Soal2 dr SN

1. CHEMISTRY TASK
SOAL & PEMBAHASAN
OLEH KEL 3 XI IA-6:
•DEBBY SIAGIAN (10)
•JOYCE ANASTASIA SETYAWAN (17)
•NAFIAH RAFIQAH R (22)
•SHABRINA NABILA MAHMUDI (27)

2. QUADRA

3. NO 1
1.According to the Law Conservation of Energy, energy…..
a. Can not be created (correction)
b. Can be transferred
c. Cannot be converted
d. Cannot form enthalpy
e. Can be distroyed

4. Solution for no 1
Law Conservation of Energy
• "Energy can neither be created nor it is destroyed,
however energy can be converted from one form
energy to any other form of energy"

5. NO 2
• To answer number 2 and 3,look at the reaction
below.
• C2H4+3O2 2CO2 + 2H2O
The enthalpy change of reaction is 1560 kJ/mole.
2. It is an….reaction,with the value of H=…..
a. balance; -1560 kJ/mole
b. Exothermic; -1560 kJ/mole
c. Exothermic; +1560 kJ/mole
d. Endothermic; -1560 kJ/mole
e. Endothermic; +1560 kJ/mole

6. Solution for no 2
Combustion=exotherm
• When a fuel reacts with Oxygen in a combustion
reaction, a large amount of heat is generally
released. (Something like a candle or a match for
example, will also release heat in smaller amounts).
The 'Release' of Heat Energy is an 'Exothermic'
reaction.
• the answer is (B) because an exotherm release heat
so -1560 kJ/mole

7. NO 3
3. The enthalpy change measured for the
reaction is 634 kJ. How much oxygen needed
for this reaction?
a. 8.91 L
b. 11.28 L
c. 22.4 L
d. 26.88 L
e. 32.4 L

8. Solution no 3
C2H4+3O2 2CO2 + 2H2O ΔH = 1560
• ΔH= 624 kJ (ΔH 1560 : 2.5)
• O2=? L
• C2H4+3O2 2CO2 + 2H2O ΔH = 1560 : 2.5
• 0.4 C2H4+ 1.2 O2 0.8 CO2 + 0.8 H2O
ΔH = 624 kJ
• 1.2 mole O2 x 22.4 = 26.88 L

9. NO 4
Look at the enthalpy cycle below
4. According to Hess’ Law,the value ΔH 2 is…
a. ΔH 1 + ΔH 3 + ΔH 4
b. ΔH 1 - ΔH 2 - ΔH 4
c. ΔH 2 + ΔH 4 - ΔH 1
d. ΔH 2 - ΔH 1 - ΔH 4
e. ΔH 1 - ΔH 3 - ΔH 4

10. ΔH 1
CaO(s) + H2O(l) Ca(s) + H2O(l)
Route 1
ΔH Route 2 Route 4 ΔH 4
Route 3
CaO(s) + H2O(l) Ca(s) + H2O(l)
ΔH 3

11. Solution for no 4 With Hess’Law
CaO(s) + H2O(l) Ca(s) + H2O(l) ΔH 1
• Ca(s) + H2O(l) CaO(s) + H2O(l) -ΔH 3
• Ca(s) + H2O(l) Ca(s) + H2O(l) -ΔH 4
CaO(s) + H2O(l) CaO(s) + H2O(l) ΔH 2

12. C6H12O6 + 6O2 6CO2 + 6H2O
ΔH = -2820 kJ/mole
C2H5OH + 3O2 2CO2 + 3H2O
ΔH = -1380 kJ/mole
Using data above,the enthalpy change for
fermentation with the equation of
5.C6H12O6 2C2H5OH + 6CO2 is…..
a. -4200 kJ/mole
b. + 1440 kJ/mole
c. -1440 kJ/mole
d. +60 kJ/mole
e. -60 kJ/mole

13. Solution for no 5
C6H12O6 + 6O2 6CO2 + 6H2O
4CO2 + 6H2O 2C2H5OH + 6O2 (reversed & x2)
C6H12O6 2C2H5OH + 2CO2 (corrected)
ΔH = ΔH 1+ ΔH 2
ΔH =-2820+ (+1380*2)
= -2820 + 2760 = +60 (D)

14. No. 6
• Given that:
MO₂ + CO MO + CO₂ ∆H⁰ = -20 kJ/mole
M₃O₄ + CO 3MO + CO₂ ∆H⁰ = +6 kJ/mole
3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ/mole
• The value of ∆H for the reactions 2MO₂ + CO
M₂O₃ +CO₂ is ….
A. -40 kJ D. -18 kJ
B. -28 kJ E. 40 kJ
C. -26 kJ

15. solution
• 2MO₂ + CO M₂O₃ +CO₂
x2 MO₂ + CO MO + CO₂ ∆H = -20 kJ
Reversed, x⅔ M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJ
Reversed, x⅓ 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ
2MO₂ + 2CO 2MO + 2CO₂ ∆H = -40 kJ
2MO + ⅔ CO₂ ⅔ M₃O₄ + ⅔ CO ∆H = -4 kJ
⅔M₃O₄ + ⅓CO₂ M₂O₃ + ⅓CO ∆H = +4 kJ
• 2MO₂ + CO M₂O₃ +CO₂ ∆H = -40

16. No. 7
• Given that
N₂ + 3H₂ 2NH₃ (g) ∆H = -92 kJ
The heat needed to dissociate 5.1 g ammonia
(Mr = 17 g/mole) is ….
A. 9.2 kJ/mole
B. 13.8 kJ/mole
C. 18.8 kJ/mole
D. 27.6 kJ/mole
E. 65.6 kJ/mole

17. solution
• A reaction with negative amount of ∆H: will
be positive when the reaction is reversed
N₂ + 3H₂ (g) 2NH₃ (g) ∆Hᵣ = kJ
-92
2NH₃ (g) N₂ + 3H₂ (g) ∆ Hᵣ= 92 kJ
• The initial enthalpy change apply when 2mole
of ammonia is reacted. In 5.1 g NH₃ there is
• mole NH₃ = mass of NH₃ = 5.1 = 0,3 mole
Mr of NH₃ 17
Therefor, the enthalpy change is
∆H: = 92:2= 46 x 0.3 = 13.8

18. No. 8
• The enthalpy change of combustion of Fe₂O is
a kJ/mole and the enthalpy of its formation is
b kJ/mole. Thus, the heat change of Fe₂O₃
formation could be represented as … kJ/mole.
A. a + b
B. a – b
C. 2a + b
D. a - 2b
E. 2b - a

19. solution
• Fe₂O + O₂ Fe₂O₃ ∆H = a
∆Hf=b ∆Hf=?
Hreaction = ∑∆Hf right - ∑∆Hf left
a = ∆Hf Fe₂O₃ - b
a + b = ∆Hf Fe₂O₃

20. No. 9
• Given that formation enthalpy of H₂O (l) and
H₂O (g) are -286 kJ/mole and -242kJ/mole. In
evaporation of 4.5 g water will … of energy.
A. release 11 kJ
B. absorb 11 kJ
C. Absorb 44 kJ
D. Release 44 kJ
E. Absorb 1332 kJ

21. solution
• ∆H: H₂O (l) is -286 kJ/mole
• ∆H: H₂O (g) is -242 kJ/mole
• H₂O (l) H₂O (g)
• ∆H = Hp - Hr
= -242 – (- 286)
= 44
n = m = 4.5 g = 0.25
Mr 18
So, the enthalpy change is 44x0.25=11 kJ

22. No.10
• H₂ + Br₂ 2HBr ∆H = -72 kJ/mole. The
amount of heat to break down 11.2 L of
hydrogen bromide into its elements is….
A. 18 kJ/mole
B. 36 kJ/mole
C. 72 kJ/mole
D. 144 kJ/mole
E. 288 kJ/mole

23. solution
• A reaction with negative amount of ∆H: will be
positive when the reaction is reversed
H₂ + Br₂ 2HBr ∆Hᵣ = kJ
-72
2HBr H₂ +Br₂ ∆Hᵣ = 72 kJ
• The initial enthalpy change apply when 2mole
HBr is reacted. In 11.2 L HBr is…
• mole HBr = Volume of HBr = 11.2 = 0.5 mole
22.4 22.4
Therefor, the enthalpy change is
∆H: = 72 : 2 = 36 x 0.5 = 18 kJ/mole

24. Number 11
The bond enthalpy in C2H5Cl is...
( C-H = 414 ; C-C = 347 ; C-Cl = 339 )
a.2840 kJ/mole
b. 2756 kJ/mole
c. 2409 kJ/mole
d.2070 kJ/mole
e. 2015kJ/mole

25. H H
| |
H – C – C – Cl (5 x C-H) + C-C + C-Cl =
| | (5x414) + 347 + 339 = 2756kJ/mole
(B)
H H

26. Number 12
The formation of hydrochloric acid, HCl, releases
183 kJ/mole energy. Using the data above, the
bond enthalpy of Cl-Cl is... (H-H=436 ; H-Cl=431)
a.67 kJ/mole
b. 124 kJ/mole
c. 201 kJ/mole
d. 268 kJ/mole
3. 312 kJ/mole

27. ½ H2 + ½ Cl2 → HCl ∆H=183kJ/mole (releases)
(½ H2 + ½ Cl2)-HCl = -183
½ (436 + Cl2) – 431 = -183
½ (436 + Cl2) = 248
436 +Cl2 = 496
Cl2 = 2Cl = 60kJ/mole

28. Number 13
The amount of energy released in the formation
of 5g HF = 20g/mole, E H-F=565kJ/mole ; E F-
F=158 kJ/mole, and E E-H=436 kJ/mole is...
a. 67
b. 124
c. 201
d. 268
e. 312

29. mHF=5gr MR HF= 20 HF=0,25 mole
½ H2 + ½ F2 → HF
218 + 79 – 565 = 268 (for 1 mole)
For 0,25 mole = 268 x 0,25 = 67 (A)

30. Number 14
Methane fuel CH4 is used to heat up 80g water. The
temperature rises from 25oC to 56oC. The amount
of used methane is 3,2 g. The value of enthalpy
change of CH4 is...
a. 1041,6 kJ
b. 10416 kJ
c. 2083,2 kJ
d. 20832 kJ
e. 5208 kJ

31. ∆H = -Q
Q = m.c.∆T
Q= 80 . 4,2 . 31 = 10416
∆H = -10416
m CH4 = 3,2 Mr = 16
3,2 : 16 = 0,2 mole
∆H for 0,2 mole = 0,2 x -10416 = -2083,2 kJ (C)

32. Number 15
Given that :
C + O2 → CO2 ∆H= -395,22kJ/mole
2H2 + O2 → 2H2O ∆H= -573,72kJ/mole
If the enthalpy change of C2H4 is +27,30kJ, then the
enthalpy change of combustion of C2H4 is...
a. -654,78kJ/mole
b. -663,18kJ/mole
c. -709,38kJ/mole
d. -887,45kJ/mole
e. -1.336,86kJ/mole

33. C + O2 → CO2 ∆H= -395,22kJ
2H2 + O2 → 2H2O ∆H= -573,72kJ
2C + 2H2 → C2H4 ∆H= 27,30kJ
C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=...?
2C + 2O2 → 2CO2 ∆H= -395,22 x 2 kJ
2H2 + O2 → 2H2O ∆H= -573,72kJ
C2H4 → 2C + 2H2 ∆H= -27,30kJ
______________________________________+
C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=-1.391,46kJ (E)

34. No 16
16. The entalphy change of reaction could be measured with the
following methods, excepts…
(a) coefficient of a reaction
(b) bomb calorimeter
(c) simple calorimeter
(d) hess law
(e) bond entalphy
Answer : A

35. No 17
16. The material needed for succesful combustion are…..
(a) carbon dyoxide, fuel , heat
(b) oxygen, Fuel, heat
(c) oxygen, carbon dyoxide, heat
(d) oxygen, water, heat
(e) carbon dyoxide, water, heat
Answer : E

36. No 18
16. The value of ΔH° could be calculated with the following
formulas, except…
(a) ΔH° = - Ccalorimeter . Δt
(b) ΔH° = E + P Δv
(c) ΔH° = ∑ (nproduct x ΔHf product) - ∑ (nreactant x ΔHf reactant)
(d) ΔH° = ∑ (bond energy of reactant)- ∑ ( bond energy of product)
(e) ΔH° = ∑ (nreactant x ΔHf reactant) - ∑ (nproduct x ΔHf product)
Answer = E

37. No 19
19. the standart entalphy change of formation ΔH°f of CH4
using the below experiment result, with condition of
temperature 25°C and 1 atm is…..
H2 (g) + ½ O2 H2O ΔH° = -285 kJ
C(s,graphite) + O2 CO2 ΔH° = -393,5 kJ
CH4 + 2 O2 CO2 ΔH° = -890,4 kJ
(a) +182,4 kJ/mole
(b) + 211,1kJ/mole
(c) - 211,1 kJ/mole
(d) + 74,7 kJ/mole
(e) - 74,7kJ/mole

38. No 20
20. A glass of water with the volume of 200ml is heated in the
microwave. The amount of absorbed heat by water if the
temperature is raised up up to 60°C is….
(Cwater = 4,18 J/g°C, pwater = 1000 g / l)
(a) 50,16 kJ
(b) 501,6 kJ
(c) 5016 kJ
(d) 50160 kJ
(e) 50160 kJ

39. No 20
m=p.v
= 1000. 200
= 200.000
Q = m . C . Δt
= 200.000 4,18 60
= 50.160.000 J
= 50.160 kJ
answer = D

40. FOTOCOPY SHEET

41. 21.Data:
2C2H2(g)+ 5O2(g) 4CO2(g) + 2H2O(l)
ΔH= -2,372.4kJ
• The correct statement for the combustion
reaction of 5.6 L C2H2 (STP) is….
a. Releasing 593.1 kJ of heat
b. Requiring 593.1 kJ of heat
c. The enthalpy of the system rises 296.55kJ
d. Absorbing 296.55 kJ of heat
e. Releasing 296.55 kJ of heat

42. Solution for no 21
ΔH 5.6 L of C2H2 = ?
2C2H2(g)+ 5O2(g) 4CO2(g) + 2H2O(l)
ΔH= -2,372.4 kJ
ΔH = -2,372.4 kJ : 2
= - 1,186.2 kJ
mole C2H2 = 5.6 L /22,4 L = 0.25mole
ΔH = 0.25 x (-1,186.2 )
= 295.55 kJ (E)

43. • Calculating ΔH Value from Simple Experiment
22. Data equation:
NaOH (aq) + HCl (aq) NaCl (aq) + H2O(l)
ΔH = -56 kJ/mole
A 100 cm3 of 0.25 M HCl solution is mixed with 200
cm3 of 0.15 M NaOH.The enthalpy change for the
equation is….
a. -0.56 kJ
b. -3.06 kJ
c. -1.68 kJ
d. -1.40 kJ
e. -2.80 kJ

44. Solution for no 22
• Acid + base → Salt + water + heat (neutralisation reaction).
HCl=100ml x 0.25 = 0.025 mole
NaOH=200ml x 0.15 = 0.03 mole
• HCl + NaOH → NaCl + H2O
The balanced equation above states that 1 mole of HCl will react
with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of H2O
There are 0.025 moles of HCl and 0.03 moles of NaOH.
The coefficients indicate the mole ratios
Since the balanced equation states that 1 mole of HCl will react
with 1 mole of NaOH, the mole ratio = 1 : 1.
So 0.025 moles of HCl will react with 0.025 moles of NaOH, to
produce 0.025 moles of NaCl and 0.025 moles of H2O, and (0.03 –
0.025) 0.05 moles of NaOH will not react.

45. NaOH (aq) + HCl (aq) NaCl (aq) + H2O(l)
ΔH = -56 kJ/mole
0.025 NaOH (aq) + 0.025 HCl (aq) 0.025 NaCl
(aq) + 0.025 H2O(l)
1mole : 0.25 = 40
ΔH = -56 : 40
ΔH = -1.4 kJ

46. 23. When 100 mL of 1M NaOH is mixed with 100
mL of 1M HCl in a container,the temperature
rises from 29C to37.5C (assume that specific
heat is the same of that of pure water, 4.2
J/C). The enthalpy value is….
a. -45.9 kJ
b. -54.6 kJ
c. +54.6 kJ
d. -71.4 kJ
e. -82.3 kJ

47. Solution for no 23
• number of moles of each NaOH and HCl that
reacted is = 100 x 1 = 100mmole = 1/10 mole
• Δt = 37.5-29=8.5 C
• m = 100 + 100 = 200 mL = 200 gram
• Q = m x c x Δt
= 200 x 4,2 x 8.5 = 7140 kJ = 7.14 kJ
• If each NaOH and HCl reacted one mole,so
• Q = 7.14 x 10 = 71.4 kJ/mole ΔH =-71.4
kJ/mole

48. • Calculating H Value According to Hess’ Law
and Standard Enthalpy
24. Look at this following diagram.
a. ΔH 1 + ΔH 2 = ΔH 3 + ΔH 4
b. ΔH 2 + ΔH 3 = ΔH 1 + ΔH 4
c. ΔH 1 + ΔH 3 = ΔH 2 + ΔH 4
d. ΔH 1 = ΔH 2 + ΔH 3 + ΔH 4
e. ΔH 4 = ΔH 1 + ΔH 2 + ΔH 3

49. ΔH 1
A+B C+D
ΔH 2
ΔH 4
P+Q R+S
ΔH 3

50. • See direction of the arrows !
• Find the equation that starts and ends with
the same reaction
• Here starts with A+B and ends with R+S
• ΔH 1 + ΔH 4 = R+S
• ΔH 2 + ΔH 3 = R+S
• So ΔH 2 + ΔH 3 = ΔH 1 + ΔH 4 (B)

51. 25. According to the diagram,the value of ΔH 3
is…
a. ΔH 1 + ΔH 2 - ΔH 4
b. ΔH 2 + ΔH 4 - ΔH 1
c. ΔH 1 - ΔH 2 + ΔH 4
d. ΔH 1 - ΔH 2 - ΔH 4
e. ΔH 1 + ΔH 4 - ΔH 2

52. Mg
ΔH 2
MgO
ΔH 1 ΔH 3
Mg(OH)2
ΔH 4

53. See direction of the arrows !
Because ΔH 1 = ΔH 2 + ΔH 3 + ΔH 4
So ΔH 3 = ΔH 1 - ΔH 2 - ΔH 4 (D)

54. No. 26
• Data:
MO₂ + CO MO + CO₂ ∆H = -20 kJ
M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJ
3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ
• The value of ∆H for the reactions 2MO₂ + CO
M₂O₃ +CO₂ is ….
A. -40 D. -18
B. -28 E. +18
C. -26

55. solution
• 2MO₂ + CO M₂O₃ +CO₂
x2 MO₂ + CO MO + CO₂ ∆H = -20 kJ
Reserved, x⅔ M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJ
Reserved, x⅓ 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ
2MO₂ + 2CO 2MO + 2CO₂ ∆H = -40 kJ
2MO + ⅔ CO₂ ⅔ M₃O₄ + ⅔ CO ∆H = -4 kJ
⅔M₃O₄ + ⅓CO₂ M₂O₃ + ⅓CO ∆H = +4 kJ
• 2MO₂ + CO M₂O₃ +CO₂ ∆H = -40

56. NO. 27
N₂ + 3H₂ (g) 2NH₃ (g) ∆H = -92 kJ
The heat required to decompose 5.1 g
ammonia (Mr = 17 g/mole) is ….
A. 4.6 kJ
B. 9.2 kJ
C. 13.8 kJ
D. 18.8 kJ
E. 27.6 kJ

57. solution
• A reaction with negative amount of ∆H: will
be positive when the reaction is reserved
N₂ + 3H₂ (g) 2NH₃ (g) ∆Hᵣ = kJ
-92
2NH₃ (g) N₂ + 3H₂ (g) ∆ Hᵣ= 92 kJ
• The initial enthalpy change apply when 2mole
of ammonia is reacted. In 5.1 g NH₃ there is
• mole NH₃ = mass of NH₃ = 5.1 = 0,3 mole
Mr of NH₃ 17
Therefor, the enthalpy change is
∆H: = 92:2= 46 x 0.3 = 13.8

58. No. 28
• Data:
• ∆Hс C₂H₅OH (g) is -728 kJ/mole
• ∆Hf CO₂ (g) is -394 kJ/mole
• ∆Hf H₂O (l) is -286 kJ/mole
• The enthalpy change for the formation of
C₂H₅OH (g) in kJ/mole is ….
A. -238 D. -952
B. -478 E. -714
C. -918

59. solution
• The chemical equation of C₂H₅OH formation is
• C₂H₅OH + 3O₂ 2CO₂ + 3H₂O
• Applying the formula of calculating the enthalpy
change of formation from the enthalpy change
combustion, we have:
• ∆H = ∑∆Hf:product ‒∑∆Hf:reactan
= (2xCO₂ + 3xH₂O) ─ (C₂H₅OH + 3O₂)
= (2x-394 + 3x-286) ‒ (-728 + 3x0)
= -788 + -858 + 728
= -918

60. No. 29
• H₂ (g) + Br₂ (g) 2HBr (g) ∆H = -72 kJ
• The heat required to decompose 11.2 L HBr
(STP) to H₂ and Br₂ is ….
A. 9 kJ
B. 18 kJ
C. 36 kJ
D. 72 kJ
E. 144 kJ

61. solution
• A reaction with negative amount of ∆H: will be
positive when the reaction is reserved
H₂ (g) + Br₂ (g) 2HBr (g) ∆Hᵣ = kJ
-72
2HBr (g) H₂ (g) +Br₂ (g) ∆Hᵣ = 72 kJ
• The initial enthalpy change apply when 2mole
HBr is reacted. In 11.2 L HBr is…
• mole HBr = Volume of HBr = 11.2 = 0.5 mole
22.4 22.4
Therefor, the enthalpy change is
∆H: = 72 : 2 = 36 x 0.5 = 18 kJ

62. No. 30
• The enthalpy for formation of H₂O and NH₃ are a
kcal/mole and b kcal/mole respectively. The
enthalpy of combustion of 4NH₃ + 7O₂
4NO₂ + 6H₂O is c. The enthalpy change of
formation of NO₂ is ….
A. a – 3b + ½ c
B. c + b – a
C. c + b -1½ a
D. 1½ a – b ─ ½ c
E. 1½ a + b + c

63. solution
• ∆Hf H₂O = a kcal
• ∆Hf NH₃ = b kcal
• ∆Hf NO₂ = ?
• 4NH₃ + 7O₂ 4NO₂ + 6H₂O ∆Hc = c
• ∆H = ∑∆Hf:product ‒∑∆Hf:reactan
c = (4x + 6a) ─ (4b + 0 )
c = 4x + 6a -4b
4x = -c +6a -4b
x = -c + 3a –b
4 2

64. Number 31
When the amount of magnesium is burned to
produce 1 gram of MgO, 14,4kJ of heat is
released. The heat of formation of MgO is ....
kJ
a. 14,4
b. -14,4
c. 288
d. -288
e. 576

65. Mg + ½ O2 → MgO
m MgO = 1 gram Mr MgO = 40 gram
So 1 gram of MgO = mole
If ∆H for mole MgO is 14,4 kJ
then the ∆H0f is :
40 x 14,4 = 576kJ (E)

66. Number 32
The heat of the combustion of Fe2O is a kcal/mole
The heat of the formation of Fe2O is b kcal/mole
The heat of the formation of Fe2O3 is...kcal
a. (a+b)
b. (a-b)
c. (a+2b)
d. (2a+b)
e. (2a+2b)

67. Fe2O + O2 → Fe2O3 a kcal/mole
Fe2 + ½ O2 → Fe2O b kcal/mole
_______________________________ +
Fe2 + O2 → Fe2O3 a+b kcal (A)

68. Number 33
The enthalpy of formation of H2O(l) and H2O(g) is -
286kJ/mole and -242kJ/mole respectively. When
we vaporize 4,5 grams of water, the process
will....kJ of heat
a. Release 11
b. Absorb 11
c. Release 44
d. Absorb 132
e. Absorb 198

69. H2 + O2 → H2O(l) ∆H = -286kJ/mole
H2 + O2 → H2O(g) ∆H = -242kJ/mole
H2O(l) → H2 + O2 ∆H = -286kJ/mole
H2 + O2 → H2O(g) ∆H = -242kJ/mole
_______________________________________+
H2O(l) → H2O(g) ∆H = -44kJ/mole
4,5 grams of H2O = 0,25 mole
0,25 x 44 = 11
From liquid to gas needs heat so it absorbs 11 kJ of
heat (B)

70. Number 34
C + O2 → CO2 ∆H= -395,22kJ
2H2 + O2 → 2H2O ∆H= -573,72kJ
If the enthalpy change for the formation of C2H4 is
+27,30kJ, the enthalpy of combustion of C2H4
is...kJ
a. -654,78
b. -663,18
c. -709,38
d. -1.336,86
e. -1.391,46

71. C + O2 → CO2 ∆H= -395,22kJ
2H2 + O2 → 2H2O ∆H= -573,72kJ
2C + 2H2 → C2H4 ∆H= 27,30kJ
C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=...?
2C + 2O2 → 2CO2 ∆H= -395,22 x 2 kJ
2H2 + O2 → 2H2O ∆H= -573,72kJ
C2H4 → 2C + 2H2 ∆H= -27,30kJ
______________________________________+
C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=-1.391,46kJ (E)

72. Number 35
The enthalpy change for the formation of CO2 is -
395kJ. The enthalpy change for the combustion of
glucose is -28,20kJ. The enthalpy change for the
combustion of ethanol is -1.368 kJ. What is ∆H
calue for the following equation?
C6H12O6 → 2C2H5OH + 2CO2
a. -874kJ
b. +706kJ
c. -1.057 kJ
d. +1.057 kJ
e. -4.583 kJ

73. C + O2 → CO2 ∆H= -395kJ
C6H12O6 + 6O2 → 6CO2 + 6H2O ∆H= -28,20kJ
C2H5OH+ 3O2 → 2CO2 + 3H2O ∆H= -1.368kJ
C6H12O6 → 2C2H5OH + 2CO2 ∆H= ...?
C6H12O6 + 6O2 → 6CO2 + 6H2O ∆H= -28,20kJ
4CO2 + 6H2O → 2C2H5OH+ 6O2 ∆H= 1.368kJ
____________________________________________+
C6H12O6 → 2C2H5OH + 2CO2 ∆H= 2707,8kJ

74. 36.
H2 + ½O2 H2O ΔH=-242kJ
The bond energy of H-H and O=O is 436 kj/mole and 500
kj/mole respectively.
The mean bond entalpy of H-O is…… kj
(a)121 (d) 464
(b)222 (e) 589
(c)363

75. No 36
H2 + O₂ H20 ΔH=-242kJ
ΔH = energy separation – energy formation
-242=(H-H + (O=O) ) - 2 (O-H)
-242=(436 +250) – 2(O-H)
O-H =(786+ 242)/2
= (1028)/2 = 464

76. 37. Bond energy data
C=C = 611 kj/mole C-Cl = 339 kj/mole
C-H = 414 kj/mole C-C = 347 kJ/mole
H-Cl = 431 kj /mole
Using the data, the entalpy change for the following reaction is ..kJ
C2H4 + HCl C2H5Cl
(a) +46 (d) -92
(b) -46 (e) -138
(c) -58

77. No 37
C=C = 611 kj/mole C-Cl = 339 kj/mole
C-H = 414 kj/mole C-C = 347 kJ/mole
H-Cl = 431 kj /mole
• C2H4 + HCl C2H5Cl
ΔH = energy separation – energy formation
=(2(C=C) + 4(C-H) +H-Cl ) -((C-C) + 5(C-H) +C-Cl)
=(611) + 4(414) +431) – ((347) + 5(414) + 339)
=(611+1656+431)-(347+2070+339)
=(2698)-(2756)
= -58 kJ

78. 38 .the enthalpy of formation of NO is +90kJ/mole. If the bond
energy of N=N is 418 kJ/mole and O=O is 498 kJ/mole.
Energy required to breakdown 2 mole NO
bond is…..
(a)413 (d) 826
(b)765 (e) 911
(c)720

79. solution NO 38
• ΔHf NO = +90kJ/mole ;N=N 418 kJ/mole ; O=O 498kJ/mole
Answer:
NO ½ N2 + ½ O2 ΔHd = -90 kJ/mole
½N2 N ΔH = 209 kJ/mole
½O2 O ΔH = 249 kJ/mole ₊
NO N +O 368 kJ/mole
ΔH atomisasi = 368 kJ/mole
N-O = 368 kJ/mole
2 N-O = 2 x 368 = 736 kJ

80. 39. Bond energy data
H-Cl = 431 kj/mole
H-H = 436 kJ/mole
Cl-Cl = 243 kj /mole
According to the data, the heat required to decompose 73
grams of HCl (Mr = 36,5 kJ/mole) to its element is
(a) 336 kJ (d) 139,5 kJ
(b) 69,75 kJ(e) -183 kJ
(c) 100 kJ

81. solution NO 39
H-Cl = 431 kj/mole
H-H = 436 kJ/mole
Cl-Cl = 243 kj /mole
HCl H2 + Cl2
ΔH = energy separation – energy formation
= HCl –(1/2 H2+1/2 Cl2)
= 431 -(218+121,5)
= 431 – 339.5 = -91,5 kJ/mole
mole : 73 / 36.5 = 2 mole
ΔH : 2 (-91,5)= -183 kJ

82. 40. Bond energy data
H-F = 565 kj/mole
H-H = 436 kJ/mole
F-F = 158 kj /mole
energy released for the formation of 5 grams HF (Mr=20kJ/mole) from its
elements is….
(a) -268 kJ (d) -67 kJ
(b) -201 kJ (e) -33,5 kJ
(c) -124

83. PEMBAHASAN NO 40
H-F = 565 kj/mole
H-H = 436 kJ/mole
F-F = 158 kj /mole
H2 + F2 HF
ΔH = P – R
= (1/2 H2 + ½ F2) – H-F
= (218+79) - 565
= 297 -565
=- 268 kJ (A)

84. THANK YOU