Entropy Changes & Processes Chapter 4 of Atkins: he Second Law: he Concepts Section 4.3 Entropy of Phase ransition at the ransition emperature Expansion of the Perfect Gas Variation of Entropy with emperature Measurement of Entropy Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-1
Entropy at a Phase ransition Changes in molecular order occur when a substance freezes or boils. Consider the phase transitions of water, at transition temperatures trs. For ice water, trs = 273 K, ice in equilibrium with liquid water at 1 atm; and for boiling water, trs = 373 K, liquid water in equilibrium with vapor at 1 atm he external pressure is constant for a glass of ice water, and in order to match attractive forces between ice molecules, energy must come from kinetic energy of the water molecules or the surroundings At trs, any transfer of heat between the system and surroundings is reversible since the two phases in the system are in equilibrium (the forces pushing the ice towards melting are equal to those pushing the water towards freezing) - so phase transition is reversible It does not matter how the ice melts (what path it takes) since entropy is a state function. What does matter for this particular expression is that the system be isothermal. If it was not isothermal, one would have a problem examining the process in steps as we shall see Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-2
Entropy at a Phase ransition, 2 At constant pressure, q = trs H, and the change in molar entropy is For exothermic and endothermic phase transitions: trs S = H trs trs H > 0 trs S > 0 disordered freezing, condensing Exothermic phase transition Endothermic melting, boiling trs H < 0 trs S < 0 ordered Consistent with decreasing disorder: gases > liquids > solids Example (but BE CAREFUL): when compact condensed phase vaporizes into a widely dispersed gas, one can expect an increase in the disorder of the molecules Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-3
routon s Rule routon s Rule: his empirical observation (see able 4.2) states that most liquids have approximately the same standard entropy of vaporization, vap S o = 85 J K -1 mol -1 : so, vap H O = b x 85 J K -1 mol -1 Some Standard Entropies of Vaporization vap H O (kj mol -1 ) θ boil ( o C) vap S O (kj mol -1 ) Benzene +30.8 80.1 +87.2 CCl 4 +30.0 76.7 +85.8 Cyclohexane +30.1 80.7 +85.1 H 2 S +18.7-60.4 +87.9 Methane +8.18-161.5 +73.2 Water +40.7 100.0 +109.1 Exceptions: In water, molecules are more organized in the liquid phase (due to hydrogen bonding), so a greater change of disorder occurs upon vaporization In methane, the entropy of the gas is slightly low (186 J K -1 mol -1 at 298 K) and in light molecules very few rotational states are accessible at room temperature - associated disorder is low Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-4
Expansion of a Perfect Gas he S for an isothermally expanding perfect gas can be written as S = nrln V f (see lecture 10 notes) (equation applies for reversible or irreversible change, since S is a state function - but remember that the value of S must be computed with the integral of reversible heat divided by ) Reversible change: S tot =0 he surroundings are in thermal and mechanical equilibrium with system, so S sur = - S =-nr ln (V f /V i ) Free irreversible expansion (into vacuum): w=0 If isothermal, q=0 in this case, and S sur =0 with S tot =nr ln (V f /V i ) V i Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-5
Variation of Entropy with emperature he entropy of a system at temperature f can be calculated from knowledge of S at initial temperature and heat supplied to make : S = When the system is subjected to constant pressure (i.e., the atmosphere) during heating, from the definition of constant pressure heat capacity, if the system is not doing expansion work (w =0),then hen, at constant pressure (or constant volume, replace with C V ): If C p is invariant to temperature change S ()= f S ()+ i i f dq rev dq rev = C p d S ()= f S i f i C p ()+ C p ln f d i Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-6
Calculating the Entropy Change Calculate S when argon at 25 o C and 1.00 atm in a container of volume 500 cm 3 expands to 1000 cm 3 and is simultaneously heated to 100 o C Methodology: Since S is a state function, we can choose a convenient path from initial to final state: (1) isothermal expansion to final volume, followed by (2) reversible heating at constant volume to final temperature Amount of Ar present is n =pv/r = 0.0204 mol C p,m (Ar) = 20.786 J K -1 mol -1 (1) Expansion from 500 cm 3 to 1000 cm 3 at constant : S=nR ln 2.00 =+0.118 J K -1 (2) Reversible heating from 25 o C to 100 o C at constant V: S=(0.0204 mol) x (12.48 J K -1 ) x ln (373 K /298 K) =+0.057 J K -1 Overall entropy change: S =+0.118 J K -1 +0.057 J K -1 =+0.175 J K -1 Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-7
Entropy Change of Melting Ice Consider the entropy changes for putting an ice cube in a glass of warm water and letting it melt (assume an adiabatic container) q 1 q 2 q 3 initial fus fus final Start at i (1) Calculate S to cool the water to 0 o C by reversibly removing heat, q 1, from the system (2) At fus, calculate the amount of heat, q 2, to be added to the system to melt the ice cube (3) Calculate the difference between the two amounts of heat and add back remaining heat so that the total heat lost or gained is zero (adiabatic system)- determine the entropy change in this process At each step, the infinitesimal entropy change for the system, ds, is just dq divided by the. For the cooling and heating of water, integrate over the temperature range, since the temperature is not constant. Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-8
Entropy Change of Melting Ice, 2 Step 1: Entropy decrease for the system as the water is cooled S 1 = fus dq rev fus d = C p i = C ln fus p Step 2: Melt the ice at the temperature of fus = 0 o C (q fus = q 2 ) Step 3: Balance the heat by (in this example) adding an amount of heat -(q 1 +q 2 ) back into the glass (ice cube melted completely), remove more heat to cool the glass to 0 o C than you would have to add in order to melt the ice -- increase in entropy: i S 2 = q fus = fush S 3 = C p ln f fus i Change in notes Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-9
Entropy Change of Melting Ice, 3 How do we know f? System is adiabatic so the total heat must be zero, or q 3 = -(q 1 +q 2 ), giving f = fus +q 3 /C p Is the process spontaneous? (We know intuitively that it is - sticking an ice cube into warm water melts the ice cube!!) How do we prove this? Show that the total entropy change (the system plus the surroundings) is positive. he entropy change of the system we know is: S 1 + S 2 + S 3 Lecture Problem: ry the above steps for 5.0 g of ice at 60. o C in 110. ml of water With C p,m of water of 75.6 J K -1 mol -1 and fus H = 6.01 kj mol -1. Calculate S 1, S 2, S 3, S, f Note that since the system is adiabatic, S sur =0 (the water is part of the system). Extra Credit Lecture Problem: Use the same conditions as above, except try the calculation for an isothermal system Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-10
Useful Entropy Equations Some changes in state and associated entropy changes for the system, for an infinitessimal change in entropy, ds =dq rev /: Vary the temperature at constant volume (C V independent of ): f dq C S = = v d = C v ln f i f i Vary the temperature at constant pressure (C p independent of ): S = dq f = i C p d For isothermal expansion of a perfect gas S = nrln V f = nrln p i V i p f What if both and P or and V change? Simple, use two steps! First change the holding either V or P constant, then change V or P at constant (depending on what you are given). f i Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-11 i = C p ln f i
Measurement of Entropy Entropy of a system at temperature can be related to entropy at = 0 by measuring heat capacities at different temperatures, and evaluating S ()= f S ()+ i he entropy of transition, trs S = trs H/ trs, is added for each phase transition between = 0 and (temperature of interest) For example, if substance melts at f and boils at b, entropy above b is S ()= S 0 ()+ + b f C p f 0 C p All of the quantities can be determined from calorimetry excepting S(0), and integrals can be evaluated analytically (see next) f i C p d ( s) d + H fus () l d + H vap + b f b C p () g d Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-12
Measurement of Entropy, 2 Plot (a) shows the variation of C p / with sample temperature. he area under the curve of C p / as a function of is required. However, since d/ = d ln, can also evaluate area under C p vs. ln. Plot (b) shows the entropy of the system varying with temperature, which is equal to the area under the curve up to the corresponding temperature, plus entropy of each phase transition passed. One problem with measuring S is measuring C p at low near = 0: Debye extrapolation: It has been shown that at temperatures near = 0, the heat capacity is approximately proportional to 3 (C p = a 3 as 0) Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-13
Example: Calculating Entropy Consider standard molar enthalpy of N 2 (g) at 25 o C, calculated from the following data: S mo (J K -1 mol -1 ) Debye exptrapolation 1.92 Integration from 10-35.61 K 25.25 Phase transition at 35.61 K 6. 43 Integration from 35.61-63.14 K 23.38 Fusion at 63.14 K 11.42 Integration from 63.14-77.32 K 25.25 Vapourization at 77.32 K 72.13 Integration from 77.32-298.15 K 39.20 Correction for Gas Imperfection* 0.92 otal 192.06 S mo (298.15 K) = S mo (0) + 192.1 J K -1 mol -1 *See section 20.5 of book - failure of hard sphere approximation Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-14
Calculating Entropy at Low emperature he molar constant-pressure heat capacity of some solid material at 12 K is 0.52 J K -1 mol -1. What is the molar entropy at that temperature? Because temperature is low, we can assume that heat capacity varies with temperature as a 3 : S ()= S 0 ()+ It turns out that the final result can be expressed in terms of a heat capacity at constant pressure S m ( 12K)= S m ()+ 0 1 3 C 12K p( ) a 3 f d = S 0 0 () = S 0 ()+ 1 3 a f3 = 1 3 C p f ()+ 2 d = S m ()+ 0 1 ( 3 0.52 J K-1 mol -1 )= S m ()+ 0 0.17 J K -1 mol -1 f 0 note small change Prof. Mueller Chemistry 451 - Fall 2003 Lecture 11-15