Chemistry 114
Third Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1.(12 points) If I was doing a kinetics experiment and found the concentration of a product
at 0, 10, 20, 30, 40, 50, and 60 minutes....
A. Tell what I would plot to see if this was a zero order reaction
X axis =_time________
Y axis = _concentration
How would I get k from this plot ? k=-slope
B. Tell what I would plot to see if this was a first order reaction
X axis =_time________
Y axis = ln(concentration)
How would I get k from this plot ? k=-slope
C. Tell what I would plot to see if this was a second order reaction
X axis =__time______
Y axis = 1/concentration
How would I get k from this plot ? k=+slope
2. (12 points) Define the following terms:
elementary step A chemical equation that describes a single collision or
transformation in a chemical mechanism. The rate equation for this reaction may be written
directly from the molecularity of the step.
intermediate species A chemical species that is created in one step of a
mechanism and is then destroyed in a subsequent step.
unimolecular step A step that involves the transformation of a single species
collisional frequency Z in the equation k=zñe-Ea/RT , and gives the number of collisions
that occur per unit time.
Homogeneous catalyst A substance that increases the rate of a reaction by
lowering the activation energy of a reaction, and is in the same phase as the reactants.
rate determining step The slow step in a reaction mechanism
1
3. (12 POINTS) If it takes 4 hours to cook a turkey at 325oF (163oC) and 10 hours to cook
the same turkey at 225o (107oC), what is the activation energy for cooking a turkey? Hint:
the rate of the reaction, k=1/cooking time.
163oC = 163 + 273 = 436K k at this temp ~ 1/4 = .25 hr-1
107oC= 107 + 273 = 380K k at this temp ~1/10 = .10 hr-1
You can also express the rate in 1/minutes
4. The following 6 mechanisms have been proposed for a reaction with the stoichiometry
of A6B +C
I
II
Step 1. A62B
Slow
Step 1. A6D
Fast Equilibrium
Step 2. B6C
Fast
Step 2. D6B+C
Slow
Sum=A6B+C
Sum=A6B+C
III
IV
Step 1. A6B+D
Fast Equilibrium
Step 1. A+B6D
Slow
Step 2. D6C
Slow
Step 2. D6C
Fast
Sum=A6B+C
Sum= A+B6C
V
VI
Step 1. A+A6B+C Slow
Step 1. A+A6B+D Slow
Step 2. D6C
Fast
Sum=A+A6B+C
Sum=A+A6B+C
Can any of the above mechanisms be eliminated?, if so, which one(s) IV,V,VI__
What would be the experimental rate law corresponding to mechanism ?rate=k[A]
What would be the experimental rate law corresponding to mechanism II?rate=k1k2/k-1 [A]
What would be the experimental rate law corresponding to mechanism III?
rate =k1k2/k-1[A]/[B]
What would be the experimental rate law corresponding to mechanism IV?rate =k[A][B]
What would be the experimental rate law corresponding to mechanism V? rate=k[A]2
What would be the experimental rate law corresponding to mechanism VI?rate=k[A]2
2
5 A (4 points).Make a reaction energy diagram for a reaction with a mechanism of A6B if
A has an energy of 10 kJ, B has an energy of 15 kJ, and the reaction has an activation of
10 kJ.
Hard to make this picture on my word processor. But the reactant starts at 10 kJ,
then there is a hump up to 20 kJ, and then the curve slopes down to the product at 15 kJ
B (4 points). Make a reaction energy diagram for a reaction with a mechanism of A6B if A
has an energy of 10 kJ, B has an energy of 5 kJ, and the reaction has an activation of 15
kJ.
Starts at 10 kJ, humps up to 25 kJ, then slopes down to 5 kJ
C (4 points). Make a reaction energy diagram for the two step reaction mechanism:
Step 1: A6B Slow
Step 2: B6C Fast
In which the energy of A is 5 kJ, B is 7 kJ and C is -5 kJ.
Make your own activation energies that are consistent with this mechanism.
This was a little more advanced. Starts at 5 kJ, has an activation hump the comes to a line
at 7 kJ, then a second activation hump before it gets to the final product at -5 kJ. Since
step 1 of the reaction was the slow step the distance between the A and the first activation
hump had to be larger than the hump between B and the second activation hump
6 A(6 points). Write the equilibrium expressions that correspond to these reactions:
CO(g) + NO2(g) W CO2(g) + NO(g)
K=[CO2][NO]/[CO][NO2]
CaCO3(s) W CaO(s) + CO2(g)
K=[CO2]
(Solids don’t appear in equilibrium expressions)
B (6 points).Write the chemical reactions that correspond to these equilibrium expressions:
___2N2(g) + 6H2(g) 64NH3(g)
NH3(aq) + H2O(l)6NH4+(aq) + OH-(aq)
When you first write the equation you should see that it
is not balanced, and figure out that you need to add a
water
3
7. (12 points) PCl5(g) W PCl3(g) + Cl2(g) has KC of .0420 M at 25oC
Calculate the KC for the reactions:
PCl3(g) + Cl2(g) W PCl5(g)
Reverse of original equation so Knew =1/Kold
=1/.042M = 23.81 M-1
3Cl2(g) + 3PCl3(g) W3PCl5(g)
3 x the above equation so Knew =Kold3
=23.813 = 13,500 M-3
Calculate the KP for the reaction
2PCl5(g) W 2PCl3(g) + 2Cl2(g)
This involves both 2X the original equation and the conversion from KC to Kp
Start with squaring the original equation
K=.04202 = .001764
Now KP = KC(RT)Än, Än=(2+2)-2 =2
KP = .001764(.08206A298)2
=1.055 atm2
8. At a certain temperature 0.700 mole of SO3 is placed in a 2.00-L container and the
reaction
2SO3(g) W 2SO2(g) + O2(g)
comes to equilibrium.
If .110 mole of O2 is present in the container at equilibrium, calculate KC.
I will do my ice table in moles and convert to molarity before I calculate Kc, You can
also convert to molarity first, and then not worry about it later.
I
C
E
2SO3(g) W 2SO2(g)
.7
0
-2X
+2X
.7-2X
2X
X must be equal to .11
So
E
.7-.22
.22
E
.48
.22
+
O2(g)
0
+X
.11 = 0+X
.11
.11
converting to molarity since in a 2 L container
E
.24
.11
.055
K= (.11)2(.055)/.242
=.0115M
4