Chemistry 123: Physical and Organic Chemistry
Midterm 2
March 13, 2009
Name: ______________________
Student ID#: ________________
/50
Total Marks
1) The optical rotation of plane-polarized light of the reactant molecule
presented below is +12.1°. Show the reaction products and indicate the net
optical rotation of the products in the SN1 reaction presented below?
/4
Rotation = 0°, equal mixture
1.5 marks each structure. 1 for correct optical rotation.
If OH product indicated 1 for each structure
2) Indicate the relationship between the two molecules presented below and
indicate which is more stable and why.
Diastereoisomer or cis / trans isomers – 1 Mark;
Trans more stable due to less steric hinderance between Br.
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/2
Chemistry 123: Physical and Organic Chemistry
Midterm 2
March 13, 2009
3) If the ratio of Carbon-14 to Carbon-12 in a living tree is 1.3 x 10-12, what will
this ratio be in 2000 years if the half life (t½) of Carbon 14 is 5730 years? What
is the rate constant (k) of the carbon-14 decay?
.
[C-14] ini
t 1/2
k = (0.693)/"t 1/2"
1.30E-12
5730
0.000120942
ln[A]t = -kt + ln [A]0
ln[A]t =
years -1
-2.76E+01
t=
[A] =
2000
/4
years
1.02069E-12
Ratio = 1.02 x 10-12 in 2000 years (3 marks)
Rate constant = 1.2 x 10-4
rate at initial concentration)
(1 marks, no deduction if they calculate
4) If a “Van Hoff plot” is generated for a reaction and a linear regression of this
plot indicates a “best fit” line of Y= 9273.5X – 14.6 Calculate the values for ΔH
and ΔS. If the temperature if set to 30°C, would the reaction be spontaneous?
Support your assertion by calculating the value for ΔG.
Question 4
Plot ln(K) vs 1/t
-dH/R = slope =
dH =
dH =
9273.5
-77099.9
-77.1
dS/R=intercept =
dG = dH-TdS
-14.6
dG =
T=
j/mol
kJ/mol
dS=
303.15
121.3844
-40.3 kJ/mol
It is Spontaneous
R=
J/mol
8.314
dS =
/6
J K-1 mol-1
-0.121
KJ/mol
Note: use KJ or J for both dH &dS
(2 marks) each dG, dS and dG (spontaneous)
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Chemistry 123: Physical and Organic Chemistry
Midterm 2
March 13, 2009
5) Consider a beaker containing 18 mL of water initially at 20°C that is cooled to
to -10°C. Describe each step of the process and calculate the amount of energy
that would need to flow in or out of the system. At each step indicate if the
entropy is increasing or decreasing and under what conditions the reaction would
be spontaneous. How much total energy would need to be added or removed?
18
18
Step 1
mL
g/mol
1
1
g/mL
mol
18
g H2O
20
C to
0 C
Cooling water
Specific
dT =
20 heat
4.18 J/g
75.24 J/C
Heat for this step
1504.8 J
1505 J flows out of the sytem (-1505 J), entropy decreases, spontaneous when T< or = 0C
Step 2
0 C to
0 C
Freezing
moles
1 mol
6010 J/mol
6010 J
6010 J flows out of the sytem (-6010 J), entropy decreases, spontaneous when T< or = 0C
Step 3
0
C to
-10 C
Cooling water
Specific
dT =
10 heat
2.01 J/g
36.18 J/C
Heat for this step
361.8 J
361.8 J flows out of the sytem (-361.8 J), entropy decreases, spontaneous when T< or = -10C
Total energy out of system
Total energy out of system
7876.6
7.8766
J
KJ
(2 marks each correct energy calc per step, 1 correct entropy indication, 1
for correct spontaneous statement, ½ off per error on entropy or
spontaneous)
6) Concisely state the 3 laws of Thermodynamics preferably using formula’s.
First law) law of conservation of energy.
ΔU = q + w
Second law) All spontaneous processes produce an increase in the entropy of
the universe.
Third law) The entropy of a pure perfect crystal at 0 K is zero
@ T = 0K, S = 0
(2 marks each)
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Chemistry 123: Physical and Organic Chemistry
Midterm 2
March 13, 2009
7) If the ΔG° for the reaction below is -33.3 kJ/mol and the initial partial pressures
of each of the gasses is given below the equation, calculate the ΔG for the
reaction at 50°C. Is it spontaneous? What is the equilibrium constant for this
reaction?
N2(g) +
3 H2(g)
1.0 atm
3.0 atm
↔
/7
2NH3(g)
0.5 atm
ΔG = ΔG° + RT ln Q
Q = (0.5)2 / (1) x (3.0)3 = 0.00925 (1 mark)
dG*
T=
dG calc =
-33.3
323.15
-45879.3
KJ/mol
K
J/mol
-33300
Q=
-45.9
J/mol
0.009259259
kJ/mol
8.314
j K-1 mol-1
(3 marks)
It is spontaneous (1 mark)
The K value is;
ln(K) = dG*/(-RT)
ln(K) = dG*/(-RT)
13.43381613
12.39452972
T=
T=
298.15
323.15
K=
K=
6.83E+05
2.41E+05
The rate constant should be calculated at 323K, but also take the one at 298K if
it is calculated properly. (2 marks)
8) ΔfH° of NH3 is -80.29 kJ/mol, determine ΔS° for the reaction presented in
question 7.
dS* is at standard conditions or
25C
dH*-dG* = TdS* =
dG* =
dH* =
-33.3
-80.29
-46990
at 298.15 K
dS* =
-46990
at 323.15 K
dS* =
kJ/mol
kJ/mol
157.6052323
145.4123472
-33300
-80290
/2
J/mol
J/mol
J/mol
J/mol
The answer is at 25C or 298.15 K, only take ½ mark off if calculated at
323.15 K
(2 marks)
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Chemistry 123: Physical and Organic Chemistry
Midterm 2
March 13, 2009
/11 (1 point each)
Multiple Choice (circle best answer)
8) Which best describes this chemical structure?
A) Primary amine
C) Secondary amine
B) tertiary amine
D) none of these
9) For the chemical structure presented in question 8, which best describes the
carbon connected to the NH2 group.
A) Primary carbon
B) tertiary carbon
C) Secondary carbon
D) none of these
.
10) For the chemical structure presented in question 8, How many secondary
hydrogen atoms are there?
A) 9
B) 0
C) 2
D) 4
11) Which of the following correctly names the chemical structure presented in
question 8?
A) 2-methylbutan-2-amine
C) Tert-butamine
B) Dimethyl-propanamine
D) 2,methyl-2,amine butane
12) What of the following is most reactive in an SN1 reaction?
A) Primary alkyl halide
B) tertiary alkyl halide
C) Secondary alkyl halide
D) not enough information to determine
.
13) For graphite which would be numerically larger?
(A) ΔfG°
(B) ΔfH°
(C) ΔS°
14) Which of the following would have the greater ΔS?
(A) C(s)
(B) C(g)
(C) CO(g)
(D) CO2(g)
15) If a reaction has a half-life of one day and it is initiated on Monday, how much
remains on Friday?
(A) ~20%
(B) ~5%
(C) ~3%
(D) ~12%
if you count Monday, then 5 days = 3.25%; If 4 days 6.25%, so marks also
for B
16) For the reaction, N2(g) + 3 H2(g) ↔ 2NH3(g) , the rate would be;
A) First order
B) Second order with regard to [NH3]
C) Third order
D) not enough information to determine
Page 5 of 6
Chemistry 123: Physical and Organic Chemistry
Midterm 2
March 13, 2009
17) CO2 is a greenhouse gas because….
(A) It absorbs infrared radiation
(B) It has a high entropy
(C) It makes plants grow better in a greenhouse
(D) It has a high enthalpy of formation
18) Carbon dating is possible due to;
(A) open minded rate laws
(B) Irradiation of N2(g) by cosmic radiation
(C) Naturally occurring Carbon isotopes (D) radioactivity from nuclear bombs
sp ht H2O(l) = 4.18 J g-1 ºC-1
sp ht H2O(s) = 2.01 J g-1 ºC-1
ΔfusHº = 6.01 kJ/mol
ΔvapHº = 44.0 kJ/mol
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