Homework Week 7

Statistical MolecularThermodynamics University of Minnesota

Homework Week 7

1. Given the following experimentally determined standard molar entropies at 298.15 K and 1 bar, ◦ −1 −1 S [CH3OH] = 126.8 J · K · mol S◦[CO] = 197.7 J · K−1 · mol−1 ◦ −1 −1 S [H2] = 130.7 J · K · mol ◦ calculate the value of ∆rS for the reaction

CO(g) + 2H2(g) −→ CH3OH(l).

(a) 332.3 J · K−1 · mol−1 (b) 197.7 J · K−1 · mol−1 (c) -332.3 J · K−1 · mol−1 (d) -201.6 J · K−1 · mol−1 (e) 322.4 J · K−1 · mol−1 (f) 201.6 J · K−1 · mol−1 (g) -322.4 J · K−1 · mol−1 (h) -233.2 J · K−1 · mol−1

Answer:

◦ ◦ ◦ ∆rS =S (products) − S (reactants) =126.8 J · K−1 · mol−1 − 197.7 J · K−1 · mol−1 − 2(130.7 J · K−1 · mol−1) = − 332.3 J · K−1 · mol−1 2. The molar heat capacity of a certain gas can be expressed as

C¯ 1000 K V = 10 − R T over the temperature range 100 < T < 500 K. Calculate ∆S if one mole of this gas is heated from 100 K to 200 K at constant volume.

(a) −8 J · K−1 (b) −2 J · K−1 (c) 1 J · K−1 (d) 2 J · K−1 (e) 8 J · K−1 (f) 10 J · K−1 (g) 12 J · K−1 (h) 16 J · K−1 (i) 20 J · K−1 (j) 24 J · K−1

Answer:

Z T2 C¯ ∆S¯ = V dT (see video 7.2) T1 T Z 200 K ¯ 10 1000 K ∆S = R − 2 dT 100 K T T ! 200 K 1000 K200 K ∆S¯ = R 10 ln + 100 K T 100 K 1000 K 1000 K ∆S¯ = R (10 · ln 2) + − = R ((10 · 0.7) + (5 − 10)) 200 K 100 K ∆S¯ = R (7 − 5) = 8 J · mol−1 · K−1 (2) = 16 J · mol−1 · K−1 3. The following equation gives S in terms of the partition function, Q,

∂ ln Q S = kBT + kB ln Q ∂T N,V The partition function for a monatomic ideal gas where all of the atoms are in their ground electronic state is given by,

1 2πmk T 3N/2 Q = B V N gN N! h2 e1

Derive an expression for the molar entropy, S¯ for this system. You may need Stirling’s approximation, ln X! = X ln X − X.

h 3/2 ¯ i ¯ 3 2πmkBT V ge1 (a) S = R + R ln 2 2 h NA h 3/2i ¯ 3 2πmkBT (b) S = 2 R + R ln h2

h ¯ i ¯ 5 2πmkBT V ge1 (c) S = R + R ln 2 2 h NA h 3/2 ¯ i ¯ 7 2πmkBT V ge1 (d) S = R + R ln 2 2 h NA h 3/2 ¯ i ¯ 5 2πmkBT V ge1 (e) S = R + R 2 2 h NA h 3/2 i ¯ 3 2πmkBT (f) S = 2 R + R ln h2 − R ln NA

h 3/2 ¯ i ¯ 5 2πmkBT V ge1 (g) S = R + R ln 2 2 h NA

Answer:

" #N qN 1 2πmk T 3/2 Q = = B V g N! N! h2 e1 3 ∂ ln Q 3 1 ln Q = N ln T + ln (terms not depending on T ) −→ = N 2 ∂T N,V 2 T   " 3/2 #N 3 1  1 2πmkBT  Used in the formula for S: S = kBT N + kB ln 2 V ge1 2 T N! h  Manipulating:

" 3/2 # 3 2πmkBT S = NkB + NkB ln 2 V ge1 + kB ln 1 − kB ln N! 2 h (=0)

Converting to per mole with N → NA and NAkB = R: " # 3 2πmk T 3/2 S¯ = R + R ln B V¯ g − k ln N ! 2 h2 e1 B A Applying Stirling’s approximation, ln N! = N ln N − N, to the last term:

−kB ln NA! = −kBNA ln NA + kBNA = −R ln NA + R " # 3 2πmk T 3/2 S¯ = R + R ln B V¯ g − R ln N + R 2 h2 e1 A

" 3/2 ¯ # ¯ 5 2πmkBT V ge1 S = R + R ln 2 2 h NA 4. Starting with the total derivative of the entropy as a function of T and P ,

∂S ∂S dS = dT + dP, ∂T P ∂P T the deﬁnition of the ﬁrst law for a reversible process,

dU = δqrev + δwrev

and the the second law, δq dS = rev T derive an expression for ∂S . ∂P T

∂S 1 ∂H (a) ∂P T = T ∂P T − V ∂S 1 ∂H (b) ∂P T = P ∂T P − V ∂S 1 ∂H (c) ∂P T = T ∂P T + V ∂S ∂H (d) ∂P T = T ∂P T − V ∂S Cp (e) ∂P T = T ∂S ∂H (f) ∂P T = T ∂P T + V Answer:

First, write the total diﬀerential of H as a function of T and P :

∂H ∂H dH = dT + dP ∂T P ∂P T

We know from week 5 that dH = T dS + V dP . Equate this to the total diﬀerential above: ∂H ∂H T dS + V dP = dT + dP ∂T P ∂P T Rearrange terms to obtain an equation for dS:

∂H ∂H T dS = dT + dP − V dP ∂T P ∂P T ∂H ∂H T dS = dT + − V dP ∂T P ∂P T 1 ∂H 1 ∂H dS = dT + − V dP T ∂T P T ∂P T Now write the total diﬀerential of S as a function of T and P : ∂S ∂S dS = dT + dP ∂T P ∂P T

Compare the total diﬀerential to the other equation we obtained for dS:

1 ∂H 1 ∂H dS = dT + − V dP T ∂T P T ∂P T

Equating the coeﬃcients of dT from the two equations above gives,

∂S 1 ∂H = ∂T P T ∂T P ∂S C = P ∂T P T ∂H where we use CP = ∂T P to obtain an equation we’ve seen before. Equating the coeﬃcients of dP gives,

∂S 1 ∂H = − V ∂P T T ∂P T 5. The molar constant volume heat capacity of ethylene is,

6085.929 K 822, 826 K2 C (T )/R = 16.4105 − + V T T 2

Use this equation to determine the change in entropy, ∆S, of one mole of ethylene heated from 300 K to 600 K at constant pressure. Assume ethylene behaves ideally (other than it’s temperature-dependent heat capacity).

(a) 742 J·K−1 (b) 2.689 ×1025 J·K−1 (c) 44.51 J·K−1 (d) 5.353 J ·K−1 (e) 47.33 J ·K−1

Answer:

Assuming that ethylene behaves ideally, we can ﬁnd the molar heat capacity at constant pressure using the following equation:

CP − CV = R

CP = R + CV

CP /R = 1 + CV /R

Now insert the equation for the molar heat capacity,

6085.929 K 822, 826 K2 C /R = 1 + 16.4105 − + P T T 2

Now we can calculate ∆S¯:

Z T2 nC ∆S¯ = P dT T1 T Z 600 K 17.4105 6085.929 K 822, 826 K2 = nR − 2 + 3 dT 300 K T T T 6085.929 K 411, 413 K2 600 K = nR 17.4105 ln T + − 2 T T 300 K = nR(5.353) = (1 mol)(8.3145 J · K−1 · mol−1)(5.353) = 44.51 J · K−1 ¯ ¯ ¯ 6. For carbon monoxide, CO, the residual entropy deﬁned as Sresidual = Scalc − Sexp is ...

(a) positive (b) negative (c) zero

Answer:

Carbon dioxide has a very weak dipole moment. As the temperature of solid carbon dioxide approaches zero kelvin, some of the dipoles will tend to be trapped in random orientations before they are able to line up in an energetically favorable way. Each carbon dioxide molecule will settle into the lattice with either dipole up, or dipole down. In this situation the lowest entropy state attainable with W = 1 is not attainable. The molecules are trapped and can not ”ﬂip” into their lowest energy state because there is not enough energy near 0 K for this to happen. Thus, the magnitude of the experimental entropy of CO will be found to be smaller than that which is calculated because the reference experimental entropy near 0 K will not be as near to 0 J·K−1 as is assumed in calculations (so less input heat will be required to reach the entropy associated with a non-zero temperature). 7. The general formula for entropy,

∂lnQ S = kBlnQ + kBT ∂T N,V was discussed in the lecture videos of week 6. In the lecture videos of week 4,

2πmk T 3/2 q(V,T ) = B V · g h2 el was presented as the partition function for a monatomic ideal gas. Which of the following is the correct expression for the change in molar entropy of an ideal gas going isobarically from (T1,V1,P1) to (T2,V2,P1).

(a) 3 Rln T2 + 3 ln V2 2 T1 2 V1 (b) 5 Rln T2 + Rln V2 2 T1 V1 (c) 3 Rln T2 + Rln V2 − 2R 3N 2 T1 V1 4T1T2 (d) Rln V2 V1 (e) 3 Rln T2 + Rln V2 2 T1 V1 (f) 3 Rln T2 2 T1 Answer: The general plan to derive an expression for the change in molar entropy of an ideal gas is to start ﬁrst with the expression for the partition function Q = qN /N!, determine the derivative with respect to temperature, substitute relevant terms into the expression for the entropy given in the question, and then determine ∆S by subtracting the entropy of the latter state from that of the former; thus,

" #N 1 2πmk T 3/2 Q = B V · g . N! h2 el

and therefore

" # 2πmk T 3/2 lnQ = Nln B V · g − lnN!. h2 el

If we designate the partition function for the initial state of the system as Q1 and that for the ﬁnal state of the system as Q2,

3 T2 V2 lnQ2 − lnQ1 = Nln + Nln . 2 T1 V1

Taking the derivative of the general logarithm of Q with respect to T gives, ∂lnQ 3N = ∂T N,V 2T We can now use the general formula for the entropy,

∂lnQ S = kBlnQ + kBT ∂T N,V to determine the entropy diﬀerence between the initial state and ﬁnal state, ∆S¯ = ¯ ¯ S2 − S1.

¯ ∂lnQ2 ∂lnQ1 ∆S = kBlnQ2 + kBT2 − kBlnQ1 − kBT1 ∂T N,V ∂T N,V 3 T2 V2 3N 3N = NkBln + NkBln + kBT2 − kBT1 2 T1 V1 2T2 2T1 3 T V = Rln 2 + Rln 2 2 T1 V1

where the ﬁnal line derives from taking N = NA. 8. Given that the entropy is deﬁned in terms of the partition function as:

∂ ln Q S = kBT + kB ln Q ∂T N,V What is the correct expression for the entropy of an ideal gas in terms of the partition function for one molecule/atom, q?

∂ ln q q (a) S = NkBT ∂T V + NkB ln N + NkB ∂ ln q q (b) S = NkBT ∂T V + NkB ln N ∂ ln q q (c) S = NkBT ∂T V + kB N! + NkB 2 ∂ ln q (d) S = NkBT ∂T V + NkB q (e) S = kB ln N + NkB Answer:

We use the expression for the partition function for an ideal gas,

qN Q(V,T ) = , N! employ Stirling’s approximation, lnN! = NlnN − N, and simplify to ﬁnd the correct answer.

∂ ln Q S =kBT + kB ln Q ∂T N,V " # ∂ qN qN =kBT ln + kB ln ∂T N! N,V N! ∂ ln q ∂ ln N! =kBT N − + kB (N ln q − ln N!) ∂T V ∂T V

Now we note that the the derivative of ln N! with respect to T is zero, and using Stirling’s approximation,

∂ ln q =NkBT + kB (N ln q − N ln N + N) ∂T V ∂ ln q q =NkBT + NkB ln + NkB ∂T V N ◦ 9. Determine the standard molar entropy change ∆rS for the reaction:

2CO(g) + O2(g) −→ 2CO2(g)

Assume all gases are behaving as though ideal. Recall that the standard molar entropy for an ideal diatomic gas,

" # S 2πMk T 3/2 V e5/2 T e Θ /T B −Θvib/T vib = ln 2 + ln − ln(1 − e ) + Θ /T + ln ge1 R h NA σΘrot e vib − 1

was discussed in the week 7 lecture videos. ◦ −1 It may also be helpful to note that S (CO2) = 213.8 J · K at 298.15 K. For diatomic oxygen, Θvib = 2256 K and Θrot=2.07 K, and gel = 3. Because O2 is a homonuclear diatomic molecule, σ = 2. For CO, Θvib = 3103 K, Θrot = 2.77 K, and ge1 = 1. Because CO is a heteronuclear diatomic molecule, σ = 1.

(a) 205.1 J · K−1 · mol−1 (b) 172.7 J · K−1 · mol−1 (c) -355.1 J · K−1 · mol−1 (d) -172.7 J · K−1 · mol−1 (e) 0.0 J · K−1 · mol−1 (f) -84.9 J · K−1 · mol−1 (g) 1.0 J · K−1 · mol−1 (h) -501.0 J · K−1 · mol−1

Answer:

We will ﬁrst determine the standard molar entropy S◦ for carbon monoxide, CO, given that Θvib = 3103 K, Θrot = 2.77 K, and ge1 = 1 and σ = 1 We will solve for the standard molar entropy one term at a time:

2πMk T 3/2 2π(4.651 × 10−26 kg)(1.381 × 10−23 J · K−1)(298.15 K)3/2 B = h2 (6.626 × 10−34 J · s)2 kg 3/2 = 1.434 × 1032 J · s2 !3/2 kg = 1.434 × 1032 kg·m2 2 s2 · s 1 3/2 = 1.434 × 1032 m2 1 = 1.434 × 1032 = 1.434 × 1032 m−3 m3 V e5/2 RT e5/2 (0.08314 dm3 · bar · mol−1 · K−1)(298.15 K)e5/2 = = = 5.015 × 10−22 dm3 23 −1 NA NAP 6.022 × 10 mol (1 bar) = 5.015 × 10−25 m3

Multiplying the two results above and taking the natural log of their product gives the ﬁrst term in the equation for S/R:

" 3/2 5/2 # 2πMkBT V e 32 −3 −25 3 ln 2 = ln (1.434 × 10 m )(5.015 × 10 m ) = 18.092 h NA

Now we ﬁnd the other terms in the equation for S/R: T e (298.15 K)e ln = ln = 5.679 σΘrot (1)(2.77 K) ln(1 − e−Θvib/T ) = ln(1 − e−3103 K/298.15 K) = −3.02 × 10−5 Θ /T 3103 K/298.15 K vib = = 3.14 × 10−4 eΘvib/T − 1 e3103 K/298.15 K − 1 ln ge1 = ln 1 = 0

Adding all of these terms gives: S = 18.092 + 5.679 − 3.02 × 10−5 + 3.14 × 10−4 + 0 = 23.77 R

Taking R = 8.3145 J · K−1 · mol−1:

S(CO) = 197.6 J · K−1 · mol−1

◦ Now we can determine the standard molar entropy S for diatomic oxygen, O2, given that Θvib = 2256 K and Θrot=2.07 K, and gel = 3 and σ = 2. We will solve for the standard molar entropy, one term at a time:

2πMk T 3/2 2π(5.313 × 10−26 kg)(1.381 × 10−23 J · K−1)(298.15 K)3/2 B = h2 (6.626 × 10−34 J · s)2 kg 3/2 = 1.751 × 1032 J · s2 !3/2 kg = 1.751 × 1032 kg·m2 2 s2 · s 1 3/2 = 1.751 × 1032 m2 1 = 1.751 × 1032 = 1.751 × 1032 m−3 m3

V e5/2 RT e5/2 (0.08314 dm3 · bar · mol−1 · K−1)(298.15 K)e5/2 = = = 5.015 × 10−22 dm3 23 −1 NA NAP 6.022 × 10 mol (1 bar) = 5.015 × 10−25 m3 Multiplying the two results above and taking the natural log of their product gives the ﬁrst term in the equation for S/R:

" 3/2 5/2 # 2πMkBT V e 32 −3 −25 3 ln 2 = ln (1.751 × 10 m )(5.015 × 10 m ) = 18.291 h NA

Now ﬁnd the other terms in the equation for S/R:

T e (298.15 K)e ln = ln = 5.277 σΘrot (2)(2.07 K) ln(1 − e−Θvib/T ) = ln(1 − e−2256 K/298.15 K) = −5.18 × 10−4 Θ /T 2256 K/298.15 K vib = = 3.92 × 10−3 eΘvib/T − 1 e2256 K/298.15 K − 1 ln ge1 = ln 3 = 1.099

Adding all of these terms gives:

S = 18.291 + 5.277 − 5.18 × 10−4 + 3.92 × 10−3 + 1.099 = 24.67 R

−1 −1 S(O2) = 205.1 J · K · mol

We can now calculate the entropy change for the reaction.

◦ ◦ ◦ ◦ ∆rS = 2S [CO2(g)] − S [O2(g)] − 2S [CO(g)] = 2(213.8 J · K−1 · mol−1) − (205.1 J · K−1 · mol−1) − 2(197.6 J · K−1 · mol−1) = −172.7 J · K−1 · mol−1

Note that the large, negative entropy change is consistent with the reduction in the number of molecules on going from left to right in the reaction. Thus, the total number of translational and rotational degrees of freedom, which dominate in contributions to entropy, are reduced and changed into vibrational degrees of freedom, which do not contribute much to entropy at these temperatures. 10. Identify the molecule in each of the pairs below which would be expected on qualitative grounds to have the largest molar entropy under the same gaseous conditions:

chesches

Figure 1: Comparison molecules

(a) i. a; ii. a; iii. a; iv. a (b) i. a; ii. b; iii. a; iv. a (c) i. b; ii. a; iii. b; iv. a (d) i. a; ii. b; iii. a; iv. b (e) i. a; ii. b; iii. b; iv. b (f) i. b; ii. b; iii. b; iv. b (g) i. b; ii. b; iii. b; iv. a (h) i. a; ii. a; iii. a; iv. b (i) none of the above Answer: i. Ethanol (a) has more atoms (and so more degrees of freedom) and moreover does not constrain any bond rotations through ring closure, as is the case for oxirane (b). ii. n-pentane (b) is ”long and ﬂoppy” while neopentane (a) is very compact and has high symmetry. iii. CO2 (b) has larger mass and a larger moment of inertia compared to O2 (a) which will lead to greater translational and rotational entropy. iv. 1-aminopentane (b) has more atoms (and so more degrees of freedom) and moreover does not constrain any bond rotations through ring closure, as is the case for pyrrole (a).