Name:
Chemical Principles II
March 8, 2011
First Midterm
There are 100 points in this exam. (-1pt for each incorrect or missing multiple choice)
R = 0.08206 L amt/mol K,
R = 8.314 J/mol K
Part 1 Multiple choice: CIRCLE the letter for each correct choice. Correct ans in boldface
1.
Why does entropy tend to increase when the temperature of a system decreases?
a.
Because the heat capacity changes with temperature change.
b.
Because the heat capacity increases with spontaneous temperature change.
c.
Because the heat capacity decreases with spontaneous temperature change.
d.
It does not; rather entropy of a system increases when temperature increases.
2.
a.
b.
c.
d.
Why does entropy of a system increase when a solid becomes a liquid or gas?
Because heat is added to the system to induce the phase change.
Because there are more energy states in the liquid or gaseous state than in the solid.
Because there are less energy states in the liquid or gaseous state than in the solid.
It does not; rather, entropy decreases with such phase changes.
3.
Why do the standard molar entropies of substances increase as the complexity of the
substance increases?
Because of increased enthalpy of the system.
Because of increased modes and degrees of freedom of movement within more complex
molecules.
Because of an overall increase in temperature of the system.
They do not; rather, standard molar entropies decrease as a substance’s complexity
increases.
a.
b.
c.
d.
4.
a.
b.
c.
d.
Under what conditions is a standard reaction entropy positive?
When there is a net production of a gas in the course of a reaction.
When a gas is consumed in the course of a reaction.
When a liquid or solid forms from gaseous reactants.
When a system undergoes a net change in temperature.
5.
a.
b.
c.
d.
How is the entropy change of the surroundings of a system computed?
By the formula: ΔS / T
By the formula: ΔH / T
By the formula: - ΔS / T
By the formula: - ΔH / T
6.
a.
b.
c.
d.
How does a negative ΔS impact the free energy change of a reaction?
It decreases ΔG.
It increases ΔG.
It does not change.
You cannot tell without knowing ΔH.
Chemical Principles II 1st Midterm 1
7.
a.
b.
c.
d.
By what biological mechanism are non-spontaneous reactions made to run at body
temperature?
By heating the bodily system in question.
By coupling them with spontaneous reactions.
By forcing enzymatic changes.
By increasing the internal energy of the body.
8.
a.
b.
c.
d.
How are Q and K related?
They are the same.
They have the same form, except that K can be used to monitor the progress of a reaction.
They have the same form, except that Q can be used to monitor the progress of a reaction.
They are unrelated, except for using the same reactants.
9.
a.
b.
c.
d.
When the value of an equilibrium constant is less than one, what does that indicate about
the position of the equilibrium?
That it lies predominantly on the reactants side.
That it lies predominantly on the products side.
That it lies roughly in the zone of equal concentrations of reactants and products.
The constant does not give any information about the position of the equilibrium.
10.
a.
b.
c.
d.
What does a large equilibrium constant indicate about the spontaneity of a reaction?
That it is spontaneous.
That it is not spontaneous.
That it may or may not be spontaneous, depending on the equilibrium temperature.
The constant does not give any information about the spontaneity of the reaction.
11.
When an equilibrium reaction is written in reverse, how can the equilibrium constant be
best expressed?
As a different K.
As the inverse of the original K, meaning 1/K.
As the negative of the original K, meaning - K.
None of the above.
a.
b.
c.
d.
12.
a.
b.
c.
d.
13.
a.
b.
c.
Why does a reaction tend to form reactants when Q > K?
Because the concentrations for K have not yet reached equilibrium.
Because Q is measured in this case when reactant concentrations are higher than at
equilibrium.
Because Q is measured in this case when product concentrations are higher than at
equilibrium.
Because the concentrations for Q are in equilibrium.
How does a catalyst affect the equilibrium constant of a reaction?
It increases the value of K.
It decreases the value of K.
It increases the value of K for an endothermic, and decreases the value of K for an
exothermic, reaction.
Chemical Principles II 1st Midterm 2
d.
It does not affect the K.
14.
a.
b.
c.
d.
e.
What happens to the value of the equilibrium constant after the temperature is raised?
It increases.
It decreases.
It increases if ΔH < 0.
It increases if ΔH > 0. endothermic, LeChâtelier and van’t Hoff : ln(KT2/KT1)
It stays the same because it is an equilibrium constant.
15.
a.
b.
c.
d.
How is vapor pressure related to equilibrium?
It is the equilibrium between the liquid and solid phase of a material.
It occurs at the boiling point of a material.
It is the equilibrium between a material’s vapor and condensed phase.
It occurs at the condensation point or dew point of a material.
16.
a.
b.
c.
d.
How do intermolecular forces affect vapor pressure?
Weak intermolecular forces can be discounted in determining vapor pressure.
Strong intermolecular forces tend to correspond to higher vapor pressure.
Strong intermolecular forces tend to correspond to lower vapor pressure.
They do not affect it at all.
17.
a.
b.
c.
d.
Why does the melting point of water decrease with pressure?
Because of its extensive hydrogen bonding network.
Because of weak, intermolecular forces.
Because it has a high vapor pressure.
It does not; the melting point of water decreases with pressure.
18.
a.
b.
c.
d.
Why is the solubility of a gas proportional to its partial pressure?
Because higher partial pressure increases the internal energy of the system.
Because increased partial pressure creates an increased number of in-solvent cavities.
Because increasing pressure increases the rate at which gas molecules contact a solvent
surface.
It is not; solubility is a state function independent of partial pressure.
19.
a.
b.
c.
d.
Raoult’s law equates what two factors?
Mole fraction of a solvent in a solution and the vapor pressure of the pure solvent.
Mole fraction of a solute in a solution and the vapor pressure of the pure solvent.
Mole fraction of a solvent in a solution and the vapor pressure of the solute.
None of the above.
20.
a.
b.
c.
d.
In the process of osmosis, what flows, and in what direction?(Use answer given in class.)
Solute, through a membrane, into a more concentrated solution.
Solute, through a membrane, into a less concentrated solution.
Solvent, through a membrane, into a less concentrated solution.
Solvent, through a membrane, into a more concentrated solution.
Chemical Principles II 1st Midterm 3
Part2
1. State the factors that affect the rate of chemical reactivity. (4 pts)
Subdivision of the physical state (or surface area)
Temperature
Concentration
Presence of catalysts
2. What is the necessary condition for a chemical reaction to be in equilibrium? (3 pt)
The RATE of the forward reaction must equal the RATE of the reverse.
3. Give three definitions of entropy that are scientifically more precise than a “measure of
disorder” and give equations which illustrates each definition. (6 pts for all)
i) At equilibrium, it is a measure of the heat passed at the equilibrium temperature
ΔS = ΔH/T
ii) It is a measure of the distance from equilibrium where theentropy will be at a maximum in an
isolated system.
ΔS = CV ln(T2/T1) or similar equations for P or V
iii) It is a measure of the available energy states
S = k ln 
4. State any correct version of the Second Law of thermodynamics. (3pts)
In an isolated system the entropywill increase to a maximum value
or
The entropy of an isolated system increases in the course of any spontaneous change.(page 267)
or
The entropy can not decrease in an isolated system (page 288)
5. Demonstrate the Second Law of Thermodynamics by calculating the total entropy when two
metal blocks of equal mass are put in contact with each other and allowed to come to
equilibrium. They are isolated from everything else. Take the heat capacities, C = 1J/K.
Calculate ΔS for each curved arrow and the total ΔS for each change in T (20, 50, and 80K).
Plot the total ΔS for each change in T. (6 pts)
Chemical Principles II 1st Midterm 4
See “Playing with entropy blocks”.
22
Chemical Principles II 1st Midterm 5
6. a) Calculate the standard molar entropy of vaporization of water at its normal boiling point
when its standard molar enthalpy of vaporization is 40.7 kJ/mol at its normal boiling point. (4 pt)
𝑜
∆𝑆 𝑜 = ∆𝐻𝑣𝑎𝑝
/𝑇𝑏𝑝
𝑜
∆𝑆𝑣𝑎𝑝
kJ
mol
=
= 109. Jmol−1 𝐾 −1
373K
40.7
6. b) Calculate the standard molar entropy of evaporation of water at 310K when the molar heat
capacity of liquid water can be taken as 75 J∙mol-1∙K-1 and of water vapor as 34 J∙mol-1∙K-1. (6 pt)
𝑜
𝑜
∆𝑆𝑡𝑜𝑡𝑎𝑙
= ∆𝑆 𝑜 (𝑙, 310 𝑡𝑜 373𝐾) + ∆𝑆𝑣𝑎𝑝
+ ∆𝑆 𝑜 (𝑔, 373 𝑡𝑜 310𝐾)
𝑇2
𝐽
373
𝐽
∆𝑆 𝑜 (𝑙, 310 𝑡𝑜 373𝐾) = 𝐶(𝑙) ln ( ) = 75
∗ ln (
) = +14.
(𝑚𝑜𝑙𝐾)
(𝑚𝑜𝑙𝐾)
𝑇1
310
𝑇2
𝐽
310
𝐽
∆𝑆 𝑜 (𝑔, 373 𝑡𝑜 310𝐾) = 𝐶(𝑔) ln ( ) = 34
∗ ln (
) = −6.3
(𝑚𝑜𝑙𝐾)
(𝑚𝑜𝑙𝐾)
𝑇1
373
𝑜
∆𝑆𝑡𝑜𝑡𝑎𝑙
= +14.
𝐽
𝐽
+ 109. Jmol−1 𝐾 −1 + −6.3
= 117Jmol−1 𝐾 −1
(𝑚𝑜𝑙𝐾)
(𝑚𝑜𝑙𝐾)
6.c) Draw a State Diagram and label the steps that demonstrate the entropy change for
evaporation at 310K in question 6.b (4pt)
Chemical Principles II 1st Midterm 6
7. (5pts) There is a temperature at (above) which any diatomic molecule will spontaneously
dissociate into two atoms. Calculate that temperature for bromine, knowing that the bond energy
for bromine is 200 kJ/mol and that the molar entropy of Br2(g) = 250 J∙K-1mol-1 and monatomic
Br (g) = 175 J∙K-1mol-1.
∆𝐺 𝑜 = 0 = ∆𝐻 𝑜 − 𝑇∆𝑆 𝑜 ,
𝑇 = ∆𝐻 𝑜 /∆𝑆 𝑜
𝑜
∆𝑆𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
= ∑ ∆𝑆 𝑜 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ ∆𝑆 𝑜 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
𝑜
∆𝑆𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
= 2 ∗ 175
𝑇 = 200000
𝐽
𝐽
𝐽
− 250
= 100
𝑚𝑜𝑙𝐾
𝑚𝑜𝑙𝐾
𝑚𝑜𝑙𝐾
𝐽
𝐽
/1𝑜0
= 2000𝐾
𝑚𝑜𝑙
𝑚𝑜𝑙𝐾
8. Nitric oxide gas reacts with chlorine gas according to the equation
2 NO + Cl2  2 NOCl
The following initial rates of reaction have been observed for certain reactant concentrations:
1.
2.
3.
Initial [NO]
mol/L
0.50
1.0
2.0
Initial [Cl2]
mol/L
0.50
0.50
1.00
Initial rate of formation of NOCl
mol/L•h
1.14
4.56
36.5
a. Calculate the rate equation describing the rate dependence on the concentration of NO
and Cl2. (6 pt)
𝑟𝑎𝑡𝑒1 [𝑘[𝑁𝑂]𝑥 [𝐶𝑙2 ]𝑦 ] 1.14 [[0.50] 𝑥 [0.50] 𝑦 ]
0.50 𝑥
=
=
=
=(
)
[[1.0]𝑥 [0.50]𝑦 ]
𝑟𝑎𝑡𝑒2 [𝑘[𝑁𝑂]𝑥 [𝐶𝑙2 ]𝑦 ] 4.56
1.00
1.14
0.50 𝑥
= (1.00)
4.56
𝑜𝑟 0.25 = 0.50𝑦 ,
log 0.25 = 𝑥 log 0.50
𝑥=2
(3 𝑝𝑡)
𝑟𝑎𝑡𝑒2 [𝑘[𝑁𝑂]2 [𝐶𝑙2 ]𝑦 ] 4.56 [[1.0]2 [0.50] 𝑦 ]
0.50 𝑦
=
=
=
= 0.25 ∗ (
)
𝑟𝑎𝑡𝑒3 [𝑘[𝑁𝑂]2 [𝐶𝑙2 ]𝑦 ] 36.5 [[2.0]2 [1.00] 𝑦 ]
1.00
0.125 = 0.25 ∗ 0.50𝑦 ,
log
0.125
= 𝑦 ∗ log 0.50 , 𝑦 = 1
0.25
Chemical Principles II 1st Midterm 7
(3𝑝𝑡𝑠)
b. Calculate the value of the rate constant. (3 pt)
1.14 mol∙L-1∙h-1 = k (0.50 mol∙L-1)2  (0.50 mol∙L-1)
Rate1 = k [NO]2  [Cl2]
Rate = 9.12 (L2 mol-2 ∙h-1
c. What is the order of the reaction in each reactant and overall order of the reaction? (1 pt)
First order in Cl2 , second order in NO , third order overall
9. Using the integrated rate law, what is plotted along the x-axis and y-axis in order to obtain a
straight line for: (6 pt)
a) A zero order reaction? x-axis time, y-axis:
[A]
b) A first order reaction? x-axis time, y-axis:
ln [A]
c) A second order reaction? x-axis time, y-axis: 1/[A]
10. The following is a proposed mechanism for the reaction of propylene oxide, C3H6O in
methanol, CH3OH containing OH ions:
Step1:
C3H6O + OH  C3H7O2
Step2:
C3H7O2
-
+ CH3OH  C4H10O2 + OH
a. What is the overall reaction? (4 pt)
C3H6O + CH3OH  C4H10O2
b. What is the molecularity of each step? (4 pt)
Step1:
bimolecular
Step2:
bimolecular
c. Write the rate equation for the overall reaction if step 1 is slower. (4 pt)
Rate = k1[C3H6O][ OH ]
(Both are species in the rate determining step, step 1)
d. BONUS: Determine the rate equation for the overall reaction if step 2 is slower and
equilibrium is established in step 1. (4 pts)
Rate = k2[C3H7O2 ][ CH3OH] , but C3H7O2 is an intermediate, so it can’t be in exp.
Chemical Principles II 1st Midterm 8
From equilibrium in first step: K = [C3H7O2 ] / [C3H6O][ OH ]
Rearranging to isolate [C3H7O2 ]
[C3H7O2 ] = K [C3H6O][ OH ]
Substituting into rate exp: Rate = k2 K [C3H6O][ OH ] [ CH3OH]
e. Identify any species that is an intermediate or a catalyst. (2pt)
Intermediate: C3H7O2 (created in one step, then used in a subsequent step)
Catalyst: OH (used in one step, then reformed in a subsequent step)
f. The overall reaction is exothermic. Draw a relative energy diagram for the reaction
mechanism where step 1 is slower. Label the levels for the reactants, products, intermediates,
and transition states; label the arrows for the first activation energy Ea and the overall
reaction energy ΔEreact. Be precise with the levels of the starting points of the arrows and the
arrowheads. (4 pts)
see fig 14.27b and 14.35b (Ed 5)
Chemical Principles II 1st Midterm 9
11. With the aid of a completely labeled diagram and at most three simple sentences, explain
why an increase in temperature T increases the rate of a reaction even though Ea is a constant at
any T. (8 pt)
Answer:
1) Draw Maxwell-Boltzmann distribution diagram, Fig. 13.26, page 560
2) At higher temperature there is more area under the curve to the right of Ea ;
therefore, there is a greater concentration of molecules with a sufficient energy to
react.
12. Biologists
have a “rule of thumb” that most reaction rates for biological systems will double
when the temperature is raised 10oC. Using the Arrhenius equation, calculate the activation
energy of these reactions from 27 to 37oC. (6 pts)
k2 = 2∙k1
k  E 1 1 
ln  1    a   =ln(k1/2k1)=ln(1/2)=-0.693=(-Ea/R)∙Δ(1/T)
k2   R T1 T2 
Solve for Ea
13. (on 2 pages) The following two reactions between gases were at equilibrium. Determine the
effect on the concentrations of all gases in order for the reactions to reestablish equilibrium after
the following changes are applied. Put an up arrow if the concentration increases for the gas, a
down arrow for a decrease and a zero if there is no change. In the “Eq” column put stl or str to
indicate the shift to the left or right to reëstablish equilibrium or zero if equilibrium is
maintained.(10 pt)
1)
2 SO3(g)
∆H = +184 kJ
2 SO2(g) + O2(g)
SO3
SO2
O2
Eq
-----
-----
-----
-----
a. Decrease the temperature
up
down
d
stl
b. Decrease the pressure
d
u
u
r
c. Decrease the volume
u
d
d
l
d. Remove some SO2(g)
d
u
u
r
e. Use Ni catalyst
all 0
Chemical Principles II 1st Midterm 10
2)
SO2(g) + NO2(g)
∆H = - 40 kJ
SO3(g) + NO(g)
SO2
NO2
SO3
------
----
----
u
d
stl
up
g. Increase the pressure
all 0 (moles of gas same on both sides)
h. Increase the volume
all 0
i. Add some SO3(g)
up
j. Use no catalyst
all 0
down
Eq
-----
f. Increase the temperature
u
down
NO
-----
d
stl
BONUS:
(1pt) What is the origin of the word “entropy”?
- in
- place or turn
“in place”, inside, in turning, transformation into
(2 pts) What is the mathematical form (the function of x) for the Maxwell-Boltzmann
distribution?
f(x) = x
2
e
2
(-x )
Chemical Principles II 1st Midterm 11