Pembahasan Soal2 termokimia

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)

QUADRA

1.According to the Law Conservation of Energy, energy…..a. Can not be created (correction)b. Can be transferredc. Cannot be convertedd. Cannot form enthalpye. Can be distroyed

NO 1

Solution for no 1Law 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"

• To answer number 2 and 3,look at the reaction below.

• C2H4+3O2 2CO2 + 2H2OThe enthalpy change of reaction is 1560 kJ/mole.2. It is an….reaction,with the value of H=…..a. balance; -1560 kJ/moleb. Exothermic; -1560 kJ/molec. Exothermic; +1560 kJ/moled. Endothermic; -1560 kJ/molee. Endothermic; +1560 kJ/mole

NO 2

Solution for no 2Combustion=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

3. The enthalpy change measured for the reaction is 634 kJ. How much oxygen needed for this reaction?

a. 8.91 Lb. 11.28 Lc. 22.4 Ld. 26.88 Le. 32.4 L

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

Solution no 3

Look at the enthalpy cycle below4. 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

NO 4

Ca(s) + H2O(l)

CaO(s) + H2O(l)

CaO(s) + H2O(l)

Ca(s) + H2O(l)

ΔH°1

ΔH°

ΔH°3

ΔH°4

Route 1

Route 2

Route 3

Route 4

CaO(s) + H2O(l) Ca(s) + H2O(l)

• Ca(s) + H2O(l) CaO(s) + H2O(l)

• Ca(s) + H2O(l) Ca(s) + H2O(l)CaO(s) + H2O(l) CaO(s) + H2O(l)

ΔH°1

-ΔH°3

-ΔH°4

ΔH°2

Solution for no 4 With Hess’Law

C6H12O6 + 6O2 6CO2 + 6H2OΔH°= -2820 kJ/moleC2H5OH + 3O2 2CO2 + 3H2OΔH°= -1380 kJ/moleUsing data above,the enthalpy change for fermentation

with the equation of5.C6H12O6 2C2H5OH + 6CO2 is…..a. -4200 kJ/moleb. + 1440 kJ/molec. -1440 kJ/moled. +60 kJ/molee. -60 kJ/mole

Solution for no 5

C6H12O6 + 6O2 6CO2 + 6H2O4CO2 + 6H2O 2C2H5OH + 6O2 (reversed & x2)

C6H12O6 2C2H5OH + 2CO2 (corrected)

ΔH°= ΔH°1+ ΔH°2

ΔH°=-2820+ (+1380*2) = -2820 + 2760 = +60 (D)

No. 6• Given that:

MO₂ + CO MO + CO₂ ∆H⁰ = -20 kJ/moleM₃O₄ + CO 3MO + CO₂ ∆H⁰ = +6 kJ/mole3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ/mole

• The value of ∆H for the reactions 2MO₂ + COM₂O₃ +CO₂ is ….A. -40 kJ D. -18 kJB. -28 kJ E. 40 kJC. -26 kJ

solution

• 2MO₂ + CO M₂O₃ +CO₂x2 MO₂ + CO MO + CO₂ ∆H = -20 kJReversed, x⅔ M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJReversed, x⅓ 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ

2MO₂ + 2CO 2MO + 2CO₂ ∆H = -40 kJ2MO + ⅔ CO₂ ⅔ M₃O₄ + ⅔ CO ∆H = -4 kJ⅔M₃O₄ + ⅓CO₂ M₂O₃ + ⅓CO ∆H = +4 kJ

• 2MO₂ + CO M₂O₃ +CO₂ ∆H = -40

No. 7• Given that

N₂ + 3H₂ 2NH₃ (g) ∆H = -92 kJThe heat needed to dissociate 5.1 g ammonia (Mr = 17 g/mole) is ….A. 9.2 kJ/moleB. 13.8 kJ/moleC. 18.8 kJ/moleD. 27.6 kJ/moleE. 65.6 kJ/mole

solution• A reaction with negative amount of ∆H⁰ will be

positive when the reaction is reversedN₂ + 3H₂ (g) 2NH₃ (g) ∆Hᵣ = -92 kJ2NH₃ (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∆H⁰ = 92:2= 46 x 0.3 = 13.8

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 + bB. a – bC. 2a + bD. a - 2b E. 2b - a

solution• Fe₂O + O₂ Fe₂O₃ ∆H = a

∆Hf=b ∆Hf=?

Hreaction = ∑∆Hf right - ∑∆Hf lefta = ∆Hf Fe₂O₃ - ba + b = ∆Hf Fe₂O₃

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 kJB. absorb 11 kJC. Absorb 44 kJD. Release 44 kJE. Absorb 1332 kJ

solution• ∆H⁰ H₂O (l) is -286 kJ/mole• ∆H⁰ 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

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/moleB. 36 kJ/moleC. 72 kJ/moleD. 144 kJ/moleE. 288 kJ/mole

solution• A reaction with negative amount of ∆H⁰ will be

positive when the reaction is reversedH₂ + Br₂ 2HBr ∆Hᵣ = -72 kJ2HBr 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 mole22.4 22.4

Therefor, the enthalpy change is∆H⁰ = 72 : 2 = 36 x 0.5 = 18 kJ/mole

Number 11

The bond enthalpy in C2H5Cl is...( C-H = 414 ; C-C = 347 ; C-Cl = 339 )a.2840 kJ/moleb. 2756 kJ/molec. 2409 kJ/moled.2070 kJ/molee. 2015kJ/mole

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

(B) H H

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/moleb. 124 kJ/molec. 201 kJ/moled. 268 kJ/mole3. 312 kJ/mole

½ H2 + ½ Cl2 → HCl ∆H=183kJ/mole

(releases)(½ H2 + ½ Cl2)-HCl = -183½ (436 + Cl2) – 431 = -183½ (436 + Cl2) = 248436 +Cl2 = 496Cl2 = 2Cl = 60kJ/mole

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. 67b. 124c. 201d. 268e. 312

mHF=5gr MR HF= 20 HF=0,25 mole½ H2 + ½ F2 → HF218 + 79 – 565 = 268 (for 1 mole)For 0,25 mole = 268 x 0,25 = 67 (A)

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 kJb. 10416 kJc. 2083,2 kJd. 20832 kJe. 5208 kJ

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

Number 15Given that : C + O2 → CO2 ∆H= -395,22kJ/mole2H2 + O2 → 2H2O ∆H= -573,72kJ/moleIf the enthalpy change of C2H4 is +27,30kJ, then the enthalpy

change of combustion of C2H4 is...a. -654,78kJ/moleb. -663,18kJ/molec. -709,38kJ/moled. -887,45kJ/molee. -1.336,86kJ/mole

C + O2 → CO2 ∆H= -395,22kJ2H2 + O2 → 2H2O ∆H= -573,72kJ2C + 2H2 → C2H4 ∆H= 27,30kJC2H4 + 3 O2 → 2CO2 + 2H2O ∆H=...?2C + 2O2 → 2CO2 ∆H= -395,22 x 2 kJ2H2 + O2 → 2H2O ∆H= -573,72kJC2H4 → 2C + 2H2 ∆H= -27,30kJ______________________________________+C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=-1.391,46kJ (E)

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

No 1716. 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

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

No 1919. 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

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

No 20m = p . v

= 1000. 200 = 200.000

Q = m . C . Δt = 200.000 4,18 60 = 50.160.000 J = 50.160 kJ

answer = D

FOTOCOPY SHEET

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 heatb. Requiring 593.1 kJ of heatc. The enthalpy of the system rises 296.55kJd. Absorbing 296.55 kJ of heate. Releasing 296.55 kJ of heat

ΔH 5.6 L of C2H2 = ?2C2H2(g)+ 5O2(g) 4CO2(g) + 2H2O(l) ΔH= -2,372.4 kJΔH°= -2,372.4 kJ : 2

= - 1,186.2 kJmole C2H2 = 5.6 L /22,4 L = 0.25moleΔH = 0.25 x (-1,186.2 )

= 295.55 kJ (E)

Solution for no 21

• Calculating ΔH° Value from Simple Experiment22. Data equation: NaOH (aq) + HCl (aq) NaCl (aq) + H2O(l) ΔH°= -56 kJ/moleA 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 kJb. -3.06 kJc. -1.68 kJd. -1.40 kJe. -2.80 kJ

• Acid + base → Salt + water + heat (neutralisation reaction).HCl=100ml x 0.25 = 0.025 moleNaOH=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 H2OThere are 0.025 moles of HCl and 0.03 moles of NaOH.

The coefficients indicate the mole ratiosSince 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.

Solution for no 22

NaOH (aq) + HCl (aq) NaCl (aq) + H2O(l) ΔH°= -56 kJ/mole0.025 NaOH (aq) + 0.025 HCl (aq) 0.025 NaCl

(aq) + 0.025 H2O(l) 1mole : 0.25 = 40ΔH°= -56 : 40ΔH°= -1.4 kJ

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 kJb. -54.6 kJc. +54.6 kJd. -71.4 kJe. -82.3 kJ

• 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

Solution for no 23

• 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

C+D

P+Q

A+B

R+S

ΔH°1

ΔH°3

ΔH°4 ΔH°2

• 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)

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

ΔH°2

ΔH°1 ΔH°3

ΔH°4

Mg

MgO

Mg(OH)2

See direction of the arrows !Because ΔH°1 = ΔH°2 + ΔH°3 + ΔH°4

So ΔH°3 = ΔH°1 - ΔH°2 - ΔH°4 (D)

No. 26• Data:

MO₂ + CO MO + CO₂ ∆H = -20 kJM₃O₄ + CO 3MO + CO₂ ∆H = +6 kJ3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ

• The value of ∆H for the reactions 2MO₂ + COM₂O₃ +CO₂ is ….A. -40 D. -18 B. -28 E. +18C. -26

solution

• 2MO₂ + CO M₂O₃ +CO₂x2 MO₂ + CO MO + CO₂ ∆H = -20 kJReserved, x⅔ M₃O₄ + CO 3MO + CO₂ ∆H = +6 kJReserved, x⅓ 3M₂O₃ + CO 2M₃O₄ + CO₂ ∆H = -12 kJ

2MO₂ + 2CO 2MO + 2CO₂ ∆H = -40 kJ2MO + ⅔ CO₂ ⅔ M₃O₄ + ⅔ CO ∆H = -4 kJ⅔M₃O₄ + ⅓CO₂ M₂O₃ + ⅓CO ∆H = +4 kJ

• 2MO₂ + CO M₂O₃ +CO₂ ∆H = -40

NO. 27

N₂ + 3H₂ (g) 2NH₃ (g) ∆H = -92 kJThe heat required to decompose 5.1 g ammonia (Mr = 17 g/mole) is ….A. 4.6 kJB. 9.2 kJC. 13.8 kJD. 18.8 kJE. 27.6 kJ

solution• A reaction with negative amount of ∆H⁰ will be

positive when the reaction is reservedN₂ + 3H₂ (g) 2NH₃ (g) ∆Hᵣ = -92 kJ2NH₃ (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∆H⁰ = 92:2= 46 x 0.3 = 13.8

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. -952B. -478 E. -714C. -918

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 = ∑∆Hf⁰product ‒∑∆Hf⁰reactan

= (2xCO₂ + 3xH₂O) ─ (C₂H₅OH + 3O₂)= (2x-394 + 3x-286) ‒ (-728 + 3x0)= -788 + -858 + 728= -918

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 kJB. 18 kJC. 36 kJD. 72 kJE. 144 kJ

solution• A reaction with negative amount of ∆H⁰ will be

positive when the reaction is reservedH₂ (g) + Br₂ (g) 2HBr (g) ∆Hᵣ = -72 kJ2HBr (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 mole22.4 22.4

Therefor, the enthalpy change is∆H⁰ = 72 : 2 = 36 x 0.5 = 18 kJ

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 + 7O₂4NO₂ + 6H₂O is c. The enthalpy change of formation of NO₂ is ….

A. a – 3b + ½ cB. c + b – aC. c + b -1½ aD. 1½ a – b ─ ½ cE. 1½ a + b + c

solution• ∆Hf H₂O = a kcal• ∆Hf NH₃ = b kcal• ∆Hf NO₂ = ?• 4NH₃ + 7O₂ 4NO₂ + 6H₂O ∆Hc = c• ∆H = ∑∆Hf⁰product ‒∑∆Hf⁰reactan

c = (4x + 6a) ─ (4b + 0 )c = 4x + 6a -4b4x = -c +6a -4b

x = -c + 3a –b 4 2

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. 14,4b. -14,4c. 288d. -288e. 576

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

Number 32

The heat of the combustion of Fe2O is a kcal/moleThe heat of the formation of Fe2O is b kcal/moleThe heat of the formation of Fe2O3 is...kcala. (a+b)b. (a-b)c. (a+2b)d. (2a+b)e. (2a+2b)

Fe2O + O2 → Fe2O3 a kcal/moleFe2 + ½ O2 → Fe2O b kcal/mole_______________________________ +Fe2 + O2 → Fe2O3 a+b kcal (A)

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 11b. Absorb 11c. Release 44d. Absorb 132e. Absorb 198

H2 + O2 → H2O(l) ∆H = -286kJ/moleH2 + O2 → H2O(g) ∆H = -242kJ/mole H2O(l) → H2 + O2 ∆H = -286kJ/moleH2 + O2 → H2O(g) ∆H = -242kJ/mole_______________________________________+H2O(l) → H2O(g) ∆H = -44kJ/mole4,5 grams of H2O = 0,25 mole0,25 x 44 = 11From liquid to gas needs heat so it absorbs 11 kJ of heat (B)

Number 34

C + O2 → CO2 ∆H= -395,22kJ2H2 + O2 → 2H2O ∆H= -573,72kJIf the enthalpy change for the formation of C2H4 is

+27,30kJ, the enthalpy of combustion of C2H4 is...kJa. -654,78b. -663,18c. -709,38d. -1.336,86e. -1.391,46

C + O2 → CO2 ∆H= -395,22kJ2H2 + O2 → 2H2O ∆H= -573,72kJ2C + 2H2 → C2H4 ∆H= 27,30kJC2H4 + 3 O2 → 2CO2 + 2H2O ∆H=...?2C + 2O2 → 2CO2 ∆H= -395,22 x 2 kJ2H2 + O2 → 2H2O ∆H= -573,72kJC2H4 → 2C + 2H2 ∆H= -27,30kJ______________________________________+C2H4 + 3 O2 → 2CO2 + 2H2O ∆H=-1.391,46kJ (E)

Number 35The 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. -874kJb. +706kJc. -1.057 kJd. +1.057 kJe. -4.583 kJ

C + O2 → CO2 ∆H= -395kJC6H12O6 + 6O2 → 6CO2 + 6H2O ∆H= -28,20kJC2H5OH+ 3O2 → 2CO2 + 3H2O ∆H= -1.368kJC6H12O6 → 2C2H5OH + 2CO2 ∆H= ...?

C6H12O6 + 6O2 → 6CO2 + 6H2O ∆H= -28,20kJ4CO2 + 6H2O → 2C2H5OH+ 6O2 ∆H= 1.368kJ____________________________________________+C6H12O6 → 2C2H5OH + 2CO2 ∆H= 2707,8kJ

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

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

37. Bond energy dataC=C = 611 kj/mole C-Cl = 339 kj/moleC-H = 414 kj/mole C-C = 347 kJ/moleH-Cl = 431 kj /mole

Using the data, the entalpy change for the following reaction is ..kJC2H4 + HCl C2H5Cl

(a) +46 (d) -92(b) -46 (e) -138(c) -58

No 37

• C2H4 + HCl C2H5ClΔH = 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

C=C = 611 kj/mole C-Cl = 339 kj/moleC-H = 414 kj/mole C-C = 347 kJ/moleH-Cl = 431 kj /mole

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

solution NO 38

• ΔHf NO = +90kJ/mole ;N=N 418 kJ/mole ; O=O 498kJ/moleAnswer:

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/mole2 N-O = 2 x 368 = 736 kJ

39. Bond energy dataH-Cl = 431 kj/moleH-H = 436 kJ/moleCl-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

solution NO 39

H-Cl = 431 kj/moleH-H = 436 kJ/moleCl-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

40. Bond energy dataH-F = 565 kj/moleH-H = 436 kJ/moleF-F = 158 kj /moleenergy 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

PEMBAHASAN NO 40

H-F = 565 kj/moleH-H = 436 kJ/moleF-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)

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