermediate species ? (Circle one)
Created in one step and destroyed in a later step
Which of the species; A, B, C, D, or E is the catalyst? (Circle one)
Present at the beginning, and regenerated before the reaction is finished
2. (12 points) If a reaction has a rate of 100 sec-1 at 100oC and an activation energy of 50
kJ/mol, at what temperature will I have to run the reaction to have the rate be 30 sec-1?
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3. (14 points) What is....
An Elementary step?
A single step in a reaction mechanism whose rate law can be written directly from the
molecularity of the reaction.
A Bimolecular reaction?
A reaction that involves the collision of two molecules
The Steric factor?
In the kinetic theory of reaction rates, the term that adjusts for the orientation of a collision.
The frequency factor?
In the kinetic theory of reaction rates the term that give the number of collisions that occur
over a given time period
A dynamic equilibrium?
Even though the concentration is not changing when a system is at equilibrium, the forward
and backward reaction are still occurring.
A reaction quotient?
An equation with the same form as the equilibrium expression, but you use your current
concentrations, not the equilibrium concentrations. (Q)
A Brosted Lowry Base?
A proton acceptor
4. (12 points) Below are six equilibrium expressions. Write the chemical equations that
correspond to these expressions. (Remember, sometimes a reactant or a product may
not appear in an equilibrium expression)
K = [HI]2/[H2][I2]
__H2(g) + I2(g) W2HI(g)___
K=[CH3OH]/[CO][H2]2
___CO(g) + 2H2(g) WCH3OH(g)
K= [HCl]2/[H2][Cl2]
K = [Ca2+][IO3-]2
K = [H3O+][OH-]
K= [H2O]2/[CO2][H2]2
__H2(g) + Cl2(g) W2HCl(g)___
__Ca(IO3)2(s) WCa2+(aq) + 2IO3-(aq)
__2H2O(l) WH3O+(aq) + OH-(aq)
_CO2(g) + 2H2(g) W2H2O(g) + C(s)
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5. (12 points) PCl3, Cl2, and PCl5 gasses are mixed together at 200oC. When the reaction
PCl3(g) + Cl2(g) W PCl5(g) has achieved equilibrium the pressures of the gasses are:
A. What is KP for this reaction at 200oC?
Note: you were given pressures so KP can be calculated directly
5.89/(.84×1.32) = 5.31atm-1
B. What is KP for the reaction 3PCl3(g) + 3Cl2(g) W 3PCl5(g) at 200oC?
Previous reaction X 3 so
5.313 = 149.9 atm-3
C. What is KP for the reaction 2PCl5(g) W2PCl3(g) +2Cl2(g) at 200oC?
Reverse of original reaction X 2 so
(1/5.31)2 = .0354 atm2
D. What is KC for the reaction 2PCl5(g) W2PCl3(g) + 2Cl2(g) at 200oC?
Convert above equation from KP to KC
KP = KC(RT)Än, Än = (2+2)-2 =+2; 200C + 273 = 473K
.0354=X(.08206A473)2 ;
X=.354/(.08206A473)2
-5
2
= 2.35x10 M
6. The reaction HF(aq) W F-(aq) + H+ has an equilibrium constant of 6.3x10-4 M.
A. (1 point)If the initial concentration of HF is .5M and the initial concentration of Fis .01M, Fill in the I(initial) entries on the ICE table below:
rxn: HF
W
FH+
I
__.5___
_.01__
_0____
C
__-X___
__+X___
__+X____
E
_.5-X
_.01+X_
__X___
B. (1 point) Will the reaction go to the right or left? ___Right___________
C. (2 points) WHY?
___Fill in missing concentration___________________
D. (1 point) Use +X and -X to fill in the C (change) entries in the above table.
See above
E. (2 point) Fill in the E(equilibrium) entries in the above table.
See above
F. (2 points) Write the equilibrium expression for this reaction
6.3x10-4 = (.01+X)X/(.5-X)
G. (2 point) Simplify the above expression by assuming that X is small compared
to [.005] and [.0001]. Write your new, simplified expression
6.3x10-4 = (.01)X/(.5)
H. (2 points) Solve for X
X = (6.3x10-4X.5)/.01 = .0315
Note1. the approximation used here was not great. In analytical chem you will see a
better way to approach this problem.
Note 2. Since I I gave you no value for [H+] I wanted you to start with this as zero. Many
people assumed it was equal to [F-]. This was not what I wanted, but I could see why you
made this assumption, so I did not mark it wrong, but it does make finishing the roblem
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difficult.
7. (10 points) The reaction C(s) + 2H2(g) WCH4(g) has a ÄHo rxn of -74.6 kJ/mol
If the reaction is initially at equilibrium.....
A.) What would happen to the concentration of CH4 if I added more C(s)
Increase
Decrease
Remain the same (Circle one)
(Adding more solid does not change the equilibrium!)
B.) What would happen to the concentration of CH4 if I added more H2(g)
Increase
Decrease
Remain the same (Circle one)
C.) What would happen to the concentration of CH4 if I added He(g)
Increase
Decrease
Remain the same (Circle one)
D.) What would happen to the concentration of CH4 if I increases the volume of the
container?
Increase
Decrease
Remain the same (Circle one)
E. What would happen to the concentration of CH4 if I cooled the container?
Increase
Decrease
Remain the same (Circle one)
8. Acid and bases
A (3 points) Name three strong acids.
Any strong acid from the table worked
B (3 points) Name three strong bases
Any strong base from the table worked, but I did take off for incorrect formulas
C. (3 points) If [H+] = 1.57x10-6 M, what it the [OH-] in that solution?
KW = [H+][OH-]; [OH-] = 1x10-14/1.57x10-6 = 6.37x10-9M
D. (3 points) If the [OH-] of a solution is 4.89x10-3, What is the pOH and pH of the
solution?
pOH=-log(4.89x10-3) = 2.31
pH + pOH = 14
pH=14-2.31 = 11.69
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