Question

For ammonia \(\left(\mathrm{NH}_{3}\right.\) ), the enthalpy of fusion is \(5.65 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of fusion is \(28.9 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

Short answer

a) Yes, NH3(s) will spontaneously melt at 200 K, as the Gibbs free energy change (∆G) is negative (-130 J/mol) at this temperature. b) The approximate melting point of ammonia is 195.5 K, as this is the temperature at which the Gibbs free energy change (∆G) is zero.

Step-by-step solution

Determine the Gibbs free energy change for the phase transition

First, we will use the given enthalpy (∆H) and entropy (∆S) values to compute the Gibbs free energy change (∆G) at the given temperature: \(\Delta G = \Delta H - T \Delta S\) where ∆H is the enthalpy of fusion, T is the temperature in Kelvin, and ∆S is the entropy of fusion. We will substitute the given values for ammonia into this equation.

Calculate ∆G at 200 K

Now we can substitute the given values and the temperature 200 K into the equation to find the Gibbs free energy change at this temperature: \(\Delta G = (5.65~ \text{kJ/mol})(1000~ \text{J/kJ}) - (200~\text{K})(28.9~\text{J/K}\cdot\text{mol})\)

Evaluate the spontaneity of the phase transition

The sign of the Gibbs free energy change (∆G) determines the spontaneity of the phase transition. If ∆G is negative, the phase transition is spontaneous; if ∆G is positive, it is not spontaneous; and if ∆G is zero, the process is at equilibrium. Now let's calculate ∆G: \(\Delta G = 5650~\text{J/mol} - (200~\text{K})(28.9~\text{J/K}\cdot\text{mol}) = 5650~\text{J/mol} - 5780~\text{J/mol} = -130~\text{J/mol}\) Since ∆G is negative, the phase transition is spontaneous at 200 K. Answer (a): Yes, NH3(s) will spontaneously melt at 200 K.

Determine the melting point

To find the melting point of ammonia, we need to find the temperature at which the Gibbs free energy change (∆G) is zero: \(\Delta G = \Delta H - T \Delta S = 0 \Rightarrow T = \frac{\Delta H}{\Delta S}\) Now we can substitute the given values for ∆H and ∆S to solve for T: \(T = \frac{5.65\times10^3~\text{J/mol}}{28.9~\text{J/K}\cdot\text{mol}} \Rightarrow T \approx 195.5~\text{K}\) Answer (b): The approximate melting point of ammonia is 195.5 K.

Key concepts

Enthalpy of Fusion

The enthalpy of fusion, often represented as \(\Delta H_{\text{fus}}\), is the amount of energy needed to change a substance from the solid phase to the liquid phase at a constant pressure. This is a crucial value in understanding the energy dynamics of melting. For example, ammonia (\(\mathrm{NH}_{3}\)) has an enthalpy of fusion of 5.65 kJ/mol, which means it requires that amount of energy to melt each mole of solid ammonia.

Understanding the enthalpy of fusion helps us comprehend the amount of heat that must be supplied or removed during the melting or freezing process of a substance. This is important in fields like metallurgy, cryogenics, and even in everyday activities like cooking, where control over phase transitions is critical.

Entropy of Fusion

Alongside enthalpy, the entropy of fusion, labeled as \(\Delta S_{\text{fus}}\), is another thermodynamic property that measures the change in disorder or randomness as a substance transitions from solid to liquid. For instance, ammonia's entropy of fusion is 28.9 J/(K·mol), suggesting an increase in the disorder of the system per mole when it melts.

The concept of entropy can sometimes be challenging to grasp because it deals with the dispersion of energy within a system and the degree of unpredictability. In the melting process, the ordered structure of a solid becomes less organized as it transitions into a liquid, thereby increasing in entropy. The understanding of entropy is vital, not just in physics or chemistry, but also helps to shed light on why certain processes, such as ice melting into water, occur more readily in nature.

Melting Point Calculation

Calculating the melting point of a substance can be done by employing the relationship between Gibbs free energy (\(\Delta G\)), enthalpy of fusion, and entropy of fusion. At the melting point, the system is in equilibrium, meaning that the Gibbs free energy change for the melting process is zero (\(\Delta G=0\)). Using the formula \(\Delta G = \Delta H - T \Delta S\), where \(T\) is the melting temperature, we can rearrange to find \(T = \frac{\Delta H}{\Delta S}\).

For ammonia, plugging in the values gives us \(T = \frac{5.65\times10^3~\text{J/mol}}{28.9~\text{J/K}\cdot\text{mol}}\), and by calculating, we find the melting point to be approximately 195.5 K. This approach to determining the melting point is widely used in both scientific research and industrial applications where the melting temperature is not yet known or needs to be confirmed.

Spontaneity of Phase Transitions

The spontaneity of a phase transition is governed by the sign of the Gibbs free energy change (\(\Delta G\)). When \(\Delta G\) is negative, the process is spontaneous; it can proceed without any energy input from the outside. If \(\Delta G\) is positive, additional energy is required to drive the process, thus it is non-spontaneous. A \(\Delta G\) of zero indicates that the system is in a state of equilibrium.

Considering the example of ammonia, by calculating \(\Delta G = \Delta H - T \Delta S\) at a certain temperature like 200 K, we determined that the sign of \(\Delta G\) was negative, indicating that the melting of solid ammonia is spontaneous at this temperature. This concept is particularly important in chemical engineering and environmental science, where the spontaneous behavior of substances can significantly impact processes and ecosystems.