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Serhud [2]
3 years ago
9

Identify each enthalpy change by name and classify each change as exothermic or endothermic.

Chemistry
2 answers:
PtichkaEL [24]3 years ago
7 0

Answer:

1. C)

2. B)

3. D)

4. A)

Explanation:

Enthalpy change is the heat of the reaction measured at constant pressure.

i.e. q_{p}=\Delta H

Chemical reactions are accompanied either by the absorption of heat (endothermic) or by the release of heat (exothermic)

1) 1 mol NaCl (s) ↔ 1 mol NaCl(l)

Ans: C) Molar heat of fusion; endothermic

Molar heat or enthalpy of fusion is the amount of energy absorbed when a solid melts into a liquid. It is an endothermic process since heat is absorbed.

2. C4H10 O(l) ↔ C4H10O (g)

Ans: B) Molar heat of vaporization; endothermic

Molar heat or enthalpy of vaporization is the amount of energy absorbed when a liquid evaporates into a gas. It is an endothermic process since heat is absorbed.

3. H2O(g) ↔ H2O (l)

Ans: D Molar heat of condensation; exothermic

Molar heat or enthalpy of condensation is the amount of energy released when a gas condenses into a liquid. It is an exothermic process since heat is released.

4. 1 mol NaCl(s) + 3.88 kJ/mol ↔ 1 mol NaCl(aq)

Ans: A) Molar heat of solution; endothermic

Heat of solution is the amount of heat absorbed when 1 mole of a substance is dissolved in water.

Len [333]3 years ago
3 0
We classify them as follows:

1. 1 mol NaCl (s) à 1 mol NaCl(l) 
C Molar heat of fusion; endothermic

2. C4H10 O(l) à C4H10O (g) 
B Molar heat of vaporization; endothermic

3. H2O(g) à H2O (l) 
D Molar heat of condensation; exothermic

4. 1 mol NaCl(s) + 3.88 kJ/molà 1 mol NaCl(aq) 
A Molar heat of solution; endothermic
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We can use this equation for boiling point elevation:
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and i is ( vant Hoff factor, the number of particles or ions per mole-clue.
and K(b) is boiling point increase constant for the solution ( and for water it is equal 0.52 C° Kg/mol)
We can assume i (vant Hoff factor ) = 1 as the sucrose is nonelectrolyte (not readily ionize).
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By substitute at:
ΔTb = i Kb M
∴  = 1 * 0.52 * 3.60 = 1.8432 C°
and when Tb = Tb° + ΔTb
∴  Tb = 100 + 1.8432 = 101.8432 C°



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