Chemical Equilibrium - Shailendra Kumar Chemistry

PHYSICAL CHEMISTRY
By: Shailendra Kumar
PHYSICAL CHEMISTRY
by: SHAILENDRA KR.
Classes at: -
Meq. Approach
SCIENCE TUTORIALS; Opp. Khuda Baksh Library, Ashok Rajpath, Patna
PIN POINT STUDY CIRCLE; House No. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Topic: Chemical Equilibrim
1. Using the listed [∆G° f
values] calculate ∆G for the reaction:
2 C 4
H 10
(g) [–17.2] + 13 O 2
(g) [0.0] → 8 CO 2
(g) [–394.4] + 10 H 2
O (g) [–228.6]
a. -605.8 b. -5406.8
c. -640.2 d. -5475.6
e. -2476.2
2. Using the listed [∆G° f
values] calculate ∆G for the reaction:
3 NO 2
(g) [51.3] + H 2
O (l) [-237.1] → 2 HNO 3
(aq) [–110.5] + NO (g) [86.6]
a. -217.6 b. -51.2
c. -24.5 d. +86.2
e. +161.9
3. Using the listed [∆G° f
values] calculate ∆G for the reaction:
Fe 2
O 3
(s) [-742.2] + 3 CO (g) [-137.1] → 2 Fe (s) [0.0] + 3 CO 2
(g) [-394.4]
a. -2337.0 b. -29.4
c. -485.0 d. +212.0
e. +485.0
4. The enthalpy and entropy change of a reaction are -29.3 kJ/mole and -2.54 J/mole K respectively
at 25°C. What is the free energy change in kJ/mole?
a. +30.1 b. -28.5
c. -31.4 d. -29.4
e. -30.1
5. The enthalpy and entropy change of a reaction are -6.1 kJ/mole and +38.9 J/mole K respectively
at 25°C. What is the free energy change in kJ/mole?
a. -5.5 b. -17.7
c. +17.7 d. +5.5
e. -11.6
6. The enthalpy and entropy change of a reaction are -26.1 kJ/mole and +3.89 J/mole K respectively
at 25°C. What is the free energy change in kJ/mole?
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
Page No.: 1
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna

PHYSICAL CHEMISTRY
By: Shailendra Kumar
a. +24.9 b. -27.3
c. -17.3 d. -24.9
e. -30.3
7. From the following ∆H° and ∆S° values predict which of reactions I, II, and III would be spontaneous
at 25°C.
∆H (kJ) ∆S (J/K)
I +10.5 +30
II -10.0 -10
III -5.0 -25
a. II b. I
c. III d. I and II
e. II and III
8. From the following ∆H° and ∆S° values predict which of reactions I, II, and III would be spontaneous
at 25°C.
∆H (kJ) ∆S (J/K)
I +10.5 +30
II +1.8 -113
III -126 +84
a. III b. I
c. II d. I and II
e. II and III
9. From the following ∆H° and ∆S° values predict which of reactions I, II, and III would be spontaneous
at 25°C.
∆H (kJ) ∆S (J/K)
I +10 -10
II -25 +20
III -5.0 -30
a. I and II b. I
c. II d. III
e. II and III
10. Using the listed information calculate ∆G (kJ) for the reaction:
2 Cds (s) + 3 O 2
(g) → 2 CdO (s) + 2 SO 2
(g)
∆H° f
(kJ/mole) -161.9, 0.0, -258.2, -296.9
S° (J/K.mole) 64.9, 205.1, 54.8, 248.2
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 2

PHYSICAL CHEMISTRY
By: Shailendra Kumar
a. -789.3 b. -832.4
c. -872.8 d. -905.4
e. -744.9
11. Using the listed information calculate ∆G (kJ) for the reaction:
2 Ag (s) + Cl 2
(g) → 2 AgCl (s)
∆H° f
(kJ/mole) 0.0, 0.0, -127.0
S° (J/K.mole) 42.7, 222.9, 96.1
a. -206.4 b. -230.5
c. -252.5 d. -241.6
e. -219.4
12. Using the listed information calculate ∆G (kJ) for the reaction:
C 6
H 12
(g) + 9 O 2
(g) → 6 CO 2
(g) + 6 H 2
O (l)
∆H° f
(kJ/mole) -156.4, 0.0, -393.5, -285.9
S° (J/K.mole) 204.4, 205.1, 213.7, 69.9
a. -3932.2 b. -3888.8
c. -4000.5 d. -4020.3
e. -3816.1
13. Calculate ∆H° f
for TeF 6
from the ∆G of reaction and the S values.
Te (s) + 3 F 2
(g) → TeF 6
(g)
∆G = –1221.5 k
s (J/K.mole) 49.7, 202.9, 337.5
a. -1317.1 b. -1280.4
c. -1415.8 d. -1338.5
e. -1435.7
14. Calculate ∆H° f
for SiF 4
from the ∆G of reaction and the S values.
Si (s) + 2 F 2
(g) → SiF 4
(g)
∆G = -1506 kJ
s (J/K.mole) 18.7, 202.9, 284.5
a. -1614.3 b. -1578.5
c. -1595.3 d. -1675.3
e. -1642.3
15. Calculate ∆H° f
for BrCl from the ∆G of reaction and the S values.
Br 2
(l) + Cl 2
(g) → 2 BrCl (g)
∆G = -2.0 kJ s (J/K.mole) 152.2, 223.1, 240.1
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 3

PHYSICAL CHEMISTRY
By: Shailendra Kumar
a. +14.6 b. +8.2
c. +29.2 d. +10.2
e. +21.0
16. The entropies of H 2
S (g), H (g) and S (g) are 205.8, and 167.8 J mole –1 K –1 respectively at 298°K.
Using the ∆G° given below calculate the bond energy of H 2
S.
H 2
S (g) → S (g) + 2 H (g)
∆G° = 677.8 kJ
a. 394.1 b. 381.2
c. 367.4 d. 389.3
e. 374.8
17. The molar entropies of NO (g), N (g), and O (g) at 298°K are 210.6, 153.2, and 161.0 J mol –1 K –1
respectively. Using the ∆G° given below calculate the bond energy of NO.
NO (g) → N (g) + O (g)
∆G° = 600.1 kJ
a. 639.0 b. 609.0
c. 631.0 d. 620.0
e. 600.0
18. The entropies of Cl 2
(g) and Cl (g) are 223.0 and 165.1 J mole –1 K –1 respectively at 298°K. Using
the data given calculate the bond energy of Cl 2
.
Cl 2
(g) → 2 Cl (g)
∆G° = +211.4 kJ
a. 241.1 b. 251.4
c. 243.3 d. 246.4
e. 249.5
19. Calculate ∆G° (kJ/mole) at 30°C for a reaction with K eq
= 4.99 x 10 5 .
a. +3.27 b. -3.27
c. -33.1 d. -14.8
e. +33.1
20. Calculate ∆G° (kJ/mole) at 275°C for a reaction with K eq
= 4.33 x 10 –5 .
a. +32.5 b. -23.0
c. +45.8 d. -45.8
e. +23.0
21. Calculate ∆G° (kJ/mole) at 58°C for a reaction with K eq
= 3.11 x 10 2 .
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 4

PHYSICAL CHEMISTRY
e. 5.9 x 10 5
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
By: Shailendra Kumar
a. +15.8 b. +2.77
c. -15.8 d. -2.77
e. -8.24
22. The K p
= 0.35 at 25°C for the following reaction. Calculate ∆G° (kJ).
H 2
(g) + I 2
(g) → 2 HI (g)
a. 7.5 b. 4.2
c. 2.6 d. 3.4
e. 5.9
23. The K p
= 9.66 x 10 –16 at 25°C for the following reaction. Calculate ∆G° (kJ).
N 2
(g) + O 2
(g) → 2 NO (g)
a. 92.3 b. 89.2
c. 85.7 d. 81.2
e. 75.0
24. The dissociation of gaseous chlorine to chlorine atoms has K p
= 0.106 atm at 1800°K. Calculate
the free energy change (kJ) for the reaction.
Cl 2
(g) → 2 Cl (g)
a. 7.16 x 10 4 b. 5.22 x 10 4
c. 3.36 x 10 4 d. 5.22 x 10 3
e. 2.12 x 10 3
25. The ∆G°f of H 2
O (l) is -237.2 kJ/mole. Calculate K f
for the following reaction at 25°C.
2 H 2
O (l) → 2 H 2
(g) + O 2
(g)
a. 2.7 x 10 –42 b. 6.8 x 10 –42
c. 2.7 x 10 –84 d. 4.5 x 10 –62
e. 7.6 x 10 –84
26. The ∆G° for the following reaction is -204 kJ. Calculate K p
for the reaction at 25°C.
CO (g) + Cl 2
(g) → COCl 2
(g)
a. 1.8 x 10 –36 b. 6.2 x 10 36
c. 4.2 x 10 32 d. 3.2 x 10 –34
e. 5.5 x 10 35
27. The ∆G° = -32.9 kJ for the following reaction. Calculate K p
at 25°C.
N 2
(g) + 3 H 2
(g) → 2 NH 3
(g)
a. 1.7 x 10 –6 b. 7.2 x 10 5
c. 8.9 x 10 5 d. 3.2 x 10 –6
Page No.: 5

PHYSICAL CHEMISTRY
By: Shailendra Kumar
28. The ∆G° and ∆H° for the following reaction are 48.5 kJ/mole and 100.6 kJ/mole respectively. The
∆S° is 174.8 J/mole K. At what temperature (°C) will the partial pressure of CO 2
equal 1.0 atm?
MgCO 3
(s) → MgO (s) + CO 2
(g)
a. 273 b. 324
c. 303 d. 391
e. 576
29. Using the listed information calculate K p
at 25°C.
2 HgO (s) → 2 Hg (l) + O 2
(g)
∆H° f
(kJ/mole) -90.8, 0.0, 0.0
∆S° (J/K.mole) 70.3, 76.0, 205.1
a. 3.12 x 10 –5 b. 1.43 x 10 –16
c. 2.48 x 10 –5 d. 2.21 x 10 –6
e. 3.12 x 10 –6
30. Using the listed information calculate K p
at 25°C.
Ag 2
O (s) → 2 Ag (s) + 1/2O 2
(g)
∆H° f
(kJ/mole) -31.0, 0.0, 0.0
∆S° (J/K.mole) 121.3, 42.6, 205.1
a. 2.22 x 10 –3 b. 2.22 x 10 –2
c. 1.09 x 10 –2 d. 4.41 x 10 –1
e. 3.41 x 10 –2
31. Consider the ∆G° f
and ∆H° f
(kJ/mole) for the following compounds. Which compound can be
most easily decomposed to from hydrogen gas and an element?
a. CaH 2
(∆G° = -149.8, ∆H° = -188.7)
b. SiH 4
(∆G° = -39.3, ∆H° = -61.9)
c. LiH (∆G° = -69.9, ∆H° = -90.4)
32. Consider the ∆G° f
and ∆H° f
(kJ/mole) for the following oxides. Which oxide can be most easily
decomposed to from the metal and oxygen gas?
a. ZnO : (∆G° = -318.4, ∆H° = -348.3)
b. Ag 2
O : (∆G° = -11.2, ∆H° = -31.1)
c. CdO : (∆G° = -228.4, ∆H° = -258.2)
d. PbO : (∆G° = -187.9, ∆H° = -217.3)
33. Consider the ∆G° f
and ∆H° f
(kJ/mole) for the following oxides. Which oxide can be most easily
decomposed to from the metal and oxygen gas?
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 6

PHYSICAL CHEMISTRY
By: Shailendra Kumar
a. ZnO : (∆G° = -318.4, ∆H° = -348.3)
b. HgO : (∆G° = -58.5, ∆H° = -90.8)
c. PbO : (∆G° = -187.9, ∆H° = -217.3)
d. Cu 2
O : (∆G° = -146.0, ∆H° = -168.8)
34. The Henry’s Law constant for the dissolution of CO 2
in water are 13.2 and 29.4 atm/M at 0°C and
25°C respectively. Calculate ∆H° (kJ) for the following process.
CO 2
(aq) → CO 2
(g)
a. -21.7 b. -9.42
c. +9.42 d. +49.9
e. +21.7
35. The equilibrium constants for following reaction at 627°C and 897°C are 4.62 x 10 1 and
1.0 x 10 –1 respectively. Calculate ∆H° for the reaction in kJ.
2 SO 2
(g) + O 2
(g) → 2 SO 3
(g)
a. -101 b. -178
c. +101 d. +149
e. +199
36. The equilibrium constants for following reaction at 847°C and 997°C are 6.0 x 10 –4 and
6.88 x 10 –3 respectively. Calculate ∆H° for the reaction in kJ.
Br 2
(g) → 2 Br (g)
a. +114 b. +149
c. -192 d. -114
e. +192
37. The equilibrium constant for the following reaction at 627°C is 46.2. The ∆H° for the reaction is -
197.8 kJ. Calculate the equilibrium constant at 897°C.
2 SO 2
(g) + O 2
(g) → 2 SO 3
(g)
a. 4.1 b. 4.1 x 10 –3
c. 3.2 x 10 –1 d. 2.0 x 10 –2
e. 1.0 x 10 –1
38. The equilibrium constant for the following reaction at 600°C is 3.9 x 10 –1 . The ∆H° for the reaction
is +35.0 kJ. Calculate the equilibrium constant at 900°C.
H 2
(g) + CO 2
(g) → CO (g) + H 2
O (g)
a. 3.3 b. 5.8
c. 2.4 d. 4.2
e. 1.3
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 7

PHYSICAL CHEMISTRY
By: Shailendra Kumar
39. The equilibrium constant for the following reaction at 1727°C is 4.1 x 10 –4 . The ∆H° for the reaction
is +181 kJ. Calculate the equilibrium constant at 2027°C.
N 2
(g) + O 2
(g) → 2 NO (g)
a. 1.5 x 10 –2 b. 1.1 x 10 –3
c. 6.0 x 10 –2 d. 4.0 x 10 –2
e. 1.7 x 10 –3
40. The equilibrium constant K a
= 3.0 x 10 –8 M at 25°C for the following reaction. Calculate ∆G° (kJ) at
25°C when [ClO – ] = [H + ] = 1.0 x 10 –6 M and [HClO] = 0.10 M.
HClO (aq) → H + (aq) + ClO – (aq)
a. -38.9 b. -62.8
c. -19.9 d. +38.9
e. +42.9
41. The K sp
= 2.0 x 10 7 M –6 for the following reaction at 25°C. Calculate the value of ∆G° (kJ) at 25°C
in a solution when [CO + 3+
3
] = 0.0050 M, [NH 3
] = 0.10 M, and [Co(NH 3
) 6
] = 1.00 M.
+ 3+
Co 3
(aq) + 6 NH 3
(aq) → Co (NH 3
) 6
(aq)
a. -41.7 b. +20.5
c. +5.7 d. -20.5
e. +47.4
42. Calculate ∆G° (kJ) at 25°C for the following reaction when the pressures are as follows:
P = 0.010 atm; P C 2H4 O2 = 0.020 atm; P CO2 = 20 atm; and P H2O = 0.010 atm
C 2
H 4
(g) + 3 O 2
(g) → 2 CO 2
(g) + 2 H 2
O (g)
∆H° (kJ/mole) 52.2, 0.0, -393.5, -241.8
S° (J/K.mole K) 219.6, 205.1, 213.7, 188.8
a. -1252 b. 1346
c. -1282 c. -1314
e. -1387
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 8

PHYSICAL CHEMISTRY
A N S W E R S
By: Shailendra Kumar
1. b 16. c 31. b
2. b 17. c 32. b
3. b 18. c 33. b
4. b 19. 34. e
5. b 20. 35. e
6. b 21. 36. e
7. a 22. c 37. e
8. a 23. c 38. e
9. a 24. c 39. e
10. e 25. e 40. c
11. e 26. e 41. c
12. e 27. e 42. c
13. 28. c
14. 29. c
15. 30. c
1. Opp. Khuda Baksh Library, Ashok Rajpath, Patna - 4
2. House no. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna
Page No.: 9