Name:_____________
(4 points)
Chemistry 114
Third Hour Exam
Remember- Show all work for partial credit. Each problem is worth 12 points.
1. In our theory of kinetics, we proposed the equation
could explain the rate of a chemical reaction.
A.)In this equation what do each of the symbols mean?
k is the rate constant you get from kinetics
p is the steric factor that deals with orientation
z is the number of collisions
e is the exponential function
Ea is the activation energy, the enegy needed to a successful reaction
R is the gas constant in J/K@mol
T is the temperature in K
B.) Explain the equation in words.
Start here: “The rate of a reaction depends on......
The rate of a reaction depends on th number of collisions x the fraction of those
collisions that have he right orientation x the number of those collisions that have an
energy equal to or greater than the activation energy.
C.) We also used the equation
What is A in this equation?
A is called the frequency factor (sometimes called the Arrhenius factor)
and represents the number of collision x the fraction of collisions with the correct
orientation.
2. If a reaction has a rate of .71 M-1@s-1 at 331o,C and the reaction has an activation
energy of 130 kJ/mol, what is the rate of the reaction at 381oC?
R = 8.314 J/K@mol; Ea in J = 130,000 J
331 + 273 = 604K, 381+273 = 654K
2
3. For the reaction O2NNH2(aq)6N2O(g) + H2O(l)
The following mechanism has been proposed:
Step 1. O2NNH2(aq)
O2NNH-(aq) + H+(aq)
Fast equilibrium
Step 2. O2NNH-(aq) 6 N2O (g) + OH-(aq)
Slow
Step 3. H+(aq) + OH-(aq) 6H2O(l)
Fast
A. Show that the proposed mechanism agrees with the overall stoichiometry of
the reaction
Sum of reactions
O2NNH2(aq)+O2NNH-(aq)+H+(aq)+OH-(aq)6O2NNH-(aq)+H+(aq)+N2O(g)+OH-(aq)+H2O(l)
= O2NNH2(aq)6N2O(g) + H2O(l) so does sum to net ionic
B. Which step ( 1, 2, or 3) determines the overall rate of the reaction?
Slow step #2
C. If you include the fast equilibrium what should the overall order of this reaction
be with respect to
Should get equation: rate = k1k2/k-1 [O2NNH2]/[H+]
O2NNH2 (aq)?
1st order
+
H (aq) ?
Actually -1 since in the denominator!
OH (aq) ?
Zero order
D. Write the rate constant(k) for this reaction in terms of K1, K-1, K2 and K3
k1k2/k-1
4. Below is an activation energy diagram of a reaction.
Hand sketched into test
A. Label following parts of the diagram
E of reactants
E of products
Hand sketched like figure 18.3 from text
ÄE of reaction
Ea
B. Is this reaction endothermic or exothermic
This test E of product < E of reactant so exothermic
C. This reaction can also be performed using a catalyst. On the graph show
what a catalyzed reaction would look like. Hand sketched lower Ea peak
5. Define the following terms:
3
Reaction quotient - The number you get when you plug your starting
concentrations into the equilibrium expression
dynamic equilibrium - The idea that when your reach equilibrium the reaction has
NOT stopped. Both forward and backward reactions continue, but they continue at the
same rate so you cannot see an overall change in the system.
elementary step - A single step in a reaction mechanism, and the only reaction
equation that you can write the rate law of a reaction from the molecularity of the
equation.
steric factor - A factor that takes into account the orientation of a collision in the
collision model of kinetics.
Heterogenous catalyst - A catalyst that is in a different phase than the reactants
or products.
enzyme - A biological catalyst, usually a protein.
substrate - The reactant in a biological catalytic reaction.
6. The reaction A(g) +B(g) W C(g) + D(g) has a KC of 1.45x10-3
The reaction C(g) + F(g) W G(g) has a KC of 4.86x10+5
A. What is KC for the reaction C(g) + D(g) W A(g) + B(g)?
Reverse of eqn 1, so Knew = 1/Kold
= 1/1.45x10-3 = 690
B. What is KC for the reaction 2C(g) + 2F(g)W 2G(g)?
We have multiplied all the coefficients in eqn 2 by 2 so
Knew = (Kold)2 = (4.86x105)2 = 2.36x1011
C. What is KC for the reaction A(g) + B(g) + F(g) W D(g) + G(g)?
This equation is what you get when you add the two equations together, so you
multiply the K’s together
K = 1.45x10-3 x 4.86x10+5 = 7.05x102
D. Assuming that the temperature is 25o C, what is KP for the reaction
A(g) + B(g) + F(g) W D(g) + G(g)?
T = 273 + 25 = 298K
KP = KC(RT)Än
Än = 2-3 = -1
using the result of C for KC
KP = 7.05x103 (.08206(298))-1
= 7.05x103 /(.08206(298))
=28.8
4
7. The reaction H2(g) + I2(g) W 2HI(g) has a KC of 53.5. If a 1 liter container is filled with
.75 moles each of H2, I2 and HI and the reaction is allowed to proceed.
A. What is Q for this set of initial conditions?
Q = .752/[.75x.75] = 1
B. Will the concentration of HI increase or decrease as the reaction moves
toward equilibrium
Q<K Not enough products, products will form.
C. What are the concentrations of all the gases at equilibrium?
H2(g) + I2(g) W 2HI(g)
Initial .75
.75
.75
Change -X -X
+2X
Equilibrium .75-X .75-X .75+2X
53.5 = (.75+2x)2 / [(.75-x)(.75-x)]
53.5 = (.75+2x)2 /(.75-X)2 = [(.75+2x)/(.75-X)]2
sqrt(53.5) = 7.314 = (.75+2x)/(.75-X)
7.314(.75-X)=.75+2x
5.486 - 7.314x = .75 + 2x
5.486-.75 = x(2+7.314)
X = (5.486-.75)/(2+7.314)
X= .508
[HI] = .75 + 2(.508) = 1.767
[H2]=[I2] = .75-.508 = .242
8. Consider the gas phase reaction: 2NH3(g) W N2 (g) + 3 H2(g) ÄHo = + 92 kJ
If the system is at equilibrium, but the following changes occur, which way will
the system shift?
Change
Shift (circle one)
NH3 removed
To Right
No Change
To Left
H2 added
To Right
No Change
To Left
N2 removed
To Right
No Change
To Left
He gas added (Does not react)
To Right
No Change
To Left
Increase container volume
To Right
No Change
To Left
Decrease Temperature
To Right
No Change
To Left