chemistry 1 thermodynamics final and solutions

7/26/2019 Chemistry 1 Thermodynamics final and solutions

1/11

1

Chemistry 1Final Exam

2011/1/14

Constants

R = 8.314 J / mol K 1 atm = 760 Torr = 1.01x105Pa

= 0.0821 L atm / K mol c = 2.99108m/s= 8.314 L kPa / K mol h = 6.6310

-34Js

In case that you need thermodyanmics data to solve the problems, please find four

tables, including Table1, Table2-1, Table2-2 and Table 3

1. An innovative engine consisting of 1 mol of monatomic perfect gas atoms is taken

through the cycle, state 1state 2state 3 as shown in the following graph.

Here 31 is an isothermal process.

(a) Determine the temperature T1, T2and T3at the state 1, 2, and 3, respectively.

(3%)(b) Calculate q, w and U for each step and for the overall cycle. (12%)

2. Answer the following questions. (2% each, 16% total)

(a) Who did make foundation of the concept of a heat and a work for

thermodynamics? Who did prove the heat and the work are a kind of the

forms of the energy? Write the name of these French engineer and British

physicist.

(b) Define the standard state

(c)

What is H for an exothermic process, H=0, H>0 or H


2/11

2

3. A sample consisting of 2.0 mol CO2occupies a fixed volume of 15.0 L at 300 K.

When it is supplied with 2.35 kJ of energy as heat its temperature increases to 341

K. Assume that CO2is described by the van der Waals equation of state, and

calculate w, U and H. a= 3.640 L2atmmol

-2, b= 0.04267 Lmol

-1. You may

treat CO2as an ideal gas to solve this problem, but only get 2 points on the

calculation of H. (2% for w , 3% for U , 4% for van der Waals solution of H, 2%

for ideal gas solution of H, 9% total)

4. The bond enthalpy in NO is 632 kJmol-1

and that of each nitric N-O bond in NO2is

469 kJmol-1

. (a) Draw Lewis structures of NO and NO2and show the bond order of

each compounds. The mean bond enthalpies given in the table below. (b) Using

Lewis structures, estimate the bond enthalpy of NO2and (c) explain the reason of

the difference between actual bond enthalpy and estimated enthalpy (3% each, 9%

total)

Table 1 Mean bond enthalpies (kJmol-1

).

5. The standard enthalpy of formation of metallocene biz(benzene)chromium was

measured in a calorimeter. It was found for the reaction Cr(C6H6)2(s)Cr(s) +

2C6H6(g) that Uro(583 K) = +8.0 kJmol

-1. Find the corresponding reaction enthalpy

at 583 K. (6%)

6.

Using the standard enthalpy of formation of the following values, calculate thestandard reaction enthalpy for each of the following reactions. (2% each, 6% total)

(a) 3NO2(g) + H2O(l)3HNO3(aq) + NO(g).

(b) B2O3(s) + 3CaF2(s)2BF3(g) + 3CaO(s).

(c) H2S(aq) + 2KOH(aq)K2S(aq) + 2H2O(l).


3/11

3

7. In an experiment, 1.0 mole of O2was compressed irreversibly from P1to P2by

driving in a piston, and in the process its temperature was increased from T1to T2.

What is the entropy change of the gas? Assuming that the gas behalves ideally and

express your answer in terms of P1, P2, T1, T2and R only. (10 %)

8. Troutons rule states that the standard entropy of vaporization is ~ 85 J K-1

mol-1

for

most substances. Based on this rule, estimate the standard enthalpy of vaporization

of liquid bromine, which boils at 59 C. (4%)

9.

Consider the super-cool water to form ice at -10 C , the molar entropy change of

the system can be estimated as

Using the principle of 2nd

law to quantitatively explain why this process is

spontaneous! Assuming that the heat capacitances are independent of temperature.

(5%)

10. Explain (a) why are almost all combustion reactions are spontaneous? (2%) (b) how

an endothermic reaction can be spontaneous? (2%) (c) how to assist an originally

non-spontaneous endothermic reaction becoming spontaneous? (2 %)

11. A scientist proposed the following two reactions to produce ethanol, a liquid fuel:

C2H4(g) + H2O(g) CH3CH2OH(l) (A)

C2H6(g) + H2O(g) CH3CH2OH(l) + H2(g) (B)

Reaction B is preferred if spontaneous, because C2H6(g) is a cheaper starting material

than C2H4(g). (a) Assume standard-state conditions and determine whether the

above reactions are spontaneous. (5%) (b) Assuming that Hand Sare

independent of temperature, estimate at what temperature ranges the above

reactions would become spontaneous. (5%)

12. You wish to construct a fuel cell to do electrical work for you based on the oxidation

of a hydrocarbon fuel. To understand how it works, you will (a) derive the relationbetween Gibbs free energy and the maximum non-expansion work at constant T

and P (5%), (b) calculate the maximum non-expansion work available through the

combustion of one mole octane fuel (C8H18, in liquid phase) at the standard-state

conditions (5%) and (c) estimate the minimum heat that must be released to the

environment which cannot be transferred to do the electrical work. (5%)


4/11

4

Table 2-1 Thermodynamic data of inorganic compounds at 25


5/11

5

Table 2-2 Thermodynamic data of inorganic compounds at 25


6/11

6

Table 3 Thermodynamic data of organic compounds at 25


7/11

7

Solutions:1.

Ans.(a) From ideal gas law, T = PV/nR.

T1= P1V1/nR = 1 x22.4/1x0.0821 ([atm][L]/[mol][atmLK-1

mol-1

]) = 273 K

T2= P1V2/nR = 1 x44.8/1x0.0821 ([atm][L]/[mol][atmLK

-1

mol

-1

]) = 546 KT3= T1= 273 K (cycle 31 is isothermal. Thus, T3= T1)

(three correct: 2, one wrong:1, two&three wrong:0)

(b) Cycle 12: w12= -PV = -P1(V2-V1) = -1 x 22.4 = 22.4 atmL = -2.27 kJ

U12= 3nRT/2 = 3P1(V2-V1)/2 = +3.40 kJ (T21= T2-T1= P1(V2-V1)/nR)

q12= U12- w12= +5.67 kJ

Another way to calculate Cycle 12 is also correct

q12= 5nRT/2 = +5.67 kJ

U12= 3nRT/2 = 3P1(V2-V1)/2 = +3.40 kJ

w12= U12- q12= -2.27 kJ

Cycle 23: w23= -PV =0, U23= 3nRT/2 = -3P1(V2-V1)/2 = -3.40 kJ

(T32= T3-T2= -P1(V2-V1)/nR), q23= U23= -3.40 kJ

Cycle 31: 13

331 1 1 1 1

1

ln ln 2 1.57V

V

VdVw nRT nRT PV kJ

V V

U31= 0, q31=w31= -1.57 kJ

For overall cycle: w = -0.70 kJ, q = +0.70 kJ, U = 0,

(one answer, 1 point)

2. Ans. (16% total, 2% each)(a) Carnot, and Joule.

(b)Pure form at P= 1 bar.

(c)H


8/11

8

3.

Ans.Sample is in the fixed volume, i.e. under constant-volume condition.

V = 0, thus, w = 0.

2.35 kJ of heat was supplied, thus q = +2.35 kJ.

FromU = q + w, H = U + PV, van der Waals equation (P+an2/V2)(V-nb)=nRT,

For initial state: (P1+an2/V2)(V-nb)=nRT1.

For final state: (P2+an2/V2)(V-nb)=nRT2.

Differential between final and initial state,

(P2- P1)(V-nb)=nr(T2- T1) and P = nRT/(V-nb).

2 1 2 0.0821 15 (341 300)2.35

15 2 0.04267

nRV T T H U V P q V P q

V nb

= 2.35 [kJ] +(2x0.0821x15x41)/(15-2x0.04267) [atm] x 1.10x102 [Jatm-1

L-1

]

= +3.03 kJ

Thus, w = 0, U = +2.35 kJ, and H = +3.03 kJ.

(2% for w , 3% for U ,

4% for van der Waals solution of H, 2% for ideal gas solution of H)

4.

Ans.(a) The Lewis structure for NO and NO2show that the bond order in NO is a double bond

and that in NO2is 1.5 on average. (2%, if both Lewis structure is correct, then 1 point. If

both bond order is correct, then 1 point)

(b) Bond enthalpy of NO at 632 kJmol-1

is close to N=O value listed in table, whereas the

N-O bond enthalpy in NO2is of 469 kJmol-1

is slightly greater than the average of a N-O

single and N=O double bond;

(630 kJ + 210 kJ) = 420 kJ.

The extra stability is due to resonance stabilization. (5%)

(c) The bond energies are the same for the two bonds, because the bonds are equivalent

due to resonance. (2%)

5. Ans.

Reaction enthalpy is obtained from H = U + ngasRT.

Thus, H = U + ngasRT = 8.0 + 2x8.314x583 =17.7 kJ. (6%)


9/11

9

6.

Ans. (6%, 2% each)(a) 3NO2(g) + H2O(l)3HNO3(aq) + NO(g).

Hor= 2Ho

f(HNO3) + Ho

f(NO, g)[3Ho

f(NO2, g) + Ho

f(H2O, l)]

= 2(-207.36)+90.25[3(+33.18) +(-285.83)]

=-138.13 kJ.

(b) B2O3(s) + 3CaF2(s)2BF3(g) + 3CaO(s).

Hor= 2Ho

f(BF3, g) +3Ho

f(CaO, s)[Ho

f(B2O3, s)+Ho

f(CaF2, s)]

= 2(-1137.0)+3(-635.09)[-1272.8+ 3(-1219.6)]

= +752.3 kJ.

(c) H2S(aq) + 2KOH(aq)K2S(aq) + 2H2O(l).

Hor= Ho

f(K2S, aq) +2Ho

f(2H2O, l)[Ho

f(H2S, aq)+Ho

f(KOH, aq)]

= -417.5+2(-285.83)[-39.7+ 2(-482.37)]

= +15.28 kJ.

7. Ans:

For this irreversible process, we may construct two (reversible) steps first, (1) and (2),and then to calculate their individual entropy changes. We then sum them up to obtain

the total entropy change of the system.

(1)Compression from P1to P2at constant temperature (T1):

(must be a reversible process)

(5%)

(2)Heating from T1to T2at constant pressure (P2):

(can be either reversible or irreversible)

(3)

The total entropy change from P1T1to P2T2:

Note that if one considers the compression in terms of volume change (V1to V2

are not specified in the problem, so they MUST be transferred to T1/P1and T2/P2

using the ideal gas law). Alternatively, one may construct the following steps

and obtain exactly the SAME answer:

(1)Compression from V1to V2at constant temperature (T1):

(must be a reversible process)

(2)Heating from T1to T2at constant volume (V2):

(can be either reversible or irreversible)

(3)The total entropy change from P1T1to P2T2:


10/11

10

8.

Ans: (4%)

Hvap= 85 (273.15+59) /1000 = 28.2 kJ mol-1

9. Ans: (5%)

Therefore,

It is a spontaneous process!

10.

Ans:The criterion for spontaneous process is G = H - TS < 0 at constant T and P.

(a) For most combustion reactions, H < 0, and H overwhelms the contribution of

theTS term for the G being always negative.

(b) For endothermic reactions, H > 0, the only way for G < 0 is TS > H, so S

must be positive to overcome the positive contribution of H.

(c) Following (b), increasing T for TS > H is a way to assist the original

non-spontaneous reaction to be spontaneous.

11.

Ans: (5% each, 10% Total)For (A) C2H4(g) + H2O(g) CH3CH2OH(l),

Hr0= -277.6952.26 + 241.82 = -88.13 kJ mol

-1< 0

Sr0= 160.7-219.56188.83 = -247.69 J K

-1mol

-1< 0

Gr0= Hr

0- TSr

0= -88.13 - 298(-247.69/1000)= -14.32 kJ mol

-1< 0

For (B) C2H6(g) + H2O(g) CH3CH2OH(l) + H2(g),

Hr0= 0 - 277.69 + 84.68 + 241.82 = +48.81 kJ mol

-1> 0

Sr0= 130.68 + 160.7229.6188.83 = -127.05 J K

-1mol

-1< 0

Gr0= Hr

0- TSr

0= 48.81 - 298(-127.05/1000)= +86.67 kJ mol

-1> 0

(a) A is spontaneous but B is non-spontaneous.

(b)

B is never spontaneous because Hr0> 0 and Sr0


11/11

11

12. Ans:(a) Starting from GH - TS

G = H - TS (at constant T and P)

= U + PV - TS (UH + PV)

= q + w + PV - TS (U = q + w)

= TS + (-PV + we,max) + PV - TS

(for reversible process, q = TS and w = -PV + we,max)

= we,max (5%)

Meaning that the maximum non-expansion work is equal to the Gibbs free

energy for a reversible process.

(b)C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(l)

we,max= Gr0= [9(-237.13) + 8(-394.36)] [(+6.4) + 0] = -5295.5 kJ mol

-1

or | we,max| = 5295.5 kJ mol-1

(5%)

(c) Hc0= -5471 kJ mol-1, the minimum heat that must be released to theenvironment (cannot be used to do the electric work) is

q TS =Hc0- Gr

0= -5471 + 5296 = -175 kJ mol

-1

or | q | = 175 kJ mol-1

(5%)

chemistry 1 thermodynamics final and solutions

Documents