1

Chemistry I Final Exam

18:00-21:00, 12 January, 2018 (Total Score: 103 points)

gas constant R = 8.314 J∙mol–1∙K–1 = 8.206×10–2 L∙atm∙mol–1∙K–1 = 8.314×10–2 L∙bar∙mol–1∙K–1

= 62.36 L∙Torr∙mol–1∙K–1

Boltzmann’s constant kB = 1.38×10–23 L∙atm∙K–1

van der Waals equation of state: (𝑃 + 𝑎𝑛2

𝑉2) (𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 , 𝑃 =

𝑅𝑇

�̅�−𝑏− 𝑎�̅�−2

1. A compound contains 64.9% carbon, 13.5% hydrogen, and 21.6% oxygen by mass. At 120°C

and 750 torr, 2.00 L of the gaseous compound has a mass of 4.50 g. What is the molecular

formula of the compound? (5%)

2. (a) Derive the relationship between the van der Waals constants (a & b) and the virial

coefficient B for a real gas.

The virial equation of state: 𝑍 =𝑃𝑉

𝑛𝑅𝑇= 1 +

𝐵

�̅�+

𝐶

�̅�2+

𝐷

�̅�3+⋯

You may use this equation: 1

1−𝑥= 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 +⋯ for |𝑥| < 1

The van der Waals constants of Ar are a = 1.35 bar∙L2∙mol-2 and b = 0.0322 L∙mol-1. Please

find (b) the second virial coefficient, B, at 200 K and (c) the Boyle temperature of Ar. Here you

can ignore the coefficients C, D, E,… (9%)

2

3. Which of the following are the central assumptions for the ideal gas and the kinetic molecular

theory of gases? (multiple answers, 複選題) (3%)

(A) molecules have negligible mass

(B) molecules have negligible volume

(C) gas is composed of very large number of molecules

(D) molecules move in straight lines in some specific directions

(E) molecules collide frequently and perfectly elastically

(F) there must be attraction and repulsion between molecules

4. Please use the Gibbs phase rule to predict the maximum number of phases that can coexist in a

system with only one component present. (3%)

5. The compounds Br2 and ICl have the same number of electrons, yet Br2 melts at -72°C,

whereas ICl melts at 27.2°C. Why? (3%)

6. Which of the following phase changes are endothermic? (multiple answers, 複選題) (3%)

(A) melting (B) vaporization (C) sublimation (D) condensation

(E) deposition (F) freezing

7. Which of the following properties indicates strong intermolecular forces in a liquid? (3%)

(A) low surface tension (B) low vapor pressure (C) low viscosity

(D) low conductivity (E) high acidity

8. Which type of semiconductor is represented by the band structure

shown on the right? (3%)

(A) An intrinsic semiconductor as in pure Si

(B) An n-type semiconductor as in Ga-doped Si

(C) An n-type semiconductor as in P-doped Si

(D) A p-type semiconductor as in Ga-doped Si

(E) A p-type semiconductor as in P-doped Si

9. A quantitative measure of how efficiently spheres pack into unit cells is called packing

efficiency, which is the percentage of the cell space occupied by the spheres. Calculate the

packing efficiencies of a simple cubic cell, a body-centered cubic cell, and a face-centered

cubic cell. (Hint: see the figure on the next page; use the relationship that the volume of a

sphere is 4πr3/3, where r is the radius of the sphere) (9%)

3

10. Classify the solid state of the following substances as ionic solids, covalent solids, molecular

solids or metallic solids. (6%)

(a) NH4Cl, (b) graphene, (c) CaI2, (d) Al, (e) glucose C6H12O6, (f) C60 (fullerene)

11. Peter builds up a special engine called as the X-engine. The X-engine is filled with 0.8 mole

CO2, which is treated as an ideal gas with molecular kinetic energies containing translational

and rotational kinetic energies only. The X-engine undergoes a cyclic process involving three

steps as shown in the figure below.

Step 1: The X-engine starts at A and undergoes an isothermal reversible expansion process

to B.

Step 2: The engine is cooled down and undergoes the constant-pressure compression process

from B to C

Step3 : Finally, it is heated at the constant volume from C to A.

Here, the X-engine is assumed to be the system. Please answer the following questions with

the SI unit.

4

(a) What is the temperature 𝑇𝐴 of the state A? (2%)

(b) What is the temperature 𝑇𝐶 of the state C? (2%)

(c) Please calculate the expansion work 𝑤(A → B) done from A to B. (2%)

(d) Please calculate ∆𝑈(B → C) from B to C. (2%)

(e) Please calculate the heat 𝑞(C → A) from C to A. (2%)

(f) Please calculate the expansion work 𝑤(𝑐𝑦𝑐𝑙𝑒) for a complete cycle. (2%)

(g) Please calculate the heat 𝑞(𝑐𝑦𝑐𝑙𝑒) for a complete cycle. (2%)

(h) Please calculate ∆𝑈(𝑐𝑦𝑐𝑙𝑒) for a complete cycle. (2%)

12. One innovative reactor is designed to be an isolated system and initially contains 1 mole

hydrogen gas and 1 mole oxygen gas at 25°C. The electric spark is used to ignite the hydrogen

inside the reactor. Here, it is assumed that all of the reactants and the products of the

combustion reaction are ideal gases. In addition, the molecular kinetic energies of these ideal

gases contain translational and rotational kinetic energy only. The standard enthalpy of

formation of liquid water and water vapor are -285.8 kJ/mol and -241.83 kJ/mol at 25°C.

(a) Write the chemical equation of the combustion reaction with the appropriate

stoichiometry coefficients and correct states. Please note that the appropriate

stoichiometry coefficients are meant to be the smallest integers. (2%)

(b) What is the standard enthalpy ∆𝐻𝑟𝑥𝑛° of the combustion reaction (kJ) at 25°C? (2%)

(c) What is the relation between ∆𝑈𝑟𝑥𝑛° and ∆𝐻𝑟𝑥𝑛

° for the combustion reaction at 25°C?

∆𝑈𝑟𝑥𝑛° > ∆𝐻𝑟𝑥𝑛

° , ∆𝑈𝑟𝑥𝑛° < ∆𝐻𝑟𝑥𝑛

° or ∆𝑈𝑟𝑥𝑛° = ∆𝐻𝑟𝑥𝑛

° ? (2%)

Please explain it by using the mathematical relation between ∆𝑈𝑟𝑥𝑛° and ∆𝐻𝑟𝑥𝑛

° . (2%)

(d) If the innovative reactor is assumed to be the system, what is the change of the internal

energy of the system after the complete combustion? (2%)

13. Predict whether the standard entropy change (∆𝑆rxn° ) is positive, negative, or about zero for

each of the following reactions. Give reasons for your predictions. (12%)

(a) 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)

(b) 2 O(g) → O2(g)

(c) NH4Cl(s) → NH3(g) + HCl(g)

(d) H2(g) + Cl2(g) → 2 HCl(g)

(e) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

(f) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

14. For the spontaneous reaction 2 Fe(s) + 3/2 O2(g) → Fe2O3(s), the standard enthalpy change

∆𝐻rxn° = −824.2 kJ and the standard entropy change ∆𝑆rxn

° = −274.7 J∙K–1 at 𝑇 = 25°C.

Please explain the apparent paradox of violating the 2nd law of thermodynamics here, where a

spontaneous reaction has a negative entropy change. (6%)

5

15. Find at which temperatures (in °C), the reactions with the following ∆𝐻rxn and ∆𝑆rxn

are at

equilibrium.

(a) ∆𝐻rxn = −15.5 kJ∙mol–1 and ∆𝑆rxn

= −27.2 J∙mol–1∙K–1 (3%)

(b) ∆𝐻rxn = +15.5 kJ∙mol–1 and ∆𝑆rxn

= +27.2 J∙mol–1∙K–1 (3%)

16. Calculate the value of ∆𝑆 for heating 7 moles of a ideal gas from 150°C to 350°C (a) at

constant pressure and (b) at constant volume. Remember that ideal gas particles can be

considered as points with no internal degrees of freedom. (6%)

6

106A Chemistry (I) Final Exam

Answers

1. 5%

PV = nRT = 750×2.00 = n×62.36×393 n = 0.0612

molar mass of this compound = 4.50/0.0612 = 73.5 g/mol

carbon = 73.5×64.9%/12 = 3.98 , hydrogen = 74.3×13.5%/1 = 9.92 ,

oxygen = 75.2×21.6%/16 = 0.99 the molecular formula is C4H10O

2. 各 3%，共 9%

(a) van der Waals equation: 𝑃 =𝑅𝑇

�̅�−𝑏−

𝑎

�̅�2

𝑍 =𝑃𝑉

𝑛𝑅𝑇=𝑃�̅�

𝑅𝑇=

�̅�

�̅� − 𝑏−

𝑎

�̅�𝑅𝑇=

1

1 −𝑏�̅�

−𝑎

�̅�𝑅𝑇

= (1 +𝑏

�̅�+ (

𝑏

�̅�)2

+ (𝑏

�̅�)3

+ (𝑏

�̅�)4

+⋯) −𝑎

�̅�𝑅𝑇

𝐵 = 𝑏 −𝑎

𝑅𝑇

(b) at 200 K, B = 0.0322 – 1.35/(0.08314×200) = -0.0490 L∙mol-1 (R用錯扣 1.5分)

(c) at Boyle temperature, B = 0 T = a/(b×R) = 504 K (R用錯扣 1.5分)

3. 3%，每個正確選項得 1%，錯誤選項扣 1%，扣至 0分為止

(B) (C) (E)

4. 3%，分段給分

f = c – r + 2 (1%), only one component c =1

maximum number of co-exist phases f = 0 (1%)

the number of co-exist phases r = 3 (1%)

5. 3%

ICl is polar and Br2 is nonpolar. The dipole-dipole interaction between polar molecules results

in higher melting point.

6. 3%，每個正確選項得 1%，錯誤選項扣 1%，扣至 0分為止

(A) (B) (C)

7. 3%

(B)

8. 3%

(C)

9. 各 3%，共 9%

(a) Simple cubic cell: cell edge a 2r

7

(b) Body-centered cubic cell:

(c) Face-centered cubic cell:

10. 各 1%，共 6%

(a) ionic solid (b) covalent solid (c) ionic solid

(d) metallic solid (e) molecular solid (f) molecular solid

11. 各 2%，共 16%

(a) 𝑇𝐴 =𝑃𝑉

𝑛𝑅=

30×1

0.8×0.08314= 451.0 (𝐾)

(b) 𝑇𝐶 =𝑃𝑉

𝑛𝑅=

10×1

0.8×0.08314= 150.3 (𝐾)

(c) 𝑤(A → B) = −n ∙ R ∙ 𝑇𝐴 ∙ ln (𝑉𝐵

𝑉𝐴) = −0.8 × 8.314 × 451 × ln (

3

1) = −8999 (𝐽)

(d) ∆𝑈(B → C) = n ∙ 𝐶𝑣̅̅ ̅ ∙ ∆𝑇 = 0.8 × 2.5 × 8.314 × (150.3 − 451) = −5000 (𝐽)

or

𝑤(B → C) = −P ∙ ∆V = −10 × (1 − 3) × 100 = +2000 (𝐽)

𝑞(B → C) = n ∙ 𝐶𝑝̅̅ ̅ ∙ ∆𝑇 = 0.8 × 3.5 × 8.314 × (150.3 − 451) = −7000 (𝐽)

∆𝑈(B → C) = 𝑞(B → C) + 𝑤(B → C) = −5000 (𝐽)

(二擇一皆算對！)

(e) 𝑞(C → A) = n ∙ 𝐶𝑣̅̅ ̅ ∙ ∆𝑇 = 0.8 × 2.5 × 8.314 × (451 − 150.3) = +5000 (𝐽)

(f) 𝑤(𝑐𝑦𝑐𝑙𝑒) = 𝑤(A → B) + 𝑤(B → C) + 𝑤(C → A) = −8999 + 2000 + 0 = −6999 (𝐽)

or

𝑤(𝑐𝑦𝑐𝑙𝑒) = ∆𝑈(𝑐𝑦𝑐𝑙𝑒) − 𝑞(𝑐𝑦𝑐𝑙𝑒) = 0 − (+6999) = −6999 (J) (2%)

(g) 𝑞(𝑐𝑦𝑐𝑙𝑒) = 𝑞(A → B) + 𝑞(B → C) + 𝑞(C → A) = +8999 − 7000 + 5000 = +6999(J)

or

𝑞(𝑐𝑦𝑐𝑙𝑒) = ∆𝑈(𝑐𝑦𝑐𝑙𝑒) − 𝑤(𝑐𝑦𝑐𝑙𝑒) = 0 − (−6999) = +6999 (J)

(h) ∆𝑈(𝑐𝑦𝑐𝑙𝑒) = 0 (J) (∵ U is the state function)

3

3

3 3

4100%

3 4 100%Packing efficiency 100%

6(2 ) 24

æ öp´ç ÷

ç ÷p ´ pè ø

= = = ´ = 52.4%

r

r

r r

4cell edge

3=

r

Packing efficiency =

2 ´4pr3

3

æ

èçç

ö

ø÷÷´ 100%

4r

3

æ

èç

ö

ø÷

3=

2 ´4pr3

3

æ

èçç

ö

ø÷÷´ 100%

64r3

3 3

æ

èçç

ö

ø÷÷

=2p 3

16´ 100% = 68.0%

cell edge 8= r

( )

3 3

3 3

4 164 100% 100%

3 3 2Packing efficiency 100%

3 88 88

æ ö æ öp p´ ´ ´ç ÷ ç ÷ç ÷ ç ÷

pè ø è ø= = = ´ = 74.0%

r r

rr

8

12. (a)(b)(d)各 2%，(d)4%，共 10%

(a) 2 H2(g) + O2(g) → 2 H2O(g)

(It is assumed that all of the reactants and the products of the combustion reaction are

ideal gases. 此理由不需寫出來)

(b) ∆𝐻𝑟𝑥𝑛° = ∆𝐻𝑓

°(𝐻2𝑂(𝑔)) × 2 − ∆𝐻𝑓°(𝐻2(𝑔)) × 2 − ∆𝐻𝑓

°(𝑂2(𝑔)) × 1

= −241.83 × 2 − 0 − 0 = −483.66 kJ

(c) ∆𝑈𝑟𝑥𝑛° > ∆𝐻𝑟𝑥𝑛

° (2%)

∆𝐻𝑟𝑥𝑛° = ∆𝑈𝑟𝑥𝑛

° + RTΔn , ∆n = (number of moles of products gases) −

(number of moles of reactants gases) = 2 − 3 = −1 < 0 (2%)

(d) ∆U = 0 J

(Because the innovative reactor is assumed to be the system and is designed be an isolated

system, the change of the internal energy of the isolated system is equal to 0 J. 此理由不

需寫出來)

13. 每小題各 2%（分段給分：正/負/接近零 1%，解釋 1%），共 12%。

(a) negative; the main reason is that gas (with higher entropy than solid) is consumed in this

reaction

(b) negative; the main reason is that 1 mole of gas is consumed in this reaction

(c) strongly positive; two moles of gas are produced from 1 mole of solid

(d) about zero, because two moles of gas undergo a conversion into two moles of gas

(the actual number (∆𝑆rxn° = 20.3 J/K) is slightly positive, but close to zero)

(e) about zero, because three moles of gas undergo a conversion into three moles of gas

(the actual number (∆𝑆rxn° = −5.1 J/K) is slightly negative, but close to zero)

(f) strongly negative, because 2 moles of gas are consumed in this reaction

14. 6%

The 2nd law of thermodynamics would be violated only if the total entropy change in this

reaction equal to (the entropy change of the system) + (the entropy change of the surrounding)

would be negative. This is not the case. The entropy change of the surrounding is given by

∆𝑆surr° =

∆𝐻surr°

𝑇=−∆𝐻rxn

°

𝑇=

824200

273.15 + 25 [𝐽

𝐾] = 2764.4 [

𝐽

𝐾]

The total entropy change ∆𝑆tot° = 2489.7 J/K in this reaction is then actually quite positive, as

can be expected for a spontaneous reaction of combustion.

15. 各 3%，共 6%

The condition for equilibrium requires that ∆𝐺rxn = ∆𝐻rxn

− 𝑇∆𝑆rxn = 0, which gives the

corresponding temperature as

𝑇 =∆𝐻rxn

∆𝑆rxn =−15500

−27.2=15500

27.2= 569.85 𝐾 = 296.7°𝐶

in both (a) & (b).

9

16. 各 3%，共 6%

For an ideal gas, �̅�𝑉 =3

2𝑛𝑅 and �̅�𝑃 =

5

2𝑛𝑅. We have

∆𝑆 = ∫𝑑𝑄

𝑇

𝑇2

𝑇1

=

{

∫

�̅�𝑉 𝑑𝑇

𝑇

𝑇2

𝑇1

= ∫

32𝑛𝑅𝑑𝑇

𝑇

𝑇2

𝑇1

=3

2𝑛𝑅 ln

𝑇2𝑇1 at constant pressure

∫�̅�𝑃 𝑑𝑇

𝑇

𝑇2

𝑇1

= ∫

52𝑛𝑅𝑑𝑇

𝑇

𝑇2

𝑇1

=5

2𝑛𝑅 ln

𝑇2𝑇1 at constant volume

Substituting in these equations 𝑛 = 7 mol, 𝑇1 = 150°C = 523.15 𝐾, and 𝑇2 = 350°C =

723.15 𝐾, and 𝑅 = 8.31447 J/mol/K, one obtains the answer:

(a) ∆𝑆 = 28.3 J

(b) ∆𝑆 = 47.1 J.