Problem 23 - Chemistry The Central Science In Si Units

Question

The normal boiling point of the element mercury (Hg) is \(356.7^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=59.11 \mathrm{~kJ} / \mathrm{mol}\) (a) When Hg boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(2.00 \mathrm{~mol}\) of \(\mathrm{Hg}\) is vaporized at \(356.7^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

(a) Entropy increases when Hg boils; (b) \(\Delta S = 0.1878\text{ kJ/K}\) for 2 mol.

Step 1: Understanding Entropy Change

When a substance boils, it changes from a liquid to a gas. This phase change involves an increase in entropy because gases have more disorder due to greater molecular movement compared to liquids.

Step 2: Entropy Change Formula

The change in entropy (  ) can be calculated using the formula: \[\Delta S = \frac{\Delta H_{\text{vap}}}{T}\]where (), , and  refers to the entropic, enthalpic, and temperature-based differences across states of matter.

Step 3: Convert Temperature

Convert the temperature from Celsius to Kelvin since the entropy equation requires temperature in Kelvin. \[T(\text{K}) = 356.7 + 273.15 = 629.85\text{ K}\]

Step 4: Calculate Entropy Change for 1 mol

Calculate entropy change for 1 mole of mercury using  in the formula:\[\Delta S_{1\text{ mol}} = \frac{59.11\text{ kJ/mol}}{629.85\text{ K}} = 0.0939\text{ kJ/(mol K)}\]

Step 5: Calculate Entropy Change for 2 mol

Since we need the entropy change for 2 moles:\[\Delta S_{2\text{ mol}} = 2 \times 0.0939\text{ kJ/(mol K)} = 0.1878\text{ kJ/K}\]

Key Concepts

Boiling Point

The boiling point is the temperature at which a liquid turns into a gas or vapor. It is a critical phase change point that depends on environmental conditions like pressure. For mercury (Hg), this boiling point is quite high at 356.7°C under normal atmospheric pressure. When mercury reaches its boiling point, the kinetic energy of its particles becomes sufficient to overcome intermolecular forces that keep the liquid intact. This energy acts as a trigger allowing mercury to transition from a liquid state to a gaseous state. It's important to note that while at the boiling point, both liquid and gas phases of mercury can exist in equilibrium. Understanding boiling points is crucial in chemistry because it helps predict and explain how and when substances transition between different states of matter. It's not just about reaching a certain temperature, but it's about understanding the energy dynamics and conditions required for such transformations.

Molar Enthalpy of Vaporization

Molar enthalpy of vaporization is the amount of energy needed to convert one mole of a liquid into gas at its boiling point. This concept helps quantify the energy required for the phase transition from liquid to gas. For mercury, the molar enthalpy of vaporization is given as 59.11 kJ/mol. This value represents the energy needed to break the attractive forces between liquid molecules and allow them to disperse as gas molecules. The higher the molar enthalpy, the more energy is required to change the state of the substance, which can indicate stronger intermolecular forces in the liquid state. In practical terms, knowing the molar enthalpy of vaporization allows chemists and engineers to calculate the energy costs associated with processes involving phase changes. It also aids in the design of equipment like distillation columns and other systems where phase changes are essential.

Entropy Change Calculation

Entropy is a measure of disorder or randomness within a system. In the context of boiling, entropy changes because gases have much higher disorder than liquids due to increased molecular movement. For the vaporization of mercury, we need to determine the change in entropy (9S) using the formula: \[\Delta S = \frac{\Delta H_{\text{vap}}}{T} \]where 9H_vap is the molar enthalpy of vaporization, and T is the absolute temperature in Kelvin. To calculate this for mercury vaporizing at its boiling point of 356.7°C, we first convert the temperature to Kelvin: \[\text{T(K)} = 356.7 + 273.15 = 629.85\text{ K}\]Using this temperature, the entropy change for 1 mole of mercury is calculated as: \[\Delta S_{1\text{ mol}} = \frac{59.11\text{ kJ/mol}}{629.85\text{ K}} = 0.0939\text{ kJ/(mol K)}\]Given that we need the entropy change for 2 moles, we simply multiply by 2:\[\Delta S_{2\text{ mol}} = 2 \times 0.0939\text{ kJ/(mol K)} = 0.1878\text{ kJ/K}\]Understanding how to calculate entropy change is fundamental in predicting the spontaneity of chemical processes as it combines with enthalpy to derive Gibbs free energy, which dictates the feasibility and direction of reactions.