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 \(\mathrm{J} / \mathrm{K} \cdot\) 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, NH$_3$(s) will spontaneously melt at 200 K. b. The approximate melting point of ammonia is 195.5 K.

Step-by-step solution

Gibbs Free Energy Formula and Notations

The Gibbs free energy (ΔG) can be determined using the following formula: ΔG = ΔH - TΔS where ΔH is the enthalpy change ΔS is the entropy change T is the temperature in Kelvin For the given problem, we have the enthalpy of fusion (ΔH) and the entropy of fusion (ΔS) as 5.65 kJ/mol and 28.9 J/K·mol, respectively. We need to find the Gibbs free energy (ΔG) to determine if the melting process is spontaneous at 200 K.

Calculate Gibbs Free Energy for Melting at 200 K

Use the provided data and the temperature of 200 K in the Gibbs free energy formula to determine the ΔG for the melting process. ΔG = ΔH - TΔS ΔG = (5.65 kJ/mol) - (200 K × 28.9 J/K·mol) First, we need to convert J to kJ: 28.9 J/K·mol = 28.9 × 10⁻³ kJ/K·mol Now, plug the values into the equation: ΔG = (5.65 kJ/mol) - (200 K × 28.9 × 10⁻³ kJ/K·mol) ΔG = 5.65 kJ/mol - 5.78 kJ/mol ΔG = -0.13 kJ/mol

Determine Spontaneity at 200 K

The process is spontaneous if ΔG is negative, non-spontaneous if ΔG is positive, and at equilibrium if ΔG is zero. Since ΔG = -0.13 kJ/mol is negative, the melting of ammonia (NH3) is spontaneous at 200 K. Answer to part (a): Yes, NH3(s) will spontaneously melt at 200 K.

Calculate the Melting Point of Ammonia

To find the melting point of ammonia, we need to determine the temperature at which the melting process is at equilibrium. At equilibrium, ΔG = 0. From the Gibbs free energy formula: ΔG = ΔH - TΔS At equilibrium, ΔG = 0, hence: 0 = ΔH - TΔS T = ΔH/ΔS Plug in the values of enthalpy and entropy of fusion: T = (5.65 kJ/mol) / (28.9 × 10⁻³ kJ/K·mol) T ≈ 195.5 K Answer to part (b): The approximate melting point of ammonia is 195.5 K.

Key concepts

Spontaneity

Spontaneity in chemical processes tells us whether a process will occur on its own without any external input. To determine spontaneity, we use Gibbs free energy ( \( \Delta G \)). The formula is:

• \( \Delta G = \Delta H - T\Delta S \)
• \( \Delta H \) represents the change in enthalpy, or heat change during the process.
• \( \Delta S \) is the change in entropy, which measures the disorder of the system.
• \( T \) is the temperature in Kelvin.

When \( \Delta G \) is negative, the process is spontaneous, meaning it can happen on its own. If positive, it needs external energy input, and if zero, the system is at equilibrium.

In our exercise, to check if ammonia (\( \text{NH}_{3}(s)\)) melts spontaneously at 200 K, we calculated \( \Delta G \) using the given \( \Delta H \) and \( \Delta S \). Our result was \( -0.13 \text{ kJ/mol} \), indicating that the melting is spontaneous under these conditions.

Melting Point Calculation

The melting point is the temperature at which a solid turns into a liquid. At this point, the solid's Gibbs free energy change (\( \Delta G \)) is zero. This equilibrium condition means there's no net energy input needed for the phase change.

To find this point, we start with the equation:

• \( \Delta G = \Delta H - T\Delta S = 0 \)

Rearranging gives:

• \( T = \frac{\Delta H}{\Delta S} \)

Using the values provided for ammonia: \( \Delta H = 5.65 \text{ kJ/mol} \) and \( \Delta S = 28.9 \times 10^{-3} \text{ kJ/K mol} \), we calculate the approximate melting point. By plugging into the formula, we find the melting point to be about 195.5 K.

Enthalpy and Entropy

Understanding enthalpy and entropy is vital in predicting chemical behavior. Enthalpy ( \( \Delta H \)) is about the heat absorbed or released in a reaction. It gives insight into the energy changes involving bonds breaking or forming. When substances change phases, \( \Delta H \) often helps determine whether heat is needed or provided.

Entropy ( \( \Delta S \)) measures a system's disorder. A high \( \Delta S \) indicates a tendency towards increased disorder or randomness. The melting of a solid generally leads to a higher entropy since the molecules in a liquid have more freedom to move.

Together, these concepts combined with temperature in the Gibbs free energy equation help us predict the spontaneity of a process. In the case of ammonia, looking at both \( \Delta H \) and \( \Delta S \) in the context of the right temperature was essential to understanding when and why its melting process is spontaneous.