On absolute values entropy of chemical compounds

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

If values S are given in the table.

At standard pressure, if warmth of formation are known

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;

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;

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

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

A) -214,78 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

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

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

A) –421,81

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)

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

A) –43,22 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

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

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

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

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

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

180. Indicate condition parameters:

A) U,H

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

A) Formations

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