Question

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mathrm{~kJ} /\) mol. (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00 \mathrm{~mol}\) of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8{ }^{\circ} \mathrm{C}\).

Short answer

(a) The entropy increases during boiling of Br2 at its normal boiling point because ΔS = ΔH / T, and both ΔH and T have positive values. (b) To calculate the value of ΔS when 1.00 mol of Br2 is vaporized at 58.8 °C, first convert the temperature to Kelvin and then use the equation ΔS = ΔH / T: T(K) = 58.8 + 273.15 = 331.95 K ΔS = (29.6 kJ/mol) / (331.95 K) = (29.6 × 10^3 J/mol) / (331.95 K) ≈ 89.16 J/(mol·K) The entropy change when 1 mol of Br2 is vaporized at 58.8 °C is 89.16 J/(mol·K).

Step-by-step solution

(Step 1: Determine if the entropy increases or decreases during boiling)

(At the normal boiling point, we know the change in Gibbs free energy (ΔG) is zero because the process occurs at equilibrium. Using the equation ΔG = ΔH - TΔS, and solving for ΔS, we get: ΔS = ΔH / T Since ΔH (enthalpy of vaporization) has a positive value and T (temperature) is also positive because it is in Kelvin, we can conclude that ΔS has a positive value. Thus, the entropy increases during boiling.)

(Step 2: Calculate the value of ΔS for vaporization of 1 mole of Br2 at 58.8 °C)

(First, we need to convert the temperature from Celsius to Kelvin: T(K) = 58.8 + 273.15 = 331.95 K Next, we use the equation ΔS = ΔH / T, where ΔH is the enthalpy of vaporization (29.6 kJ/mol) and T(K) is the temperature in Kelvin: ΔS = (29.6 kJ/mol) / (331.95 K) Now, we convert ΔH units from kJ/mol to J/mol to obtain ΔS in J/(mol·K): ΔS = (29.6 × 10^3 J/mol) / (331.95 K) Finally, we calculate the value of ΔS: ΔS = 89.16 J/(mol·K) So, the entropy change when 1 mol of Br2 is vaporized at 58.8 °C is 89.16 J/(mol·K).)

Key concepts

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. At this specific temperature, a liquid turns into vapor. It is a crucial physical property that indicates the conditions under which a liquid becomes gaseous.

For bromine (Br extsubscript{2}), the normal boiling point is given as 58.8°C. Under this condition, the liquid bromine transitions to vapor, which means the molecules have gained enough energy to overcome intermolecular forces.

During boiling, one can observe an increase in entropy—a measure of disorder or randomness in a system. Liquids have less molecular movement compared to gases. Thus, as the liquid bromine boils to become gaseous, the randomness, and thereby the entropy, increases. In conclusion, boiling points like the one for bromine provide critical information about the conditions under which entropy changes during phase transitions.

Enthalpy of Vaporization

The enthalpy of vaporization is the amount of energy required to change one mole of a substance from liquid to gas at a constant temperature and pressure. This energy is necessary to break the intermolecular forces holding the molecules in the liquid phase.

For bromine, the molar enthalpy of vaporization is 29.6 kJ/mol. This means that to convert one mole of liquid bromine to its gaseous form, 29.6 kJ of energy must be absorbed from the surroundings. This quantity is typically positive, reflecting the endothermic nature of the vaporization process.

Understanding the enthalpy of vaporization helps explain why the entropy of a system changes. When bromine vaporizes, energy is absorbed, and molecules move more freely, increasing their disorder and thus their entropy.

Gibbs Free Energy

Gibbs free energy ( abla G) is a term used in thermodynamics to describe the amount of energy available for doing work during a chemical process at constant temperature and pressure. It serves as a key indicator for spontaneity. If abla G is negative, the process is spontaneous; if positive, it is non-spontaneous.

At the boiling point where a substance transitions from liquid to gas, abla G is zero because the system is at equilibrium. This condition means that the rate of vaporization equals the rate of condensation, and the system has no overall change in free energy. Using the equation abla G = abla H - T abla S, one can understand how enthalpy and entropy relate to this phase equilibrium.

• abla H represents the enthalpy change, indicating the heat absorbed or released.
• T abla S represents the temperature (T) multiplied by the entropy change ( abla S).

In the case of bromine boiling at its normal boiling point, this equation helps calculate the entropy change by setting abla G to zero, showing a systematic increase in disorder as the liquid becomes gas.