Using standard warmth of combustion for substances to calculate thermal effect of reaction

C₂H₅OH_(ж)+СH₃СООН_(ж) → CH₃COOС₂H_5(ж)+H₂О_(ж) при 25⁰С, if warmth of combustion of substances are following -1366,9 кJ/mol, -873,8 кJ/mol, -2254,2 кJ/mol:

A) 13,5 kJ;

B) 15,4 kJ;

C) 17,8 kJ;

D) 7,9 kJ;

E) 11,5 kJ

 

168. Calculate thermal effect of reaction at 292K and 1, 0133⋅10⁵ Па for reaction: Fe₂O_3(тв)+3CO_(г)=2Fe_(тв)+3CO_2(г),

-821,32 кJ/mol, -110,5 кJ/mol, 0, -393,51 кJ/mol:

A) -27,71 кJ;

B) 13,5 кJ;

C) 141,82 кJ;

D) -354,8 кJ;

E) 0;

 

169. To define thermal effect of reaction: СН º СН + СО + Н₂О (ж) → СН₂ = СНСООН (ж)

At standard pressure, if warmth of formation are known

Substance	[kJ/mol ]
СН º СН	226,75
СО	–110,50
Н2О (ж)	–285,84
СН2 = СНСООН (ж)	–384,37

 

A) -214,78 кJ

B) 107,36 кJ

C) 1007,36 кJ

D) –623,09 кJ

E) –154,2 кJ

 

170. Calculate thermal effect of reaction (kJ) at Т = 298К: CuCl₂ + Zn = ZnCl₂ + Cu

values are resulted in the table:

Substance
ZnCl2	–478,2
CuCl2	–262,3

A) -215,9

B) 740,5

C) –903,6

D) 904,2

E) 370,2

 

171. Calculate entropy change under standard conditions for reaction: 2H₂S + SO₂ = 2 Н₂О (ж) + 3S,

If values S are given in the table.

Substance	[J/mol ×К]
Н2О	69,95
S	32,55
SO2	248,1
H2S	205,64

A) –421,81

B) –102,50

C) 513,24

D) 350,72

E) 231,11

 

Using standard warmth of combustion of substances to calculate thermal effect of reaction at 298К

С₂Н₅ОН(ж) + СН₃СООН(ж) → СН₃СООС₂Н₅(ж) + Н₂О (ж)

Substance	DНсг. [kJ/mol ]
С2Н5ОН(ж)	–1366,9
СН3СООН(ж)	–873,8
СН3СООС2Н5(ж)	–2254,2
Н2О (ж)

 

A) 13,5 кJ

B) -2254,2 кJ

C) –240,2 kJ

D) –3,21 kJ

E) –1,34 kJ

 

173. Calculate standard change of entropy at 298К for reaction: Cd (тв) + 2AgCl (тв) = 2Ag (тв) + CdCl₂ (тв)

On absolute values entropy of chemical compounds

Substance	S [J/mol ×К]
Cd	51,76
AgCl	96,07
Ag	42,69
CdCl2	115,3

A) –43,22 J/К

B) 10,16 J/К

C) 0

D) 325,82 J/К

E) –10,16 J/К

 

174. Calculate thermal effect of chemical reaction (kJ) at Т=298К, Р=1,013×10⁵ Па

Cd (тв) + 2AgCl (тв) = 2Ag (тв) + CdCl₂ (тв), if warmth of formation of substances are known

Substance	298 [КJ/mol ]
Сd
Ag
AgCl	–126,8
CdCl2	–389,0

A) –135,4

B) –263,2

C) –515,8

D) 443,2

E) 344,8

 

175. Calculate standard change of energy of Gibbs (kJ) for reaction: Cd (тв) + 2AgCl (тв) = 2Ag (тв) + CdCl₂ (тв),

if are known DН₂₉₈ = –135,4 kJ; DS₂₉₈ = –76,84 J/К.

A) –112,5

B) 1393,55

C) –141,7

D) 24,61

E) 41,42

 

176. Calculate thermal effect of reaction (kJ) at Т =298К: CuCl₂ + Zn = ZnCl₂ + Cu

values are resulted in the table:

 

Substance
ZnCl2	–478,2
CuCl2	–262,3

A) -215,9

B) 740,5

C) –903,6

D) 904,2

E) 370,2

 

177. What quantity of warmth will be allocated at dissolution of 200 g. copper oxide in hydrochloric acid at the thermo chemical reaction: CuO+2HCl = CuCl₂+H₂O - 63, 6 kJ.?

A. 160

B. 140

C. 120

D. 180

E. 110

178. To calculate quantity of warmth this will be absorbed at receiving of 12 g. of oxygen at the thermo chemical reaction: 2KNO₃= 2KNO₂+O₂+ 255 kJ.

A. 95,6

B. 80,4

C. 75,6

D. 69,6

E. 59,6

 

179. What quantity of warmth will be allocated at combustion of 4, 48 l. (n.c.) methane at the thermo chemical reaction:

CH₄+2O₂ = CO₂+2H₂O-878 kJ?

A. 175,6

B. 150,3

C. 140,6

D. 130,6

E. 165,3

180. Indicate condition parameters:

A) U,H

B) U,Q

C) Q,A

D) P,W

E) A,S

 

181. Thermal effect at formation of one mole of substance is called as warmth …

A) Formations

B) Combustion

C) Thermal effect

D) Dissolution

E) Crystallisation