課程名稱︰普通化學甲上
課程性質︰(化工系)必修
課程教師︰蘇志明
開課學院：理學院
開課系所︰化學系
考試日期（年月日）︰102/11/29
考試時限（分鐘）：130分鐘
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題 :
Second General Chemistry Examination Nov.29,2013
Θ
1. How do you pronouce ΔH ? (以下"Plimsoll" 符號皆以 "Θ" 代替)
2. (a) Consider a one-dollar coin with a mass of 3.8 g lying on a horizontal
desk top. Calculate the entropy change if the coin was dropped from a height
of 1 cm to the table top. (Note: The gravitational acceleration is
9.80 m/s^2 and the desk temperature is 298 K.)
(b) Use the Boltzmann's entropy equation to estimate the probability that
the coin as described in (a) would jump 1 cm from the desk top by absorbing
the equivalent amount of thermal energy from the table.
(c) If the nitrogen molecule was used in (a) and (b) instead, estimate the
probability that it would jump to a height of 4000 m, the altitude of the
Jade Mountain. What would be the atmosphere pressure of the Jade Mountain
if the atmospheric pressure of our classroom is 1 atm?
3. At 1 bar, calculate the molar Gibbs free energies of liquid water at 373 K,
and gaseous water (i.e. water vapor) at 400 K, respectively. (Assume that
the absolute standard molar enthalpy of liquid water at 298 K is 0.00. The
other related standard thermodynamics data are attached at the end of this
problem sheet.)
4. Consider the following equilibrium reaction of the water vapor:
H2O(g) ←→ H2(g) + 1/2O2(g)
(a) Calculate the standard Gibbs free energy changes, and the equilibrum
constants of the above reaction at 298 K and 2000 K, respectively.
(b) Assuming that the water vapor pressures in the reaction chamber were
found to be 0.01 bar at 298 K, and 1.00 bar at 2000 K, calculate the partial
pressures of H2(g) at the above two temperatures, respectively.
5. By mixing 4.0 g of pure sugar (i.e. sucrose) with 1000.0 of water at 25°C,
you got a sugar water solution. The density of the solution is 1.004 g/cm3.
Calculate the molality (m), molarity (M), and mole fraction (Xsugar) of the
sugar solution. The molecular weight of sucrose is 342 g/mol.
6. Following the sugar solution system as described in Problem (5), answer the
follwing questions:
(a) Calculate the osmotic pressure of the sugar solution with respect to
pure water.
(b) Calculate the change of the molar Gibbs free energy of water due to the
pressure of the osmotic pressure as obtained in (a) at 298 K. (Note: the
molar volume of water is 18 cm3.)
(c) Calculate the change of the molar Gibbs free energy of water as its mole
fraction varies from 1 (i.e. pure water) to Xwater (sugar solution) at
298 K and 1 bar.
7. The phase diagram of water is shown below:
│
│ \ Liquid ∕
│ \ ∕
│ \ ∕
│ \ ╱
│ \ ╱
│ Solid \ ╱
│ \╱
│ ∕
│ ∕ Gas
│ ∕
│ ∕
│ ∕
│ ∕
│
└──────────────
Consider the phase equilibrium P-T curve between the gas and liquid phase:
Θ
H2O(l) ←→ H2O(g) with ΔH vap
(a) The equilibrium constant of the above reaction can be expressed as
Keq = Pvap
Where Pvap is just the water vapor pressure. From the thermodynamics
relation between Keq and ΔG°, derive the relationship between Pvap and T
(or in terms of the relation between P1, T1 and P2, /t2. Here, the numbers
1 and 2 indicate two separate equilibrium conditions).
(b) In a different yet more general approach, one could start from the
follwing general relationship which describes the equilibrium between the
liquid and gas phases:
__ __
Gl = Gg
From here and with the help of some general thermodynamics relations, one
could obtain the differential equation which describes the function
Pvap = Pvap(T). Derive this differential equation for this approach. With
some further approximation, one could also obtain the functional form as
derived in part (a). Show this procedure.
(c) Why the slope of the P-T relation between the solid-liquid phase
equilibrium of the water is negative as shown in the above phase diagram?
Some constants and equations:
dS = dqrev/T H = E + PV G = H - TS
dG = VdP - SdT
d(G/T) H
-------- = - ---
dT T^2
ΔG°= -RT ln Keq
∫ln ax dx = x ln ax -x
∫1/x dx = ln│x│
ln(1+x) ≒ x (│x│<< 1) (Taylor's expansion)
S = kB lnQ
(一) 定壓 (1 bar) 下的莫耳熱容 (Cp) (J/mol‧K)
O2(g) = 29.4 H2(g) = 28.8 H2O(l) = 75.3
H2O(g) = 33.6 N2(g) = 29.1
(二) 壓力 1 bar, 溫度 298 K 的標準生成焓 (Hr°, standard enthalpy of
formation) (kJ/mol)
H2O(l) = -285.8
H2O(g) = -241.8
(三) 溫度 298 K 的標準莫耳熵 (S°, standard molar entropy) (J/mol‧K)
H2(g) = 130.6 O2(g) = 205.1 H2O(g) = 188.8
H2O(l) = 69.9
(四) 相關常數:
-1 -1 -1 -1
氣體常數 R = 0.0831 L bar K mol = 0.0821 L atm K mol
-1 -1
= 8.31 J K mol
kB = R/NA
1 bar = 10^5 Pa
1 J = 1 Pa m3
1 L atm = 101.3 J
1 atm = 760 torr = 1.01325 bar
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