Q12.66P - Chemistry: Molecular Nature Of Matter And Change

Question

Use Figure 12.1, p. 439 to answer the following:

(a) Does it take more heat to melt 12.0 g of ${\mathbf{CH}}_4$ or 12.0g of $\mathbf{Hg}$ ?

(b) Does it take more heat to vaporize 12.0g of ${\mathbf{CH}}_4$  or 12.0g of  ?

(c) What is the principal intermolecular force in each sample?

Step-by-Step Solution

Answer

1. It takes more heat to melt 12.0g of Hg than 12.0g of  ${\mathbf{CH}}_4$.
2. It takes more heat to vaporize 12.0g of ${\mathbf{CH}}_4$  than 12.0g of Hg.
3. The principal intermolecular force is London dispersion forces in Ar, ${\mathbf{CH}}_4$  and benzene; weak dipole-dipole interactions in Diethyl ether; H-bonding in ethanol and water; and metallic bonding in Hg.

Heat of vaporization and fusion

The heat supplied to vaporize one mole of the liquid to its gaseous vapors is known as the heat of vaporization. The heat of fusion is the energy required to fuse one mole of solid from the solid state to the liquid state.

Calculation of fusion heat required

12.0gm of  ${\mathbf{CH}}_4$ (Mol. Wt. 16.0 gm) means 0.75 mole of the compound and 12.0gm of Hg (Mol. Wt. is 200.6 gm) means 0.06 mole of the compound.

Hence, it would take 0.705 kJ/mol energy to melt  ${\mathbf{CH}}_4$ and 1.404 kJ/mol energy to melt Hg.

Thus, 12.0 g of Hg will need more heat to melt than  ${\mathbf{CH}}_4$.

Calculation of vaporization heat required

It would take 6.675 kJ/mol energy to vaporize ${\mathbf{CH}}_4$  and 3.54 kJ/mol energy to vaporize Hg.

Thus, 12.0 g of  ${\mathbf{CH}}_4$ will need more heat to melt than Hg.

Intermolecular forces

The principal intermolecular force is London dispersion forces in Ar, ${\mathbf{CH}}_4$  and benzene; weak dipole-dipole interactions in Diethyl ether; H-bonding in ethanol and water; and metallic bonding in Hg.