- Although (¶
H/¶
p)T could be derived tortuously from heat capacity expressions, there’s a much, much easier way if you use Maxwell’s equations! In fact, all you’ll need to know is the integral, G = H – TS, and total differential, dG = Vdp – SdT, versions of the Gibb’s Free Energy. Use those to derive the simple expression for (¶
H/
¶
p)T, the isothermal Joule-Thompson Coefficient, m
T.
H = G + TS
(¶H/¶p)T
= (¶G/¶p)T
+ T(¶S/¶p)T
dG = Vdp - SdT means that (¶G/¶p)T = V
and Maxwell's equation from this total differential is
(¶S/¶p)T
= - (¶V/¶T)p
by the equivalence of the cross-derivatives. So
mT
= (¶H/¶p)T
= V - T(¶V/¶T)p
- The expression you’re looking for in (1) is
mT =
V – T(¶V/¶T)p,
and we’ve already shown in class that it must be zero for ideal gases, pV=RT (V presumed molar). But it’s not zero for a Van der Waals gas. The major feature that makes m
T non-zero is VdW’s attraction term not its molecular volume term. So use p = (RT/V) – a/V2 where a = 1.408 atm L2/mol2 for N2, and obtain m
T at 1 atm and 25°
C. Experimentally, it’s – 0.072 L. (You may need the quadratic formula x = [–b±(b2–4ac)½]/2a , but only the + sign works.)
p = (RT / V) - a / V2
0 = (¶p/¶T)p
= R / V - (RT / V2)(¶V/¶T)p
- (-2a / V3)(¶V/¶T)p
(¶V/¶T)p
= (R / V) / [ (RT / V2) - (2a / V3) ]
= R / [ (RT / V) - (a / V2)
- (a / V2) ]
= R / [ p - (a / V2) ]
T(¶V/¶T)p
= RT / [ p - (a / V2) ]
p = 1 atm = (RT / V) - (a / V2)
rearrange to solve for V (/mol presumed)
pV2 = RTV - a
p V2 - RT V + a = 0
V = [ + RT + { (RT)2 - 4 pa }½ ] / 2p
RT = 0.08206 atm L mol-1 K-1 (298 K)
= 24.45 atm L (surprise)2
V = [(24.45 atm L)+{(24.45 atm L)2-4(1 atm)(1.408 atm L2)}½]/2 atm
= 24.39 L not quite 24.45 due to the attractions
mT = V - T(¶V/¶T)p
= V - { RT / [p - (a / V2)] }
= 24.39 L - { 24.45 atm L
/ [1.0 atm - 1.408 atm L2 / (24.39 L)2] }
= - 0.118 L (not bad for an estimate; only 64% high)
- Using the partial VdW equation of question (2), p = (RT/V) – a/V2, find the temperature at which N2’s fugacity is 9.8 atm when p = 10 atm. (Hint: when you integrate, substitute the ideal gas equation to simplify! But don’t do that before you integrate!) a is as before.
f / p = f = 9.8 atm / 10.0 atm = 0.98
lnf =
integral0p (Z-1)/p dp
= ln(0.98) = - 0.202
Z = pV / RT = 1 - a / (RTV)
Z-1 = -a / RTV
lnf = (-a / RT)
integral0p (1/pV) dp (substitute pV~RT)
~ (-a / RT) (1/RT) integral0p dp
= [-a / (RT)2] p
T2 = - a p / (R2 lnf)
T = [- a p / (lnf)]½ / R
= [-1.408 atm L2 × 10 atm / (-0.0202) ]½ / (0.08206 atm L K-1)
= 322 K
- Nitrogen’s normal boiling point is 77.34 K. But it doesn’t go critical until 130 K. If the critical pressure is 34 atm, what is nitrogen’s average D
vapH over that temperature range? (Hint: at its normal boiling point, D
vapH is 5.586 kJ/mol, but you don’t need that. It’s just a confirmation.)
p = p0 e-(DvapH / R)×[(1/T) - (1/T0)]
or
-(DvapH / R) = ln(p/p0) / [(1/T) - (1/T0)]
or
DvapH = R ln(p/p0) / [(1/T0 - (1/T)]
= 8.314 J/mol K-1 ln(34 atm/1 atm) / [(1/77.34) - (1/130)] K-1
= 5598 J/mol = 5.598 kJ/mol
- At 136.7°
C, chlorobenzene’s vapor pressure is 863 torr while bromobenzene’s is only 453 torr. Assume that they form an ideal mixture. (a) What’s the mole fraction of chlorobenzene in such a mixture that has a vapor pressure of 760 torr at that temperature? (b) Suppose you prepared such a mixture, raised its temperature to 136.7°
C, and it wasn’t boiling! What would that tell you about the relative (non-ideal) intermolecular forces?
Ideal mixture implies a linear relation between both partial
pressures and the corresponding liquid mole fraction. That implies a
linear relation of the total pressure (if the vapor's ideal too).
- p = p0fBr
+ XfCl(p0fCl
- p0fBr)
p = 453 + (863-453)XfCl
= 453 + 410 XfCl
all pressures in torr
XfCl = (p-453)/410 = (760-453)/410
= 0.749 hence
XfBr = 1 - 0.749 = 0.251
- Since ambient pressure is 760 torr, we expect that
mixture to be boiling. If it's not, the London and dipole-dipole forces
binding the chlorobenzene to the bromobenzene must
be stronger than those binding each molecule to its own kind, making the solution
a more attractive phase, and lowering the vapor pressures.
-
- Derive the pH expression for the stoichiometric point in the titration of a weak base, pKb, by a strong acid in terms of the concentration of the "salt," [BH+] = S.
- Hydrazine is a weak base with pKb of 5.77. If a 0.01 molar solution of it is titrated with 0.01 M HCl, at what pH should we stop the titration?
-
BH+
à
BH(aq)+ H+
Kconj acid = Kw / Kb = Ka
Ka = [BH][H+]/[BH+] = x2/(S-x)
~ x2/S = [H+]2/S
pKa = 2 pH + log10S
pH = ½pKa - ½log S
= ½pKw - ½pKb - ½log S
-
S = [hydrazine H+] = 0.01 M (½)
(since the volume doubles with the addition of the acid)
pH = ½(14) - ½(5.77) - ½log(0.005) = 5.27
- Nitrate ion can oxidize mercury(I) to mercury(II) in acid solution. What is the 25°
C equilibrium constant for that reaction? To how many significant figures are you entitled?
NO3-
+ 4 H+ + 3 e-
à
NO(aq) + 2 H2O E°
= + 0.96 V
2 Hg2+ + 2 e-
à
Hg22+ E°
= + 0.92 V
E° = + 0.96 - (+ 0.92) = + 0.04 V (one significant figure!)
represents the reaction (2× the first minus 3× the second):
2 NO3- + 8 H+ + 6 Hg2+
à
2 NO(aq) + 3 Hg22+ + 4 H2O
n = 6
ln K = - DG° / RT
= + (nF / RT)E°
= 9.34
K = eln K = e+9.34
= 1.14×104 ~ 104 (only 1 digit good)
- The limiting ionic conductivities (in mS m2 / mol) for the hydrogen and hydroxyl ions are 34.96 and 19.91, respectively. What is the resistance (R in ohms = S-
1) of pure water at 25°
C in a vessel 1 m long with cross-sectional area of 0.1 m2? (Hint: R = 1/G and G = k
A/L where k
is the conductivity and L and A are the length and cross-sectional area.)
k = 34.96 [H+] + 19.91 [OH - ] mS m2 /mol
= (34.96 + 19.91)×10-7 mol/L mS m2 /mol
= 5.487×10-9 S m2 / 10-3 m3
= 5.487×10-6 S/m
R(W) = L / (kA)
= 1 m / (5.487×10-6 S m-1 × 0.1 m2)
= 1.822×106 S-1
= 1.822 MW
- The proposed mechanism for the decomposition of acetaldehyde to form methane and CO involves the methyl radical as follows:
CH3CHO à
·CH3 + ·CHO k1
CH3CHO + ·CH3 à
CH4 + CO + ·CH3 k2
2 ·CH3 à
C2H6 k3
Ignoring the fate of the ·CHO
but not knowing which rate constant is rate-limiting, you can still obtain the overall rate of appearance of methane by presuming that the methyl radical is in steady-state, e.g., its steady-state rate is zero. Notice that methyl radical catalyzes the production of methane in reaction 2, so ·
CH3’s rate of change is not effected by reaction 2. (Hint: there are no equilibrium reactions in this scheme. Don’t get anxious about non-integer orders; happens all the time.)
d[CH4]/dt = k2
[CH3CHO][·CH3]
0 = d[·CH3]/dt
= k1 [CH3CHO]
- 2 k3 [·CH3]2
[·CH3]2
= ½(k1/k3) [CH3CHO]
[·CH3]
= (k1/2k3)½ [CH3CHO]½
d[CH4]/dt =
k2(k1/2k3)½
[CH3CHO]3/2
(if we believe the mechanism)
- Collision theory bimolecular rate constants result in k2 µ
T½ exp(-
Ea/RT). How should we correct the usual method of obtaining Ea from plots of ln k2 versus 1/T?
Normally, we'd equate Ea/R with the negative slope of
ln(k2) vs (1/T), but now there's another T-dependent term.
ln k2 = ½ ln T - Ea/RT + ln C not T-dependent
ln k2 = - ½ ln(1/T) - (Ea/R)(1/T) + ln C
d ln k2/d(1/T) = - ½ (1/T)-1 - Ea/R
= - ½T - Ea/R
Ea = - R×slope - ½RT
So while the ln k2 vs (1/T) plot won't be linear,
we can still estimate the slope at various temperatures then subtract
½RT from the usual Ea calculation to make the
modest correction.