Download Chemistry(I) Final Exam 1/11/2008

Chemistry(I)
Final Exam 1/11/2008
ANSWER
Total Score: 102 points
Time: 2 hours (18:00-20:00)
You are not allowed to ask TAs questions regarding the content of the test.
However, using non-programmable calculators is allowed.
1.
(a) Give a brief statement about “Charles’s law”. (b) Why can we have the idea
about “the lowest possible temperature (absolute zero)” according to this law? (c)
A gas occupies 506 inch3 at 10 psi/-4°F, and 313 inch3 at 20 psi/104°F. Use
Boyle’s law and Charles’s law to find the absolute zero in degrees Fahrenheit
(°F).
(4+4+4 points)
(a) For a fixed amount of gas under constant pressure (2 points), the volume
varies linearly with the temperature, V=constant x T.
(2 points)
(b) According to Charles’s law, lower temperature causes smaller volume of
a gas. The smallest volume of an ideal gas can have is 0. So the temperature
at which an ideal gas reach zero volume is the absolute zero.
(4 points)
(c) 313 inch3 at 20 psi/104°F Î 626 inch3 at 10 psi/104°F (Boyle’s law)
let x°F is equal to absolute zero
under constant pressure, V1/T1 = V2/T2 (Charles’s law)
506 / (-4 - x) = 626 / (104-x) Î x = -459.4 (°F)
2.
(4 points)
Through a series of enzymatic steps, glucose and oxygen are consumed in our
bodies to provide energy.
C6H12O6(s) + 6 O2(g) Æ
6 CO2(g) + 6 H2O(l)
Given that the partial pressure of oxygen in the air is 0.2 atm (pressure of air =
1.0 atm) and the temperature is 300 K. (a) Calculate the volume of air needed to
consume 18 g of glucose. (b) Actually, in one breathing, 500 mL of air is taken
into the lung and only 1/4 of oxygen can diffuse into the blood. Calculate the
number of breathing needed. (R = 0.082 L·atm/mol·K)
(3+3 points)
(a) M.W. of glucose = 180 ; 18 g glucose = 0.1 mol Î 0.6 mol O2 is needed
PV = nRT , 0.2xV = 0.6x0.082x300 Î V = 73.8 (L)
(b) 73.8/0.5/(1/4) = 590.4
(3 points)
(3 points)
3.
Consider the properties of H2(g) and H2O(g). (a) Assume they are ideal gases. The
root mean spead of H2O(g) is 640 m/s at 25°C. Calculate the root mean spead of
H2(g)? (b) The van der Waals a parameters of these two gases are 5.536 and
0.2476 L2·atm/mol2. Use your knowledge of the factors governing the magnitudes
of a to assign each of these two gases to it’s a value. ( (P+an2/V2)(V-nb) = nRT)
(3+4 points)
(a) M.W. of H2 = 2 , H2O = 18
Î root mean speed of H2 = 640 x (18/2)1/2 = 1920 (m/s)
(3 points)
(b) parameter a represents the attraction between gas molucules,
Î 5.536 L2·atm/mol2 Æ H2O
0.2476 L2·atm/mol2 Æ H2
(4 points)
4. Compute the fraction of the space occupied by atoms in: a) cubic closed-packed
structure, b) cubic body-centered structure, c) cubic primitive structure. (15
points)
a)
“ volume of a single atom = 4/3 π r3
“ number of atoms in a unit cell = 4
“ total volume occupied by atoms = 16/3 π r3
“ length of the diagonal of the wall of the unit cell = 4 r
“ unit cell constant = a = 2 √2 r
“ volume of the unit cell = 16 √2 r3
“ fraction of space occupied by atoms = π/(3 √2) = 74.05%
b)
“ volume of a single atom = 4/3 π r3
“ number of atoms in a unit cell = 2
“ total volume occupied by atoms = 8/3 π r3
“ length of the diagonal of the unit cell = 4 r
“ unit cell constant = a = 4/√3 r
“ volume of the unit cell = 64/√27 r3
“ fraction of space occupied by atoms = π√27/(24) = 68.02%
c)
“ volume of a single atom = 4/3 π r3
“ number of atoms in a unit cell = 1
“ total volume occupied by atoms = 4/3 π r3
“ unit cell constant = a =
2r
“ volume of the unit cell = 8 r3
“ fraction of space occupied by atoms = π/6 = 52.36%
5.
The three most popular structures of ionic crystals are: 1) rock-salt structure, 2)
cesium chloride structure, 3) zinc-blende structure. Identify these three
structures among the figures below. (6 points)
A
B
C
D
E
F
Answer: 1C, 2B, 3E
6.
Rank the following liquids in order of increasing viscosity at 50º C: C6H5OH,
C6H5SH, C6H6. (6 points)
Question 5.23 (5.25 in old edition) from the textbook. Correct answer is
C6H6 < C6H5SH < C6H5OH
7.
Calculate the lattice enthalpy of silver chloride from the following data. (8 points)
enthalpy of formation of Ag(g): +284 kJ⋅mol−1
first ionization energy of Ag(g): +731 kJ⋅mol−1
enthalpy of formation of Cl(g): +122 kJ⋅mol−1
electron affinity of Cl(g): +349 (ΔH = −349) kJ⋅mol−1
enthalpy of formation of AgCl(s): −127 kJ⋅mol−1
-(−127 kJ⋅mol−1)+(122 kJ⋅mol−1)+(731 kJ⋅mol−1)+(284 kJ⋅mol−1)-(349 kJ⋅mol−1)
= 915 kJ⋅mol−1
8.
Calculate the standard reaction enthalpy for the following reaction. (7 points)
N2H4(l) + H2(g) → 2NH3(g)
Given:N2H4(l) + O2(g) → N2(g) + 2H2O(g) ΔH° = −543 kJ⋅mol−1
2H2(g) + O2(g) → 2H2O(g)
ΔH° = −484 kJ⋅mol−1
N2(g) + 3H2(g) → 2NH3(g)
ΔH° = −92.2 kJ⋅mol−1
+
N2H4(l) + O2(g) → N2(g) + 2H2O(g)
ΔH° = −543 kJ⋅mol−1
2H2O(g) → 2H2(g) + O2(g)
ΔH° = +484 kJ⋅mol−1
N2(g) + 3H2(g) → 2NH3(g)
ΔH° = −92.2 kJ⋅mol−1
----------------------------------------------------------------------------------N2H4(l) + H2(g) → 2NH3(g)
ΔH° = −151.2 kJ⋅mol−1
Ans: −151.2 kJ⋅mol−1
9. Use the bond enthalpies in following tables to estimate the reaction enthalpy for
the reactions:
(A) CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
(B) N2(g) + 3F2(g) → 2NF3(g) (5 points)
(5 points)
(A) ΔH(C-H) + ΔH(Cl-Cl) - ΔH(C-Cl) - ΔH(H-Cl)
= 412 + 242 – 338 – 431 = –115 kJ. mol-1
(B)ΔH(N2) + 3ΔH(F2) - 6ΔH(N-F)
= 944 + 3*158 – 6*195 = 248 kJ. mol-1
10.
(12%) At 00C, ∆Hfusion(H2O)= 6.01kJmol-1 and ∆Sfusion(H2O)= 22.01 JK-1mol-1.
Calculate the ∆Ssys for each of the following processes.
(1) H2O(l), -100C Æ
H2O(l), 00C
∆Hsys= 750 Jmol-1
(2) H2O(l),
00C Æ
H2O(s), 00C
∆Hsys= -6010 Jmol-1
(3) H2O(s),
00C Æ
H2O(s), -100C ∆Hsys= -370 Jmol-1
(4) H2O(l), -100C Æ
H2O(s), -100C ∆Hsys= -5630 Jmol-1
Given: Cp(s)= 37 JK-1mol-1 & Cp(l)= 75 JK-1mol-1 for H2O
.
(1) ∆Ssys= Cp(l)ln(273/263)= 2.80 JK-1
(2) ∆Ssys= -∆Sfusion = -22.01 JK-1
(3) ∆Ssys= Cp(s)ln(263/273)= -1.38 JK-1
(4) ∆Ssys= 2.80 -22.01 -1.38= -20.59(JK-1)
11. (6%) Calculate the change in molar entropy of the surroundings (∆Ssur) and the
change in molar Gibbs free energy of the system (∆Gsys) when water freezes at
-100C & 1 atm.
∆Ssur=
-∆Hsys/T= 5630/263= 21.41(JK-1)
∆Gsys= ∆Hsys -
T∆Ssys =
-5630 – 263 x (-20.59)= -214.83(J)
Or ∆Gsys= -T∆Stotal= -263 x (21.41 – 20.59)= -215.66(J)
12. (7%) At low temperatures, heat capacities are proportional to T3 (the Debye T3
law: C= aT3). Show that, near T= 0K, the entropy of a substance is equal to oneT
third of its heat capacity at the same temperature. Note: S(T) = ∫
0
T
∵ S(T)= ∫
∴ S(T)= ∫
0
T
0
dq/T= ∫
T
0
CdT/T
aT3dT/T= aT3/3 = C/3
1
dq.
T