Using The Following Data Calculate Δsfus and Δsvap for Na

This guide explains how to calculate the standard free energy of fusion (δsfus) and standard free energy of vaporization (δsvap) for sodium (Na) using provided thermodynamic data. We'll cover the formulas, assumptions, practical applications, and common pitfalls.

Introduction

The standard free energy of fusion (δsfus) and vaporization (δsvap) are fundamental thermodynamic properties that describe the energy changes during phase transitions of a substance. For sodium (Na), these values are crucial in understanding its behavior under different conditions.

Key Concept: Free energy changes (δG) are calculated using the formula δG = δH - TδS, where δH is enthalpy change, T is temperature in Kelvin, and δS is entropy change.

In this guide, we'll focus on calculating these values for sodium using standard thermodynamic data. The calculations involve:

Standard enthalpy of fusion (δHfus)
Standard enthalpy of vaporization (δHvap)
Standard entropy of fusion (δSfus)
Standard entropy of vaporization (δSvap)

Formulas

The standard free energy changes are calculated using the following formulas:

Standard Free Energy of Fusion (δsfus)

δsfus = δHfus - TδSfus

Where:

δHfus = standard enthalpy of fusion (kJ/mol)
T = temperature in Kelvin (typically 298.15 K for standard conditions)
δSfus = standard entropy of fusion (J/mol·K)

Standard Free Energy of Vaporization (δsvap)

δsvap = δHvap - TδSvap

Where:

δHvap = standard enthalpy of vaporization (kJ/mol)
T = temperature in Kelvin (typically 298.15 K)
δSvap = standard entropy of vaporization (J/mol·K)

These formulas show that free energy changes depend on both enthalpy and entropy contributions. The temperature term (TδS) represents the entropy contribution to the free energy change.

Calculation

To calculate δsfus and δsvap for sodium, you'll need the following standard thermodynamic data:

Property	Value	Units
δHfus (Na)	2.59	kJ/mol
δSfus (Na)	9.8	J/mol·K
δHvap (Na)	97.7	kJ/mol
δSvap (Na)	76.0	J/mol·K

Using these values and assuming standard temperature (T = 298.15 K), we can calculate:

Example Calculation for δsfus

δsfus = δHfus - TδSfus

= 2.59 kJ/mol - (298.15 K × 9.8 J/mol·K)

= 2.59 - 2923.67

= -2921.08 kJ/mol

Example Calculation for δsvap

δsvap = δHvap - TδSvap

= 97.7 kJ/mol - (298.15 K × 76.0 J/mol·K)

= 97.7 - 22647.2

= -22549.5 kJ/mol

These negative values indicate that both fusion and vaporization processes are spontaneous under standard conditions, releasing free energy.

Interpretation

The calculated values have several important implications:

Spontaneity: The negative values confirm that both fusion and vaporization are spontaneous processes for sodium.
Energy Requirements: The large negative value for δsvap (-22549.5 kJ/mol) indicates that vaporizing sodium requires significantly more energy than fusing it.
Thermodynamic Stability: The results show that sodium is more stable in its solid form than in its liquid or gaseous forms.

Practical Consideration: These values are important in industrial processes involving sodium, such as metal production and chemical reactions.

Understanding these free energy changes helps in predicting the behavior of sodium under different conditions and designing efficient processes for its handling.

FAQ

What is the difference between δsfus and δsvap?

δsfus represents the free energy change when sodium changes from solid to liquid, while δsvap represents the free energy change when sodium changes from liquid to gas.

Why are the values negative?

Negative values indicate that the processes are spontaneous and release free energy under standard conditions.

What temperature is used in these calculations?

Standard temperature (298.15 K or 25°C) is used for these calculations.

How accurate are these calculations?

The accuracy depends on the precision of the input thermodynamic data. The calculations follow standard thermodynamic principles.