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Chapter 6
Heat capacity, enthalpy, & entropy
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6.1 Introduction
In this lecture, we examine the heat capacity as a function of temperature, compute the
enthalpy, entropy, and Gibbs free energy, as functions of temperature.
We then begin to assess phase equilibria constructing a phase diagram for a single component
(unary) system.
• By eq. 2.6 & 2.7
(2.6)
(2.7)
(2.6a)
(2.7a)
Integration of Eq. (2.7a) between the states (𝑇𝑇2 , 𝑃𝑃) and (𝑇𝑇1 , 𝑃𝑃) gives the
difference between the molar enthalpies of the two states as
(6.1)
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6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
- Empirical rule by Dulong and Petit (1819) : Cv ≈ 3R
(classical theory: avg. E for 1-D oscillator, 𝜀𝜀𝑖𝑖 = kT, E = 3N0kT = 3RT)
- Calculation of Cv of a solid element as a function of
T by the quantum theory: First calculation by Einstein (1907)
- Einstein crystal – a crystal containing n atoms, each of which
behaves as a harmonic oscillator vibrating independently
discrete energy 𝜀𝜀𝑖𝑖 = 𝑖𝑖 +
1
2
ℎ𝑣𝑣
(6.2)
– a system of 3n linear harmonic oscillators
(due to vibration in the x, y, and z directions)
(6.3)
The Energy of Einstein crystal
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6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
Using, 𝜀𝜀𝑖𝑖 = 𝑖𝑖 +
1
2
ℎ𝑣𝑣 & eq. 4.13
Into
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6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
Taking
where 𝑥𝑥 = 𝑒𝑒 −ℎ𝜈𝜈⁄𝑘𝑘𝑘𝑘 , gives
and
in which case
(6.4)
• Differentiation of eq. with respect to temperature at constant volume
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6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
• Defining ℎ𝜐𝜐⁄𝑘𝑘 = 𝜃𝜃𝐸𝐸 : Einstein characteristic temperature
(6.5)
𝐶𝐶𝑉𝑉 ≈ 𝑅𝑅 𝑎𝑎𝑎𝑎 𝑇𝑇 → ∞
𝐶𝐶𝑉𝑉 ≈ 0 𝑎𝑎𝑎𝑎 𝑇𝑇 → 0
the Einstein equation good at higher T,
the theoretical values approach zero more rapidly than do the actual values.
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6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
• Problem: although the Einstein equation adequately represents actual heat capacities at higher
temperatures, the theoretical values approach zero more rapidly than do the actual values.
• This discrepancy is caused by the fact that the oscillators do not vibrate with a single
frequency.
• In a crystal lattice as a harmonic oscillator, energy is expressed as
𝐸𝐸𝑛𝑛 =
ℎ𝑣𝑣𝐸𝐸
+ 𝑛𝑛𝑛𝑣𝑣𝐸𝐸
2
(n = 0,1,2,….)
Einstein assumed that 𝑣𝑣𝐸𝐸 is const. for all the same atoms in the oscillator.
• Debye’s assumption (1912) : the range of frequencies of vibration available to the oscillators is
the same as that available to the elastic vibrations in a continuous solid.
: the maximum frequency of vibration of an oscillator
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6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
• Integration Einstein’s equation in the range, 0 ≤ 𝑣𝑣 ≤ 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚
obtained the heat capacity of the solid
which, with x=hυ/kT, gives
(6.6)
• Defining 𝜃𝜃𝐷𝐷 = ℎ𝜐𝜐𝑚𝑚𝑚𝑚𝑚𝑚 ⁄𝑘𝑘 = ℎ𝜐𝜐𝐷𝐷 ⁄𝑘𝑘 : Debye characteristic T
• 𝑉𝑉𝐷𝐷 (Debye frequency)=𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚 =
𝜃𝜃𝐷𝐷 �𝑘𝑘
ℎ
• Debye’s equation gives an excellent fit to the experimental data at
lower T.
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6.2 THEORETICAL CALCULATION OF THE HEAT CAPACITY
• The value of the integral in Eq. (6.6) from 0 to infinity is 25.98, and thus, for very low
temperatures, Eq. (6.6) becomes
(6.7)
: Debye 𝑇𝑇 3 law for low-temperature heat capacities.
Debye’s theory: No consideration on the contribution made to the heat capacity by the uptake of
energy by electrons (∝ absolute temperature)
• At high T, where the lattice contribution approaches the Dulong and Petit value, the molar Cv
should vary with T as
in which bT is the electronic contribution.
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6.3 THE EMPIRICAL REPRESENTATION OF HEAT CAPACITIES
• By experimental measurements,
: Normally fitted
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
For a closed system of fixed composition, with a change in T from T1 to T2 at the const. P
T
ⅰ) ∆H = H T2 , P − H T1 , P = ∫T 2 Cp dT (6.1) : ∆H is the area under a plot of 𝐶𝐶𝑃𝑃 𝑣𝑣𝑣𝑣 𝑇𝑇
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ⅱ) A + B = AB chem. rxn or phase change at const. T, P
∆H T, P = HAB T, P − HA T, P − HB T, P
(6.8) : Hess′ law
∆H < 0 exothermic
∆H > 0 endothermic
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
• Enthalpy change
Consider the change of state
where ∆𝐻𝐻(𝑎𝑎 → 𝑑𝑑) is the heat required to
increase the temperature of one mole of solid
A from 𝑇𝑇1 to 𝑇𝑇2 at constant pressure.
(𝒾𝒾)
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
∴
or
(6.9)
where
convention assigns the value of zero to H of elements in their stable states at 298 K.
ex) M(s) + 1/2O2 g = MO s at 298K
1
∆𝐻𝐻298 = 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 ,298 − 𝐻𝐻𝑀𝑀 𝑠𝑠 ,298 − 𝐻𝐻𝑂𝑂2 𝑔𝑔 ,298
2
= 𝐻𝐻𝑀𝑀𝑀𝑀 𝑠𝑠 ,298 as 𝐻𝐻𝑀𝑀 𝑠𝑠 ,298 &
𝐻𝐻𝑂𝑂2 𝑔𝑔 ,298 =0 by convention
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
1
1
Fig 6.7 : For the oxidation Pb + O2 = PbO with H of mole of
2
2
O2 gas , 1mole of Pb(s) at 298K (=0 by convention)
T
ab : 298 ≤ T ≤ 600K, where HPb(s) = ∫298 Cp,Pb(s) dT
1
T
ac
� : 298 ≤ T ≤ 3000K, where H1O (g) = ∫298 Cp,O2 (g) dT ;
2
2 2
∆HPbO s ,298K = -219,000 J
T
de : 298 ≤ T ≤ 1159K where HPbO s ,T = 219,000 + ∫298 Cp,PbO(s) dT J
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
1
With H of mole of O2(g) and 1mole of Pb(s) at
2
298K(=0 by convention)
1
2
f : H of mole of O2(g) and 1mole of Pb(s) at T.
g : H of 1mole of PbO(s) at T.
Thus
where
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
From the data in Table 6.1,
and, thus, from 298 to 600 K (𝑇𝑇𝑚𝑚,𝑃𝑃𝑃𝑃 )
With T=500K, ∆H500K = −217,800 J
In Fig. 6.7a, h: H of 1 mole of 𝑃𝑃𝑏𝑏(𝑙𝑙) at 𝑇𝑇𝑚𝑚 of 600K and
600 to 1200K, given as
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
• In Fig. 6.7b, ajkl: H of 1 mole of Pb and 1 mole of O2(g) , and
hence ∆HT′ is calculated from the cycle
where
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
Thus
This gives ∆𝐻𝐻1000 = −216,700 𝐽𝐽 at 𝑇𝑇 ′ =1000K
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
If the T of interest is higher than the Tm of both the
metal and its oxide, then both latent heats of
melting must be considered.
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
• If the system contains a low-temperature phase in equilibrium with a high-temperature phase at
the equilibrium phase transition temperature then introduction of heat to the system (the
external influence) would be expected to increase the temperature of the system (the effect) by
Le Chatelier’s principle.
• However, the system undergoes an endothermic change, which absorbs the heat introduced at
constant temperature, and hence nullifies the effect of the external influence. The endothermic
process is the melting of some of the solid. A phase change from a low- to a high-temperature
phase is always endothermic, and hence ∆H for the change is always a positive quantity. Thus
∆Hm is always positive. The general Eq. (6.9) can be obtained as follows:
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6.4 ENTHALPY AS A FUNCTION OF TEMPERATURE AND COMPOSITION
Subtraction gives
or
(6.10)
and integrating from state 1 to state 2 gives
(6.11)
Equations (6.10) and (6.11) are expressions of Kirchhoff’s Law.
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE
THIRD LAW OF THERMODYNAMICS
• The 3rd law of thermodynamics
: Entropy of homogeneous substance at complete internal
equilibrium state is ‘0’ at 0 K.
For a closed system undergoing a reversible process,
(3.8)
At const. P,
As T increased,
(6.12)
the molar S of the system at any T is given by
(6.13)
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE
THIRD LAW OF THERMODYNAMICS
T. W. Richards (1902) found experimentally that ΔS → 0 and ΔCp → 0 as T → 0. (Clue for the 3rd law)
Nernst (1906)
→ 0 as T → 0.
Why? by differentiating Eq. (5.2) G = H – TS
with respect to T at constant P:
From Eq. (5.12)
dG = -SdT + VdP
thus
→0
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE
THIRD LAW OF THERMODYNAMICS
(i) ΔCp = ΣνiCp i → 0 means that each Cp i → 0 (solutions)
by Einstein & Debye (T → 0, Cv → 0)
(ii) ΔS = ΣνiSi → 0 means that each Si → 0
thus, Ω th = Ω conf = 1
i.e., every particles should be at ground state at 0 K, (Ω th = 1)
every particles should be uniform in concentration (Ω conf = 1).
Thus, it should be at internal equilibrium. Plank statement
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE
THIRD LAW OF THERMODYNAMICS
• If (𝜕𝜕∆G⁄𝜕𝜕T)P and (𝜕𝜕∆H⁄𝜕𝜕T)P → 0 as T → 0, ∆S & ∆CP → 0 as T → 0
• Nernst’s heat theorem states that “for all reactions involving substances in the condensed
state, ΔS is zero at the absolute zero of temperature”
• Thus, for the general reaction A + B = AB,
∆𝑆𝑆 = 𝑆𝑆𝐴𝐴𝐴𝐴 − 𝑆𝑆𝐴𝐴 − 𝑆𝑆𝐵𝐵 = 0 𝑎𝑎𝑎𝑎 𝑇𝑇 = 0 and if 𝑆𝑆𝐴𝐴 and 𝑆𝑆𝐵𝐵 are assigned the value of zero at 0 K,
then the compound AB also has zero entropy at 0 K.
• The incompleteness of Nernst’s theorem was pointed out by Planck, who stated that “the
entropy of any homogeneous substance, which is in complete internal equilibrium, may be
taken to be zero at 0 K.”
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE
THIRD LAW OF THERMODYNAMICS
the substance be in complete internal equilibrium:
① Glasses
- noncrystalline, supercooled liquids
liquid-like disordered atom arrangements
→ frozen into solid glassy state → metastable
- 𝑆𝑆0 ≠0, depending on degree of atomic order
② Solutions
- mixture of atoms, ions or molecules
- entropy of mixing
- atomic randomness of a mixture determines its degree of order
: complete ordering : every A is coordinated only by B atoms and vice versa
: complete randomness : 50% of the neighbors of every atom are A atoms and 50% are B atoms.
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE
THIRD LAW OF THERMODYNAMICS
③ Even chemically pure elements
- mixtures of isotopes → entropy of mixing
ex)Cl35 − Cl37
④ Point defects
- entropy of mixing with vacancy
Ex) Solid CO Structure
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6.5 THE DEPENDENCE OF ENTROPY ON TEMPERATURE AND THE
THIRD LAW OF THERMODYNAMICS
• Maximum value if equal numbers of molecules were oriented in opposite directions and random
mixing of the two orientations occurred. From Eq. (4.18) the molar configurational entropy of mixing
would be
using Stirling’s approximation,
measured value: 4.2 J/mole K : requires complete internal equilibrium
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6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
• The Third Law can be verified by considering a phase transition in an element such as α → β
where α & β are allotropes of the element and this for the case of sulfur:
For the cycle shown in Fig. 6.11
For the Third Law to be obeyed, 𝑆𝑆Ⅳ =0, which requires that
where
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6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
• In Fig 6.11, a monoclinic form which is stable above 368.5 K and an orthorhombic form which is
stable below 368.5 K
• The measured heat capacities give
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6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
Assigning a value of zero to S0 allows the absolute value of the entropy of any material
to be determined as
and molar entropies are normally tabulated at 298 K, where
With the constant-pressure molar heat capacity of the solid expressed in the form
the molar entropy of the solid at the temperature T is obtained as
When T>𝑇𝑇𝑚𝑚
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6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
Richard’s rule (generally metal)
∆𝐻𝐻𝑚𝑚
�𝑇𝑇𝑚𝑚 ≈ ∆𝑆𝑆𝑚𝑚 ≈ 9.6J/K(FCC) ,
8.3J/K(BCC)
From FCC
Trouton’s rule (generally metal)-more useful!!
∆𝐻𝐻𝑉𝑉
�𝑇𝑇𝑏𝑏 ≈ ∆𝑆𝑆𝑏𝑏 ≈ 88J/K(for both FCC and BCC)
From BCC
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6.6 EXPERIMENTAL VERIFICATION OF THE THIRD LAW
Because of the similar molar S of the condensed
phases Pb and PbO, it is seen that ∆S for the reaction,
1
2
is very nearly equal to − 𝑆𝑆𝑇𝑇,𝑂𝑂2 at 298K
∆S is of similar magnitude to that caused by the
1
disappearance of the gas, i.e., of mole of O2(g)
2
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6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY
(i) For a closed system of fixed composition, with a change of P at const. T,
(dH =TdS+VdP)
Maxwell’s equation (5.34) gives (𝜕𝜕𝑆𝑆⁄𝜕𝜕𝑃𝑃) 𝑇𝑇 = −(𝜕𝜕𝑉𝑉⁄𝜕𝜕𝑇𝑇)𝑃𝑃
and
Thus
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6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY
The change in molar enthalpy caused by the change in state from (P1 , T) to (P2 , T) is thus
(6.14)
For an ideal gas, 𝛼𝛼 = 1⁄𝑇𝑇 an Eq. (6.14) = 0, H of an ideal gas is independent of P.
• The molar V and α of Fe are, respectively, 7.1𝑐𝑐𝑐𝑐3 and 0.3 × 10−4 𝐾𝐾 −1 .
the P increase on Fe from 1 to 100 atm at 298 K causes the H to increase by
The same increase in molar H would be obtained by heating Fe from 298
to 301 K at 1 atm P.
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6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY
(ii) For a closed system of fixed composition, with a change of P at const. T,
Maxwell’s equation (5.34) gives (𝜕𝜕𝑆𝑆⁄𝜕𝜕𝑃𝑃) 𝑇𝑇 = −(𝜕𝜕𝑉𝑉⁄𝜕𝜕𝑇𝑇)𝑃𝑃 &
Thus, for the change of state from (P1 , T) to (P2 , T)
For an ideal gas, as 𝛼𝛼=1/T, Eq. (6.15) simplifies to
(6.15)
Same as decreasing
temperature
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6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY
- Solid : An increase in the pressure exerted on Fe
Fe: from 1 to 100 atm at 298K
⇒ ΔS = -0.0022 J/K
Al: from 1 to 100 atm at 298K
⇒ ΔS = -0.007 J/K
- For same ΔS, how much is the temperature change?
Fe → 0.29K required
Al → 0.09K required
∴ very insignificant effect
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6.7 THE INFLUENCE OF PRESSURE ON ENTHALPY AND ENTROPY
(iii) For a closed system of fixed composition with changes in both P and T,
combination of Eqs. (6.1) and (6.14) gives
(6.16)
and combination of Eqs. (6.12) and (6.15) gives
(6.17)
For condensed phases over small ranges of P, these P dependencies can be ignored.
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