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Chapter 17: Problem 43

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

Expert verified
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

01

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.
02

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
03

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.
04

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.

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