Problem 14 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Calculate the entropy changes for the system and surroundings when \(1.00 \mathrm{mol}\) of \(\mathrm{NaCl}\) melts at \(1100 \mathrm{K}\). Calculate \(\Delta_{\mathrm{us}} \mathrm{G}\) and estimate the melting point of NaCl. (Section 14.5 ) (For the melting of sodium chloride (NaCl), \(\Delta_{\text {ivs }} H=+30.2 \mathrm{kJmol}^{-1}\) and \(\Delta_{\text {fus }} S=+28.1 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).)

Step-by-Step Solution

Answer

Entropy change for the system is 28.1 J/K; for the surroundings is -27.45 J/K. \( \Delta_{\text{fus}}G \) is -0.71 kJ/mol. Estimated melting point is 1075 K.

Step 1: Calculate Entropy Change of the System

The entropy change for the system during the melting process is given as \( \Delta_{\text{fus}} S = +28.1 \, \mathrm{J\,K^{-1}\,mol^{-1}} \). Thus, the entropy change of the system when 1 mol of NaCl melts at 1100 K is \( 28.1 \, \mathrm{J\,K^{-1}} \).

Step 2: Calculate Entropy Change of the Surroundings

The entropy change for the surroundings \( \Delta_{\text{surroundings}} S \) can be determined using the formula: \( \Delta_{\text{surroundings}} S = -\frac{\Delta_{\text{fus}} H}{T} \). Substituting the given values, we get \( \Delta_{\text{surroundings}} S = -\frac{30.2 \, \mathrm{kJ\,mol^{-1}}}{1100 \, \mathrm{K}} = -27.45 \, \mathrm{J\,K^{-1}} \). We converted the enthalpy from kJ to J by multiplying by 1000.

Step 3: Calculate Gibbs Free Energy Change

The Gibbs free energy change during the melting process is given by the formula \( \Delta_{\text{fus}}G = \Delta_{\text{fus}} H - T\Delta_{\text{fus}} S \). Substituting the values, \( \Delta_{\text{fus}}G = 30.2 \, \mathrm{kJ\,mol^{-1}} - 1100 \, \mathrm{K} \times 28.1 \, \mathrm{J\,K^{-1}\,mol^{-1}} \). Converting \( T\Delta_{\text{fus}} S \) into kJ by dividing by 1000, we find \( \Delta_{\text{fus}}G = 30.2 - 30.91 = -0.71 \, \mathrm{kJ\,mol^{-1}} \).

Step 4: Estimate the Melting Point

The melting point is the temperature at which \( \Delta_{\text{fus}}G \) becomes zero. Therefore, set \( \Delta_{\text{fus}} H = T \Delta_{\text{fus}} S \) and solve for \( T \): \( T = \frac{\Delta_{\text{fus}} H}{\Delta_{\text{fus}} S} = \frac{30.2 \, \mathrm{kJ\,mol^{-1}}}{28.1 \, \mathrm{J\,K^{-1}\,mol^{-1}}} \). Converting enthalpy to J, \( T = \frac{30200 \mathrm{Jmol^{-1}}}{28.1 \mathrm{JK^{-1}mol^{-1}}} \approx 1075 \mathrm{K} \).

Key Concepts

Entropy Change

Entropy, often symbolized as \( S \), represents the degree of disorder or randomness in a system. When sodium chloride (NaCl) melts, it undergoes a phase transition from solid to liquid. This transition is marked by an increase in entropy because the atoms in the liquid phase are more disordered.

For the system during the melting process of NaCl, the given entropy change \( \Delta_{\text{fus}} S \) is \(+28.1 \, \mathrm{J\,K^{-1}\,mol^{-1}} \). This positive value indicates that the system has gained disorder. As the lattice structure of solid NaCl breaks to form a liquid, the ions move more freely, increasing the entropy.
To calculate the change in entropy for the surroundings, we need to consider that energy is absorbed as heat. The equation \( \Delta_{\text{surroundings}} S = -\frac{\Delta_{\text{fus}} H}{T} \) helps determine this value, factoring in both enthalpy change \( \Delta_{\text{fus}} H \) and temperature \( T \). Here, \(-27.45 \mathrm{J\,K^{-1}} \) was calculated for the surroundings when \(1 \mathrm{mol} \) of NaCl melts at \(1100 \mathrm{K} \), showing that the surroundings lose some degree of randomness by releasing heat.

Gibbs Free Energy

Gibbs Free Energy, denoted by \( G \), is fundamental in predicting the direction of a chemical reaction and determining if it is spontaneous under constant pressure and temperature. The change in Gibbs Free Energy during a reaction or process is calculated using the equation:

• \( \Delta_{\text{fus}}G = \Delta_{\text{fus}} H - T\Delta_{\text{fus}} S \)

This equation combines both the enthalpy \( \Delta_{\text{fus}} H \) and the entropy \( T\Delta_{\text{fus}} S \) considerations, providing a holistic view of a process.

For the melting of NaCl, the change in Gibbs Free Energy was found to be \(-0.71 \, \mathrm{kJ\,mol^{-1}} \) at \(1100 \mathrm{K} \). A negative \( \Delta_{\text{fus}}G \) signifies that the melting process is spontaneous at this temperature. This negativeness reflects that the energy cost of order breaking is compensated by the disorder created, making the process' spontaneity favorable.

Enthalpy

Enthalpy, \( H \), represents the total heat content of a system. It is a crucial concept in thermodynamics because it accounts for both internal energy and the product of pressure and volume. In the context of phase changes, like melting, enthalpy change \( \Delta_{\text{fus}} H \) becomes particularly relevant.

For sodium chloride, the enthalpy change \( \Delta_{\text{fus}} H \) is \(+30.2 \, \mathrm{kJ\,mol^{-1}} \). This positive value indicates that energy is consumed to disrupt the orderly solid structure of NaCl, allowing it to turn into a liquid. Hence, the process is endothermic, meaning it absorbs heat.
Since \( 1 \, \mathrm{mol} \) of NaCl requires \( 30.2 \, \mathrm{kJ} \) to melt, the process reflects the energy constraints of overcoming ionic interactions within the solid structure to enhance particle mobility.

Melting Point Estimation

Melting Point is defined as the temperature where the solid and liquid state of a substance is in equilibrium, meaning \( \Delta_{\text{fus}} G = 0 \). For this situation, we use the relationship:

• \( \Delta_{\text{fus}} H = T \Delta_{\text{fus}} S \)

Solving for \( T \) gives the melting point. Hence, \( T = \frac{\Delta_{\text{fus}} H}{\Delta_{\text{fus}} S} \), which leads to calculating the melting point for NaCl.

Upon solving using the provided values, \( T \approx 1075 \, \mathrm{K} \) was derived. This gives a reasonable estimation for the melting point of NaCl, where the energy absorbed (enthalpy) and the increase in disorder (entropy) balance each other's effects, allowing for a stable transition between solid and liquid phases.