4.1 Which compound in the table is a gas at room temperature?
4.2 Define the term homologous series - NSC Physical Sciences - Question 4 - 2016 - Paper 2
Question 4
4.1 Which compound in the table is a gas at room temperature?
4.2 Define the term homologous series.
4.3 A type of van der Waals force exists between molecules of ... show full transcript
Worked Solution & Example Answer:4.1 Which compound in the table is a gas at room temperature?
4.2 Define the term homologous series - NSC Physical Sciences - Question 4 - 2016 - Paper 2
Step 1
Which compound in the table is a gas at room temperature?
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The compound which is a gas at room temperature is Butane (A), as it has a boiling point of -1°C.
Step 2
Define the term homologous series.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
A homologous series is a group of organic compounds that have the same functional group, where one member differs from the next by a -CH₂- group.
Step 3
A type of van der Waals force exists between molecules of compound A and also between molecules of compounds B, C, and E. Write down the NAME of the Van der Waals force.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The type of van der Waals force present is the London dispersion force.
Step 4
Refer to the TYPE and STRENGTH of intermolecular forces to explain the difference in the boiling points between:
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Step 5
Compounds A and B
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Compounds A (Butane) and B (Butan-2-one) show different types of intermolecular forces. Compound A has only London dispersion forces, whereas compound B contains dipole-dipole interactions in addition to these London forces, which makes the boiling point of compound B (79.5°C) higher than that of compound A (-1°C).
Step 6
Compounds C and D.
97%
121 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Compound C (Butanol) and D (Butanoic acid) have different boiling points due to the presence of hydrogen bonding in compound D, while compound C experiences dipole-dipole and London forces. This hydrogen bonding in compound D leads to a significantly higher boiling point of 163.5°C compared to compound C's 117.4°C.
Step 7
Which compound has a HIGHER vapor pressure?
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Compound D (Butanoic acid) has a lower vapor pressure compared to compound E (Pentanolic acid) because pentanoic acid has weaker intermolecular forces due to its larger molecular structure and presence of only London forces.
Step 8
Refer to MOLECULAR STRUCTURE, TYPE and STRENGTH of intermolecular forces, to explain the answer to QUESTION 4.5.1.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Molecular structure plays a key role in the vapor pressure of a compound. Compound D has strong hydrogen bonding due to the presence of a carboxylic group which gives it a lower vapor pressure compared to compound E that has weaker intermolecular forces allowing it to vaporize more readily, hence having a higher vapor pressure.
