Ag ca In so sb Te I 12260 Fr Ra ItAr 105 12601 Dr. Haouari Srping 2015 Exam 1 Po

Question

Ag ca In so sb Te I 12260 Fr Ra ItAr 105 12601 Dr. Haouari Srping 2015 Exam 1 Po At Ru February 11, 2015 3. Identify the type or types of intermol ecular forces present in each substance and the select the substance in each pair that has the higher boiling point: (a) propane C3Hs or n-butane C4H10 (b) diethyl ether CHaCH20CH2CH3 or 1-butanol CHaCH2CH2CH20H (c) sulfur dioxide Soz or sulfur trioxide Sob (d) phosgene clzco or formaldehyde H2co 4. Ethanol (caHsoH melts at-114 °c and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 KU/mol, and its enthalpy of vaporization is 38.56 K/mol.The specific heat of a. How much heat is required to convert 42.0g of ethanol at 35°C to the vapor phase at 78 b How much heat is required to convert the same amount of ethanol at -155°Cto the vapor phase at 78 °C?

Explanation / Answer

#3

a) Propane & n-butane

London dispersion forces for both

In case of non polar molecules, the predominant forces are London dispersion forces.The strength increases as the size of chain increases, so n-butane has stronger london dispersion forces than propane

stronger the force, higher is the boiling point

So n-butane has higher boiling point

b) diethyl ether & 1-butanol

CH3CH2-O-CH2CH3       & CH3CH2CH2CH2-O-H

CH3CH2-O-CH2CH3 - London dispersion forces

CH3CH2CH2CH2-O-H - Hydrogen bond

In case of 1-butanol, there is direct attachment between O & H due to which butanol can form intermolecular Hydrogen bond which is absent in diethylether.

Hydrogen bonding is the strongest intermolecular force of attraction, therefore n-butanol has higher boiling point

c)SO2 & SO3

SO2 - dipole dipole

SO3 - London dispersion force

If we look at the the geometry of SO2, it has a bent structure which makes the molecule polar

whereas SO3 has trigonal planar geometry which makes the molecule non polar

In general, we know that dipole is a stronger force than LDF, but this is not applicable in this case because of the large size of SO3 molecule. The strength of LDF increases as the size of molecule increases & in case of SO3 since it has planar structure, there is large surface area for contact, which increases the strength of LDF

Therefore SO3 has higher boiling point

d)

Phosgene & Formaldehyde

Both of them have dipole -dipole forces but again Phosgene has higher molecular mass than formaldehyde, so requires a greater force to pass in vapor phase

So Phosgene has higher boiling point

#4

a) Q = m x C x delta T

at 35 degree C, ethanol is in liquid state, so we use C, specific heat of liquid ethanol

=42.0 g x 2.3 J/g-K x 43 K

4153.8 J

b)

Here we are converting solid ethanol to vapor phase

so there are total 3 stages we will encounter in this conversion.

Q1 is the heat required to melt ice from -155 C to its melting point, -114 C

at this temp, ethanol will be solid, so we will use C of solidethanol

Q1 = m x C x delta T = 42.0 x 0.97 x 41 = 1670.34 J

Q2, is the heat required to melt all the ethnol at its melting point

it will be equal to enthaply of fusion x mol of ethanol

Q2 = 5.02 kJ/mol x 42 / 46.07 = 4.577 kJ

Q3, is the amount of heat to raise the temp from -114 to boiling point 78

Q3 = 42 g x 2.3 J/g-K x 192 K = 18547.2 J

Qtotal = Q1 + Q2 + Q3 = 1670.34 J + 4.577 kJ + 18547.2 J = 24.795 kJ = 25 kJ

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