QUESTION 4 NOV 2014 The boiling points of four compounds of hydrogen, represented by the letters P, Q, R and S, are given in the table below.

Compound P ($\text{CH}_4$) is a non-polar molecule and only experiences weak London dispersion forces. Compound Q ($\text{NH}_3$) is a polar molecule and can form strong hydrogen bonds, in addition to dipole-dipole forces and London dispersion forces.

Hydrogen bonds are significantly stronger intermolecular forces than London dispersion forces.

More energy is required to overcome the strong hydrogen bonds in $\text{NH}_3$ compared to the weak London dispersion forces in $\text{CH}_4$. Therefore, $\text{NH}_3$ has a higher boiling point ($-33 \text{ °C}$) than $\text{CH}_4$ ($-164 \text{ °C}$).

4.2.2 Compound S ($\text{SiH}_4$)

Both compound P ($\text{CH}_4$) and compound S ($\text{SiH}_4$) are non-polar molecules with tetrahedral shapes, meaning they only experience London dispersion forces.

The molecular mass of $\text{CH}_4$ is approximately $16 \text{ g/mol}$, while the molecular mass of $\text{SiH}_4$ is approximately $32 \text{ g/mol}$. $\text{SiH}_4$ has a higher molecular mass and more electrons.

As the molecular mass and number of electrons increase, the electron cloud becomes more polarizable, leading to stronger London dispersion forces between molecules.

More energy is required to overcome the stronger London dispersion forces in $\text{SiH}_4$ compared to $\text{CH}_4$. Therefore, $\text{SiH}_4$ has a higher boiling point ($-112 \text{ °C}$) than $\text{CH}_4$ ($-164 \text{ °C}$).

4.3 Explain why the boiling points of compounds Q ($\text{NH}_3$) and R ($\text{H}_2\text{O}$) differ by referring to ELECTRONEGATIVITY and DEGREE OF POLARITY:

Compound R ($\text{H}_2\text{O}$) has a boiling point of $100 \text{ °C}$, while compound Q ($\text{NH}_3$) has a boiling point of $-33 \text{ °C}$. Both molecules are polar and can form hydrogen bonds.

Oxygen is more electronegative than nitrogen. This makes the O-H bonds in $\text{H}_2\text{O}$ more polar than the N-H bonds in $\text{NH}_3$.

Due to the greater electronegativity of oxygen and the presence of two lone pairs, $\text{H}_2\text{O}$ can form more and stronger hydrogen bonds per molecule (up to four) compared to $\text{NH}_3$ (which typically forms fewer and weaker hydrogen bonds).

The stronger and more extensive network of hydrogen bonds in $\text{H}_2\text{O}$ requires significantly more energy to overcome during boiling, resulting in a much higher boiling point than $\text{NH}_3$.

4.4 Write down the letter from the table that represents the following:

4.4.1 ONE polar compound:

4.4.2 ONE non-polar compound:

QUESTION 3 NOV 2015

3.1 The boiling point is the temperature at which the vapor pressure of a liquid equals the external atmospheric pressure, causing the liquid to change into a gas.

3.2 Calculate the molecular mass of compound D ($\text{C}_9\text{H}_{20}$):

Molecular mass = $(\text{number of C atoms} \times \text{atomic mass of C}) + (\text{number of H atoms} \times \text{atomic mass of H})$

Atomic mass of C $\approx 12.01 \text{ g/mol}$
Atomic mass of H $\approx 1.01 \text{ g/mol}$
Molecular mass of $\text{C}_9\text{H}_{20} = (9 \times 12.01) + (20 \times 1.01)$

Molecular mass $= 108.09 + 20.20 = 128.29 \text{ g/mol}$

3.3 In what phase is compound B at 25 °C?
Compound B ($\text{C}_2\text{H}_6$) has a boiling point of $-89 \text{ °C}$. Since $25 \text{ °C}$ is above its boiling point, compound B will be in the gaseous phase.

3.4 Name the type of intermolecular force present in compound A ($\text{CH}_4$):
Compound A ($\text{CH}_4$) is a non-polar molecule.

3.5 Explain why the boiling point increases from compound A to E:

Compounds A to E are alkanes with increasing chain length and molecular mass ($\text{CH}_4$, $\text{C}_2\text{H}_6$, $\text{C}_7\text{H}_{16}$, $\text{C}_9\text{H}_{20}$, $\text{C}_{20}\text{H}_{42}$).

As the molecular mass increases, the number of electrons in the molecule increases. This leads to a greater polarizability of the electron cloud, resulting in stronger London dispersion forces between molecules.

More energy is required to overcome these stronger London dispersion forces, causing the boiling point to increase from compound A to E.

3.6 How does the vapour pressure of compound B compare to the vapour pressure of compound C? Write down only HIGHER THAN, LOWER THAN or EQUAL TO.
Compound B has a boiling point of $-89 \text{ °C}$. Compound C has a boiling point of $36 \text{ °C}$. A lower boiling point indicates weaker intermolecular forces, which means molecules can escape into the gas phase more easily, leading to a higher vapor pressure. Therefore, compound B will have a higher vapor pressure than compound C.