Question

For the vaporization of benzene, \(\Delta H_{\text {vap }}=30.7 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S_{\text {vap }}=87.0 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol}) .\) Calculate \(\Delta S_{\text {surr }}\) and \(\Delta S_{\text {total }}\) at: (a) \(70^{\circ} \mathrm{C}\) (b) \(80^{\circ} \mathrm{C}\) (c) \(90^{\circ} \mathrm{C}\) Does benzene boil at \(70^{\circ} \mathrm{C}\) and 1 atm pressure? Calculate the normal boiling point of benzene.

Short answer

Benzene does not boil at 70°C. Its normal boiling point is approximately 79.72°C.

Step-by-step solution

Convert Temperature to Kelvin

First, convert the given Celsius temperatures to Kelvin using the formula:\[T(K) = T(°C) + 273.15\]For each temperature:- At \(70^{\circ} \mathrm{C}\), \(T = 70 + 273.15 = 343.15 \; \mathrm{K}\)- At \(80^{\circ} \mathrm{C}\), \(T = 80 + 273.15 = 353.15 \; \mathrm{K}\)- At \(90^{\circ} \mathrm{C}\), \(T = 90 + 273.15 = 363.15 \; \mathrm{K}\)

Determine \(\Delta S_{\text{surr}}\) for Each Temperature

The change in entropy of the surroundings can be calculated using the equation:\[\Delta S_{\text{surr}} = - \frac{\Delta H_{\text{vap}}}{T}\]Since \(\Delta H_{\text{vap}} = 30.7 \; \mathrm{kJ/mol} = 30,700 \; \mathrm{J/mol}\):- At \(343.15 \; \mathrm{K}\),\[\Delta S_{\text{surr}} = - \frac{30,700}{343.15} = -89.47 \; \mathrm{J/(K \cdot mol)}\]- At \(353.15 \; \mathrm{K}\),\[\Delta S_{\text{surr}} = - \frac{30,700}{353.15} = -86.93 \; \mathrm{J/(K \cdot mol)}\]- At \(363.15 \; \mathrm{K}\),\[\Delta S_{\text{surr}} = - \frac{30,700}{363.15} = -84.55 \; \mathrm{J/(K \cdot mol)}\]

Calculate \(\Delta S_{\text{total}}\) for Each Temperature

The total change in entropy \(\Delta S_{\text{total}}\) for the vaporization process is given by:\[\Delta S_{\text{total}} = \Delta S_{\text{vap}} + \Delta S_{\text{surr}}\]Given \(\Delta S_{\text{vap}} = 87.0 \; \mathrm{J/(K \cdot mol)}\):- At \(343.15 \; \mathrm{K}\),\[\Delta S_{\text{total}} = 87.0 - 89.47 = -2.47 \; \mathrm{J/(K \cdot mol)}\]- At \(353.15 \; \mathrm{K}\),\[\Delta S_{\text{total}} = 87.0 - 86.93 = 0.07 \; \mathrm{J/(K \cdot mol)}\]- At \(363.15 \; \mathrm{K}\),\[\Delta S_{\text{total}} = 87.0 - 84.55 = 2.45 \; \mathrm{J/(K \cdot mol)}\]

Analyze \(\Delta S_{\text{total}}\) to Determine Boiling at 70°C

Benzene boils when \(\Delta S_{\text{total}}\) is zero or positive, which indicates that the process is spontaneous.At \(70^{\circ} \mathrm{C}\) (343.15 K), \(\Delta S_{\text{total}} = -2.47 \; \mathrm{J/(K \cdot mol)}\), which is negative. Therefore, benzene does not boil at this temperature.

Calculate Normal Boiling Point of Benzene

The normal boiling point occurs when \(\Delta S_{\text{total}} = 0\), meaning:\[\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_{\text{boil}}}\]Substitute \(\Delta H_{\text{vap}} = 30,700 \; \mathrm{J/mol}\) and \(\Delta S_{\text{vap}} = 87.0 \; \mathrm{J/(K \cdot mol)}\):\[87.0 = \frac{30,700}{T_{\text{boil}}}\]Solving for \(T_{\text{boil}}\), we get:\[T_{\text{boil}} = \frac{30,700}{87.0} \approx 352.87 \; \mathrm{K}\]Convert to Celsius by subtracting 273.15:\[T_{\text{boil}} = 352.87 - 273.15 \approx 79.72^{\circ} \mathrm{C}\]

Key concepts

Vaporization of Benzene

Vaporization is the process where a liquid turns into a gas. For benzene, the energy needed to vaporize, called the enthalpy of vaporization \(\Delta H_{\text{vap}}\), is about \(30.7 \text{ kJ/mol}\).
During vaporization, molecules overcome intermolecular forces when they enter the gas phase.
The change in entropy for the system, known as the entropy of vaporization \(\Delta S_{\text{vap}}\), quantifies this disorder increase.
This is \(87.0 \text{ J/(K mol)}\) for benzene and reflects how the gas phase is more disordered than the liquid phase.
When studying vaporization, two main energy concepts come into play:

• Enthalpy Change \(\Delta H\): the heat absorbed at constant pressure.
• Entropy Change \(\Delta S\): the degree of disorder or randomness increase.

These changes are essential for determining conditions like boiling points or spontaneity of phase changes.

Entropy of Surroundings

The entropy of the surroundings \(\Delta S_{\text{surr}}\) is an important factor to consider in thermodynamic processes. It is calculated by the formula: \(\Delta S_{\text{surr}} = - \frac{\Delta H_{\text{vap}}}{T}\).
This equation shows how energy absorbed by the system affects the surroundings.
Key points about \(\Delta S_{\text{surr}}\):

• Negative Sign: The negative sign indicates that the system's absorption of heat from the surroundings decreases their entropy.
• Direct Relationship: The change depends inversely on temperature; lower temperatures amplify the entropy decrease in the surroundings.
• Unit Interaction: Typical units here are J/(K mol), harmonizing energy input (J) and temperature (K).

Understanding \(\Delta S_{\text{surr}}\) helps predict spontaneous processes, as it works alongside the system's entropy change to influence the total entropy change of a given process.

Normal Boiling Point Calculation

The normal boiling point of benzene is the temperature at which its vapor pressure equals atmospheric pressure resulting in boiling.
It is derived where the total entropy change \(\Delta S_{\text{total}}\) is zero or positive, as it means equilibrium or a spontaneous boiling process.
To find this point, solve \(\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_{\text{boil}}}\):

• Calculate: \(T_{\text{boil}} = \frac{30,700}{87.0} \approx 352.87 \text{ K}\).
• Convert to Celsius: \(T_{\text{boil}} - 273.15 \approx 79.72^{\circ} \text{C}\).

This means under 1 atm, benzene's normal boiling point is nearly \(79.72^{\circ} \text{C}\).
When \(\Delta S_{\text{total}}\) turns zero, the system reaches an energy balance conducive for continuous phase change.
This calculation is crucial for understanding how benzene transitions from liquid to vapor at standard conditions.