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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 (螖G) can be determined using the following formula: 螖G = 螖H - T螖S where 螖H is the enthalpy change 螖S is the entropy change T is the temperature in Kelvin For the given problem, we have the enthalpy of fusion (螖H) and the entropy of fusion (螖S) as 5.65 kJ/mol and 28.9 J/K路mol, respectively. We need to find the Gibbs free energy (螖G) 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 螖G for the melting process. 螖G = 螖H - T螖S 螖G = (5.65 kJ/mol) - (200 K 脳 28.9 J/K路mol) First, we need to convert J to kJ: 28.9 J/K路mol = 28.9 脳 10鈦宦 kJ/K路mol Now, plug the values into the equation: 螖G = (5.65 kJ/mol) - (200 K 脳 28.9 脳 10鈦宦 kJ/K路mol) 螖G = 5.65 kJ/mol - 5.78 kJ/mol 螖G = -0.13 kJ/mol
03

Determine Spontaneity at 200 K

The process is spontaneous if 螖G is negative, non-spontaneous if 螖G is positive, and at equilibrium if 螖G is zero. Since 螖G = -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, 螖G = 0. From the Gibbs free energy formula: 螖G = 螖H - T螖S At equilibrium, 螖G = 0, hence: 0 = 螖H - T螖S T = 螖H/螖S Plug in the values of enthalpy and entropy of fusion: T = (5.65 kJ/mol) / (28.9 脳 10鈦宦 kJ/K路mol) T 鈮 195.5 K Answer to part (b): The approximate melting point of ammonia is 195.5 K.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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