Name:____________
Chemistry 114
Second Hour Exam
1. (10 points) I am investigating the reaction A + B 6 C using the method of initial rates,
and I have gotten the following data:
[A]
[B]
0.1
0.2
0.2
Initial rate
(mol/l@min)
0.1
0.000143
0.1
0.000306
0.2
0.001226
What is the order of this reaction with respect to A, with respect to B, what is the k of the
reaction?
.000306 k (.2 ) x (.1) y
 .2 
=
. =   ; 214
. = 2 x ; ln(214
. ) = x ln(2 );
x
y ; 214


.000143 k (.1) (.1)
.1
.761 = x × .693; x = .761/.693 = 11
.
X
.001226 k (.2 ) x (.2 ) y
 .2 
=
 ; 4.0 = 2 y ; ln(4) = y ln( 2 );
x
y 4.0 = 

.000306 k (.2 ) (.1)
.1 
1386
.
= y × .693; y = 1386
. /.693 = 2
Y
.000143
.000143
1.1
2 ; k =
(.1) (.1)
(.07943)(.01)
= .18 units are poorlydefined
.000143 = k (.1) 1.1 (.1) 2 ; k =
2. (10 points) I have a reaction that is second order with respect to O2. I observe that when
the concentration of O2 in the reaction is .05 mol/l, the t1/2 of the reaction is 7 minutes.
What is the k for this reaction?
For a second order reaction T1/2 = 1/[A o]k; 7 minutes=1/(.05)k
k = 1/[(.05 mole/l) ×7 min] = 2.857 l mole -1 min-1
2
3. (10 points) I want to analyze the reaction of 2N2O(g) 6 2N2(g) + O2(g) using the
integrated rate law approach. I run the reaction once and obtain the following data.
[N2O ](mol/l)
Time(sec)
5
5
4.5
10
20
4.0
3.5
40
80
3.0
2.5
160
How will I manipulate this data to obtain the order and the rate of this reaction? (Describe
the steps used - do not actually do the analysis)
Plot [N2O] vs time. If this plot is linear, then you have zero order kinetics, and the rate of the
reaction (k) is equal to the negative of the slope. If the plot was not linear, try the next plot.
Plot ln[N2O] vs time. If this plot is linear, then you have first order kinetics, and the rate of
the reaction (k) is equal to the negative of the slope. If the plot was not linear, try the next
plot.
Plot 1/[N2O] vs time. If this plot is linear, then you have second order kinetics, and the rate
of the reaction (k) is equal to the slope. If the plot was not linear, you are out of luck.....
4. (10 points) Define Each of the following:
Elementary StepA reaction whose rate can be written from its molecularity. It defines the chemicals
that are actually reacting a step of a chemical reaction.
Reaction MechanismA series of elementary steps that describe the reactions that occur in the pathway of
a chemical reaction
Rate Limiting stepThe elementary step of a chemical reaction that has the slowest rate. This slow
reaction limits the rate of the overall reaction and defines the overall kinetics of a reaction
5. (10 points) The balanced equation for the gas phase reaction of NO2 and F 2 is:
2NO2(g) + F 2(g) 6 2NO2F(g)
3
The experimentally determined mechanism for this reaction is rate =k [F] [NO2 ]
A suggested mechanaism for this reaction is :
NO2 + F2 6 NO2F + F (Slow)
F + NO2 6 NO2F (Fast)
Prove that this is either an acceptable or an unacceptable mechanism for this reaction.
To prove that a reaction mechanism is acceptable you must show that the sum of the
elementary steps is equal to the overall stoichiometric equation for the reaction, and the
experimental kinetics must match the kinetics proposed in the mechanism.
Summing the elementary steps we have:
NO2 +F + NO2 + F2 6 NO2F + F + NO2F =
2 NO2 + F2 6 2NO2F which matches with the stoichiometric reaction so we are OK
there. However the slow elementary step in this reaction would have kinetics: rate =
k[F 2][NO2], and this does not match the experimental kinetics so this reaction is not
acceptable.
6. (10 points) I have a reaction that has a rate of .10 min-1 at 25oC and .20 min-1 at 35oC.
Calculate the activation energy for this reaction and define the term ‘activation energy’.
k  E  1 1
X  1
1 
 .2 
ln 2  = a  −  ; ln  =

−

 .1  8.3145  298 308 
 k1 
R  T1 T2 
.693 =
X
(.000109)
8.3145
X = .693(8.3145)/.000109
X = 52,890 J / mol
Activation energy is the energy that is needed to start a reaction.
4
7. (10 points) For the reaction N2(g) + 3Cl2(g) W 2NCl3 (g), K C = 2.30x1010
Calculate the following K C values
1/3 N2(g) + Cl2(g)W 2/3 NCl3
K this reaction = [K original reaction]1/3 = 2.844x103
NCl3(g) W ½ N2(g) + 3/2 Cl2(g)
K this reaction = [1/K original reaction]1/2 = 6.59x10-6
Calculate the K P for the reaction N2(g) + 3Cl2(g) W 2NCl3 (g)
KP=K C×(RT))n; = 2.30x1010 [(.08206)298]-2 = 2.85x107
8. (10 points) I am going to mix 10 ml of 0.2M Na 2SO4 with 5 ml of 0.05 M CaCl2. Use the
Q calculation, the K sp of CaSO4 and these concentrations to predict wether or not a
precipitate will form when these solutions are mixed. [K sp is the equilibrium expression for
the reaction for CaSO4(s) WCa2+(aq) + SO42-(aq) ]
Q= [Ca2+][SO42-]
[Ca2+] = .005 l (05 mol/l)/.015 l = .01667 M
[SO42-] = .01 L (.2 mol/l)/.015 l = .133M
Q = .01667(.133) = 2.2x10-3
Q>K so you have too many products, that means reactants will form, and, in this case, the
reactant is the solid CaSO4, so a precipitate will form.
5
9. (10 points) At 25oC, K P = 5.4x10-3 for the reaction
NH4OCONH2(s) W 2NH3(g) + CO2(s)
In an experiment carried out at 25oC 1 g of NH4OCONH2 is placed in an evacuated
container and allowed to come to equilibrium. Calculate the equilibrium partial pressures
of NH3 and CO2
KP= (Pressure NH3)2 (pressure CO2)
Assume pressure CO2 =X
Pressure NH3 = 2X
5.4x10-3 = (2X)2 (X)
=4X3
X3 = 5.4x10-3 /4
X= .11 = pressure CO2
Pressure NH3 = 2X = .22
10. (10 points) In which direction will the position of the equilibrium
2HI(g) W H2(g) + I2(g)
be shifted for each of the following changes (toward products, toward reactants, or
no change)
a. H2(g) is added
toward reactants
b. I2(g) is removed
toward products
c. HI(g) is removed
toward reactants
d. Some Ar(g) is added
no change
e . The volume of the contained is doubled
no change
f. The temperature is decreased (the reaction is exothermic)
toward products