Use the following information to answer questions 2 through 8: Diethyl ether, CH

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?

Flag this Question

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?

Flag this Question

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?

Flag this Question

Question 61 pts

6. Which one of the following compounds has a higher normal boiling point than diethyl ether?

Flag this Question

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?

Flag this Question

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.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.