Posted: July 16th, 2022

Which is generally stronger, intermolecular or intramolecular forces? Intramolecular forces include covalent bonds, ionic bonding, and metallic bonding.

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Question 1:
a. Which is generally stronger, intermolecular or intramolecular forces? Intramolecular forces include covalent bonds, ionic bonding, and metallic bonding.
b. Which of these kinds of interactions are broken when a liquid is converted to a gas?
Question 2:
Which type of intermolecular forces accounts for each of these differences:
a. CH3OH boils at 65 °C and CH3SH boils at 6 °C.
b. Xe is a liquid at atmospheric pressure and 120 K and Ar is a gas under the same conditions.
c. Acetone boils at 56 °C and 2-methylpropane boils at -12 °C. (Lewis structures below)
Question 3:
Select which molecule will have the stronger intermolecular dispersion forces?
a. Br2 or O2
b. CH3CH2CH2SH or CH3CH2CH2OH
c. SiH4 vs GeH4
Question 4:
Using your knowledge of intermolecular forces, rationalize the difference in boiling points for each pair:
a. HF (20 °C) vs HCl (-85 °C)
b. CHCl3 (61 °C) and CHBr3 (150 °C)
c. Br2 (59 °C) and ICl (97 °C)
Question 5:
Based on the type or types of intermolecular forces, predict the substances in each pair that has the higher boiling point:
a. Propane (C3H8) or n-butane (C4H10)
b. Diethyl ether (CH3CH2OCH2CH3) or 1-butanol (CH3CH2CH2CH2OH)
c. Sulfur dioxide (SO2) or sulfur trioxide (SO3)
Question 6:
Carbon tetrachloride CCl4 and chloroform CHCl3, are common organic liquid. Carbon tetrachloride’s normal boiling point is 77 °C; chloroforms normal boiling point is 61 °C. Which statement is the best explanation of this data?
a. Chloroform can hydrogen bond, but carbon tetrachloride cannot.
b. Carbon tetrachloride has a larger dipole moment than chloroform.
c. Carbon tetrachloride’s large size and mass results in a more polarizable molecule and results in stronger dispersion forces.
Question 7:
Based on their composition and structure list, CH2Cl2, CH3CH2CH3, and CH3CH2OH in order of:
a. Increasing intermolecular forces
b. Increasing viscosity
c. Increasing surface tension
Question 8:
The fluorocarbon compound C2Cl3F3 has a normal boiling point 47.6 °C. The specific heats of and C2Cl3F3
(l) C2Cl3F3 (g) are 0.91 J/g·K and 0.67 J/g·K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol. Calculate the heat required to convert 35.0 g of C2Cl3F3 from a liquid at 10.00 °C to a gas at 105.00 °C.

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