Question

The molar heats of fusion and sublimation of lead are 4.77 and 182.8 kj/mol, respectively. estimate the molar heat of vaporization of molten lead.

Answer 1

Answer:The molar heat of vaporization of molten lead is 178.03 kJ/mol.

Explanation:

The molar heats of fusion of lead

$Pb(l)\rightarrow Pb(s),\Delta H_f=-4.77 kJ/mol$...(1)

The molar heats of sublimation of lead

$Pb(s)\rightarrow Pb(g),\Delta H_s=182.8 kJ/mol$...(2)

Adding (1) and (2)

$Pb(l)\rightarrow Pb(g),\Delta H_v=?$

$\Delta H_v=\Delta H_f+\Delta H_s=(-4.77) kJ/mol+182.8 kJ/mol=178.03 kJ/mol$

The molar heat of vaporization of molten lead is 178.03 kJ/mol.

Answer 2

The molar heat of vaporization of molten lead is$\boxed{{\text{178}}{\text{.03 kJ/mol}}}$ .

Further Explanation:

Phase change:

The phase change is defined as the change from one state to another state without altering its chemical composition. It is also known as a phase transition, state change or physical change.

Sublimation:

When substance gets converted from its solid to gas state without going through liquid state, sublimation is said to take place. Some substances like naphthalene and dry ice undergo sublimation.

Vaporization:

This phase transition involves conversion of substance from its liquid state to vapor or gaseous state.

Fusion:

When substance undergoes a change from its liquid to solid state, it is said to have undergone fusion.Molar heat of fusion is the amount of heat required to melt one mole of substance in its solid state.

The chemical equation for fusion of Pb is as follows:

${\text{Pb}}\left( s \right) \to {\text{Pb}}\left( l \right)$  

The chemical equation for sublimation of Pb is as follows:

${\text{Pb}}\left( s \right) \to {\text{Pb}}\left( g \right)$  

The chemical equation for vaporization of Pb is as follows:

${\text{Pb}}\left( l \right) \to {\text{Pb}}\left( g \right)$  

Since vaporization involves conversion of liquid lead to gaseous lead, molar heat of vaporization can be calculated by the sum of negative of molar heat of fusion and molar heat of sublimation as given below. This is because heat of fusion is heat required to convert solid into liquid, but in vaporization we are going from liquid to vapor state.

$\Delta {H_v} = - \Delta {H_f} + \Delta {H_s}$

                                                      …… (1)

Here,

$\Delta {H_v}$ is molar heat of vaporization.

$\Delta {H_f}$ is molar heat of fusion.

$\Delta {H_s}$ is molar heat of sublimation.

Substitute 4.77 kJ/mol for $\Delta {H_f}$ and for $\Delta {H_s}$ in equation (1).

$\begin{aligned} \Delta {H_v} &= - \left( {4.77{\text{ kJ/mol}}} \right) + 182.8{\text{ kJ/mol}} \\ &={\text{178}}{\text{.03 kJ/mol}} \\ \end{aligned}$  

Learn more:

1. Identify the phase change in which crystal lattice is formed: brainly.com/question/1503216
2. The main purpose of conducting experiments: brainly.com/question/5096428

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Phase transition

Keywords: fusion, sublimation, vaporization, liquid, solid vapor, Pb, 178.03 kJ/mol, 182.8 kJ/mol, 4.77 kJ/mol, molar heat of vaporization, molten lead.