Use the following information to answer questions 2 through 8: Diethyl ether, CH
ID: 1011485 • Letter: U
Question
Use the following information to answer questions 2 through 8:
Diethyl ether, CH3CH2OCH2CH3 ,
specific heat solid:
1.89 J/g K
heat of fusion:
7.19 kJ/mol
normal melting point:
–116.3 °C
Molar mass:
74.1 g/mole
specific heat liquid:
2.32 J/g K
heat of vaporization:
27.2 kJ/mol
normal boiling point:
34.6 °C
specific heat gas:
1.61 J/g K
3. What is the vapor pressure at 25°C?
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Question 41 pts
Use the following information to answer questions 2 through 8:
Diethyl ether, CH3CH2OCH2CH3 ,
specific heat solid:
1.89 J/g K
heat of fusion:
7.19 kJ/mol
normal melting point:
–116.3 °C
Molar mass:
74.1 g/mole
specific heat liquid:
2.32 J/g K
heat of vaporization:
27.2 kJ/mol
normal boiling point:
34.6 °C
specific heat gas:
1.61 J/g K
4. What would be the vapor pressure of diethyl ether at –30.0°C?
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Question 51 pts
Use the following information to answer questions 2 through 8:
Diethyl ether, CH3CH2OCH2CH3 ,
5. What is the strongest intermolecular force present in diethyl ether?
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Question 61 pts
6. Which one of the following compounds has a higher normal boiling point than diethyl ether?
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Question 71 pts
Use the following information to answer questions 2 through 8:
Diethyl ether, CH3CH2OCH2CH3 ,
specific heat solid:
1.89 J/g K
heat of fusion:
7.19 kJ/mol
normal melting point:
–116.3 °C
Molar mass:
74.1 g/mole
specific heat liquid:
2.32 J/g K
heat of vaporization:
27.2 kJ/mol
normal boiling point:
34.6 °C
specific heat gas:
1.61 J/g K
7. A 30.0 g sample of diethyl ether at 25°C is cooled to –130°C. What is the enthalpy change?
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Question 81 pts
8. Is this process exothermic or endothermic?
specific heat solid:
1.89 J/g K
heat of fusion:
7.19 kJ/mol
normal melting point:
–116.3 °C
Molar mass:
74.1 g/mole
specific heat liquid:
2.32 J/g K
heat of vaporization:
27.2 kJ/mol
normal boiling point:
34.6 °C
specific heat gas:
1.61 J/g K
Explanation / Answer
3. Normal boiling point = 34.6 C , VaPOUR pressure = 1atm
so that, at 25C = VaPOUR pressure = <1atm
use calsius -cleyperon equation
ln(p2/p1) = DHvAP/R[1/T1 - 1/T2]
p2 = VaPOUR pressure at 34.6 C = 1 atm
P1 = VaPOUR pressure at 25 c = ?
T1 = 298 k , T2 = 307.75 k
DHvap = heat of vaporization = 27.2 kj/mol
ln(1/x) = ((27.2*10^3)/8.314)((1/298)-(1/307.75))
x = VaPOUR pressure at 25 c = 0.706 atm
answer: VaPOUR pressure = <1atm
4.
use calsius -cleyperon equation
ln(p2/p1) = DHvAP/R[1/T1 - 1/T2]
p2 = VaPOUR pressure at 34.6 C = 1 atm
P1 = VaPOUR pressure at -30 c = ?
T1 = 243.15 k , T2 = 307.75 k
DHvap = heat of vaporization = 27.2 kj/mol
ln(1/x) = ((27.2*10^3)/8.314)((1/243.15)-(1/307.75))
x = VaPOUR pressure at -30 c = 0.06 atm
= 45.6 torr
answer: 44.9 trr
5. dipole-dipole forces
6. hydrogen peroxide, H2O2
7. q = mliquidsDT+n*DHfus + msolidsDT
= 30*2.32*(25--116.3)+(30/74.1)*7.19*10^3+30*1.89*(-116.3--130)
= 13.52 kj
DH = -13.52 kj
8. it is exothermic process.