Question

Thenormal boiling point of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is \(64.7{ }^{\circ} \mathrm{C}\), and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=\) \(71.8 \mathrm{~kJ} / \mathrm{mol} .\) (a) When \(\mathrm{CH}_{3} \mathrm{OH}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00\) mol of \(\mathrm{CH}_{3} \mathrm{OH}(l)\) is vaporized at \(64.7{ }^{\circ} \mathrm{C}\).

Short answer

(a) The entropy of methanol increases during boiling as it undergoes a phase transition from a closely packed liquid state to a more dispersed gaseous state. (b) To calculate the change in entropy, first convert the temperature to Kelvin: \(T= 64.7 + 273.15 = 337.85\, \mathrm{K}\). Then, use the equation \(\Delta S = \frac{\Delta H}{T}\) with the given enthalpy of vaporization, ∆H = 71.8 kJ/mol: \(\Delta S = \frac{71.8 \times 1000\, \mathrm{J/mol}}{337.85\, \mathrm{K}} = 212.70\, \mathrm{J/(mol\cdot K)}\). So, the change in entropy when 1.00 mol of methanol is vaporized at 64.7°C is 212.70 J/(mol·K).

Step-by-step solution

Identify the direction of entropy change

When a substance undergoes a phase transition, such as boiling, the process is accompanied by an increase in entropy. This is because boiling leads to an increase in the disorder of the system, as the molecules move from a more closely packed liquid state to a more dispersed gaseous state. So, during the boiling process of methanol, its entropy increases. (b) Calculating the value of ∆S when 1.00 mol of methanol is vaporized at 64.7°C

Convert given temperature to Kelvin

To work with the equation mentioned earlier, we need to express the given temperature in Kelvin instead of Celsius. We can convert the temperature using the following formula: \(T(K) = T(°C) + 273.15\) The given temperature is 64.7°C, so in Kelvin, it would be: \(T= 64.7 + 273.15 = 337.85\, \mathrm{K}\)

Use the equation to calculate ∆S

Now that we have the temperature in Kelvin, we can calculate the change in entropy using the given enthalpy of vaporization, ∆H = 71.8 kJ/mol. The equation for entropy change is: \(\Delta S = \frac{\Delta H}{T}\) Plugging in the values: \(\Delta S = \frac{71.8\, \mathrm{kJ/mol}}{337.85\, \mathrm{K}}\) To get the answer in \(\mathrm{J/(mol\cdot K)}\), we need to multiply the result by 1000. \(\Delta S = \frac{71.8 \times 1000\, \mathrm{J/mol}}{337.85\, \mathrm{K}} = 212.70\, \mathrm{J/(mol\cdot K)}\) Therefore, the change in entropy when 1.00 mol of methanol is vaporized at 64.7°C is 212.70 J/(mol·K).

Key concepts

Enthalpy of Vaporization

The enthalpy of vaporization is the amount of energy required to transform a given quantity of a substance from a liquid into a gas. This transformation is essential in many natural and industrial processes, such as when water boils to become steam.
In the case of methanol, the enthalpy of vaporization is specified as 71.8 kJ/mol.
This value signifies that when 1 mol of methanol is heated to its boiling point, it requires 71.8 kJ of energy to overcome the intermolecular forces keeping the molecules together as a liquid.
The enthalpy of vaporization can vary significantly between different substances due to differences in intermolecular forces. More robust forces result in higher enthalpy values because more energy is needed to break these interactions.
Understanding enthalpy of vaporization provides insights into the energy demands of phase transitions and the nature of liquid-gas interactions in a substance.

Phase Transition

Phase transition refers to the process where a substance changes from one state of matter to another. Common phase transitions include melting, freezing, condensation, and vaporization.
These transitions involve energy changes, either requiring energy (endothermic) or releasing energy (exothermic).
Boiling, a type of vaporization, is an endothermic process where liquid molecules gain enough energy to escape into the gaseous state.
This involves an increase in the system's internal energy as well as an increase in its entropy, given that the gas phase is more disordered than the liquid phase.
One important concept during phase transitions is that temperature remains constant while the change occurs. For example, methanol boils at a fixed temperature of 64.7°C, even though energy continues to be supplied to the system.
Understanding these processes is fundamental in thermodynamics as they illustrate the principles of energy conversion and molecular behavior during thermal changes.

Entropy Change

Entropy is a measure of the disorder or randomness in a system. When methanol or any other substance undergoes boiling, entropy increases because gas molecules spread out and move more freely than in a liquid state.
This transition results in a more disordered state and thus a higher entropy.
We can quantify this change in entropy using the formula:

• \( \Delta S = \frac{\Delta H}{T} \)

where \( \Delta H \) is the enthalpy of vaporization, and \( T \) is the temperature in Kelvin.
For methanol, with an enthalpy of vaporization of 71.8 kJ/mol and a boiling temperature of 337.85 K, the change in entropy \( \Delta S \) is calculated to be 212.70 J/(mol·K).
Such calculations help understand the energetic and molecular dynamics involved in phase changes, as they provide a numeric value to the increase in disorder associated with the transformation from liquid to vapor.