Chemistry 114
Third Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. (12 points) The reaction portrayed by the stoichiometric equation:
2NO2(g) + Br2(g) 6 2NO2Br
has an experimentally determined rate law of:
Rate =k[NO2][Br2]
One proposed mechanism for this reaction is:
(Step 1)
NO2(g) + Br2(g) WNO2Br2(g)
(Step 2)
NO2Br2(g) 6 NO2Br(g) + Br(g)
(Step 3)
Br(g) + NO2(g) 6NO2Br(g)
fast equilibrium
slow
fast
A. Prove to me that this mechanism matches the reaction’s overall stoichiometry
NO2(g) + Br2(g) +NO2Br2(g) +Br(g) + NO2(g)6NO2Br2(g) +NO2Br(g) + Br(g) +NO2Br(g)
=2NO2(g) + Br2(g) 6 2NO2Br
B. Write the three different rate laws that would be found for the three different
individual steps
Step 1: rate = k1 [NO2][Br2]
Step 2: rate = k2[NO2Br2]
Step 3: rate = k3[Br][NO2]
C. What is the true rate law for this mechanism when the fast equilibrium for Step
1 is included with the rate law of the slow second step.
From fast equilibrium k1[NO2][Br2] = k-1 [NO2Br2]
[NO2Br2] = k1/k-1 ×[NO2][Br2]
Substitute in for [NO2Br2] in the rate equation for step 3
Rate = k2× k1/k-1 ×[NO2][Br2]
2. (12 points) In this chapter we had the equation
Tell me what each symbol in the equation means, and then tell me what the
equation as a whole means.
z= collisional frequency
ñ= steric or orientation factor
e is the exponential function
Ea is the activation energy
R is the gas constant in units of J/KAmol
T is the temperature in K
As a whole this equation says that the rate of a reaction can be broken down into the # of
collisions times a factor that says if the collision has the right alignment time a factor that
determines what fraction of the collisions have a high enough energy at a given
temperature.
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3. (12 points)
A. I have a reaction with a rate of 12 sec-1 at 25oC. If the reaction is 100 times faster
at 100oC, what is the activation energy for this reaction?
Point 1 rate = 12, T = 25+273 = 298K
Point 2 rate = 12x100 = 1200, T = 100+273 = 373K
B. If I use a catalyst to decrease the activation energy to ½ of its original value, what
is the reaction rate for the catalyzed reaction at 100oC
(If you don’t trust your answer for part A use can use 35 kJ/mol for the activation energy
of the catalyzed reaction) So Ea = 28,150J
4.(8 points) A. I have a reaction where the energy of the reactants is -2 kJ, the ÄH for the
reaction is +3 kJ, and the reaction in the forward direction has an activation energy of 10
kJ. Draw the Activation Energy Diagram for this reaction. (Diagram with the hill in the
middle)
Diagram starts at -2, the hump in the middle is at +8 and then end of the graph is
at +1 kJ
(2 points) B. What is the energy of the products in this reaction?
-2+3 = +1
(2 points) C. What is the activation energy for the REVERSE reaction?
+8-1 = +7 kJ
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5. (12 points)
A. What is a plot of ln(k) vs 1/T used for?
This is used to find the activation energy for a reaction
B. What is the reaction quotient, Q, used for?
This is used to determine if a reaction will create products and use of reactants, or
create reactants and use up products as it moved toward equilibrium
C. What is an ICE table used for?
An ICE table is used in equilibrium calculations to keep track of how the initial
conditions will change as a system approaches equilibrium.
D. How does one relate a KC to a KP ?
KP = Kc(RT)Än
E. How do you use the ÄH of a reaction to predict how a change in temperature will
shift an equilibrium constant?
A positive ÄH means you add heat like a reactant, a negative ÄH mean you add
heat like a product, and they you apply LeChatlier’s principle for the chang ein temperatur
either increasing or decreasing the heat.
6. (12 points) The reaction PCl5(g) 6 PCl3(g) + Cl2(g) has a Kc of 1.8M at 250o C
A. Calculate Kc for the reaction Cl2(g) + PCl3(g) 6 PCl5 (g)
This is reverse so K = 1/1.8 or .556 M-1
B. Calculate Kc for the reaction 3PCl5(g) 6 3PCl3(g) + 3Cl2(g)
This multiplies all the stoichiometirc coefficeints by 3 so
Knew = Koriginal3
=1.83
=5.832 M3
C. Calculate Kp for the reaction 2Cl2(g) + 2PCl3(g) 6 2PCl5 (g)
First, this is 2 times the reaction in A so KC = .5562 = .309M-2
Next we need to change from KC to KP
KP = KC(RT)Än; Än = 2-4 = -2
.309M-2(.08206×(250+273))-2
.309/(42.92)2
=1.67x10-4atm-2
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7. Given that the equation H2(g) + CO2(g) W H2O(g) + CO (g) has a Kp of 1.44, what are the
equilibrium pressures of all gases if you start a reaction with 5 atm pressures for both H2
and CO2 and 1 atm pressures for both H2O and CO?
(2 points for Q, 2 Points for the direction the equilibrium shifts, 2 points for the ICE table,
2 points for an equilibrium expression that includes X, 2 points for solving for X, 2 points
for your final answer)
Q=12/52 = .04
Q<K so reaction moves toward products
Initial
Change
Equilb
H2(g) + CO2(g) W H2O(g) + CO (g)
5
5
1
1
-X
-X
+X
+X
5-X 5-X
1+X 1+X
1.44 = (1+X)(1+X)/((5-X)(5-X)
1.44 = (1+X)2/((5-X)2
sqrt(1.44) = (1+X)/((5-X)
1.2= (1+X)/((5-X)
1.2(5-X)=1+X
6-1.2X=1+X
6-1=1X+1.2X
5=2.2X
X=5/2.2 = 2.27
Pressure H2 = pressure CO2 = 5-2.27 = 2.23 atm
Pressure H2O = Pressure CO = 1+2.27 = 3.27 atm
8. (12 points) The reaction N2 + 3H2 W2NH3 has a ÄH of -92 kJ/mol. If the reaction is at
equilibrium at 25oC in a 5 L container that also contains CO2, but I want to decrease the
amount of N2 in the container.
Will it matter if I change the amount of NH3 in the container? Yes or No (Circle one)
Should I increase or decrease the amount of NH3 in the container? (Circle one)
Will it matter if I change the amount of H2 in the container? Yes or No (Circle one)
Should I increase or decrease the amount of H2 in the container? (Circle one)
Will it matter if I change the amount of CO2 in the container? Yes or No (Circle one)
Should I increase or decrease the amount of CO2 in the container? (Circle one)
Will it matter if I change the volume of the container? Yes or No (Circle one)
Should I increase or decrease the volume of the container? (Circle one)
Will it matter if I change the temperature of the container? Yes or No (Circle one)
Should I increase or decrease the temperature of the container? (Circle one)
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