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Chapter 18: Problem 55

Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) of the system for the following processes at 1 atm: (a) ammonia melts at \(-60^{\circ} \mathrm{C},(\mathrm{b})\) ammonia melts at \(-77.7^{\circ} \mathrm{C},(\mathrm{c})\) ammonia melts at \(-100^{\circ} \mathrm{C}\). (The normal melting point of ammonia is \(-77.7^{\circ} \mathrm{C}\).)

Short Answer

Expert verified
(a) \(\Delta H>0\), \(\Delta S>0\), \(\Delta G<0\); (b) \(\Delta H>0\), \(\Delta S>0\), \(\Delta G=0\); (c) \(\Delta H>0\), \(\Delta S>0\), \(\Delta G>0\).

Step by step solution

01

Understanding Process (a)

For part (a), ammonia melts at -60掳C. The melting point of ammonia is -77.7掳C, meaning -60掳C is above the melting point. Thus, at this temperature, melting is a spontaneous process.- **\(\Delta H\)**: Positive, because energy is required to break the intermolecular forces during the phase change from solid to liquid.- **\(\Delta S\)**: Positive, as the entropy increases when a solid becomes a liquid.- **\(\Delta G\)**: Negative, since melting is spontaneous at a temperature above the melting point.
02

Understanding Process (b)

In part (b), ammonia melts at its normal melting point of -77.7掳C.- **\(\Delta H\)**: Positive, as energy input is necessary for the phase change.- **\(\Delta S\)**: Positive, because entropy increases from solid to liquid.- **\(\Delta G\)**: Zero, since the process is at equilibrium at the melting point.
03

Understanding Process (c)

For part (c), ammonia melts at -100掳C, below its melting point of -77.7掳C. Melting is not spontaneous at this temperature.- **\(\Delta H\)**: Positive, as energy is needed to overcome intermolecular forces.- **\(\Delta S\)**: Positive, due to increased randomness in a liquid.- **\(\Delta G\)**: Positive, as melting is non-spontaneous below the melting point.

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Key Concepts

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

Enthalpy
Enthalpy (\( \Delta H \)) represents the heat content of a system. It reflects how much energy is needed for a substance to undergo a phase change, such as melting or boiling. In thermodynamics, when we talk about melting, energy must be added to break the intermolecular forces holding the molecules in a solid. For the processes in our exercise:
  • **Positive \( \Delta H \)**: Energy is absorbed during melting, indicating it takes energy to transform solid ammonia to liquid.
This energy required to change from a structured solid to a free-flowing liquid is why \( \Delta H \) is always positive for melting. Think of it like breaking a bond; you need to put in a little effort!
When predicting enthalpy changes, remember:
  • **Endothermic**: \( \Delta H > 0 \)
  • **Exothermic**: \( \Delta H < 0 \)
Entropy
Entropy (\( \Delta S \)) is a measure of disorder or randomness. When ammonia melts, the ordered solid structure becomes a less ordered liquid. This increase in randomness means:
  • **Positive \( \Delta S \)**: The system becomes more disordered.
Imagine shaking a box of toys. If they start all aligned in rows and then scatter, the randomness increases. Melting increases the potential states a molecule can occupy, which is what entropy is all about! Higher entropy signifies greater molecular disorder. When dealing with entropy in phase changes:
  • **Increase in disorder**: \( \Delta S > 0 \)
  • **Decrease in disorder**: \( \Delta S < 0 \)
Gibbs Free Energy
Gibbs Free Energy (\( \Delta G \)) determines if a process is spontaneous, meaning it can occur without external intervention. This is crucial for predicting reactions.
In our case of melting ammonia, \( \Delta G \) provides insight:
  • **Negative \( \Delta G \)**: The process is spontaneous, as seen when melting above the melting point.
  • **Zero \( \Delta G \)**: Indicates equilibrium, like melting at the exact melting point.
  • **Positive \( \Delta G \)**: The process is non-spontaneous, needing additional energy, as with melting below the melting point.
Keep in mind, \( \Delta G \) combines enthalpy, entropy, and temperature to assess spontaneity: \( \Delta G = \Delta H - T \Delta S \). This equation ties together energy changes and disorder to comprehensively predict reaction feasibility!
Phase Changes
Phase changes involve transitions between solid, liquid, and gas phases. Each transition, such as melting, requires specific conditions of temperature and pressure. During a phase change:
  • **Melting**: Solid to liquid, increasing both \( \Delta H \) and \( \Delta S \).
Understanding how substances like ammonia behave during phase changes helps us predict and control reactions. For instance:
  • **Above melting point**: Substances melt spontaneously.
  • **At melting point**: System is at equilibrium; neither spontaneous nor non-spontaneous.
  • **Below melting point**: Requires external energy; non-spontaneous melting.
Such knowledge is key in fields like chemistry and material science, assisting in the design of processes and materials tailored to specific thermal conditions.

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Most popular questions from this chapter

The enthalpy change in the denaturation of a certain protein is \(125 \mathrm{~kJ} / \mathrm{mol}\). If the entropy change is \(397 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol},\) calculate the minimum temperature at which the protein would denature spontaneously.

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s)\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g)\). Give the reasons for your arrangement.

For reactions carried out under standard-state conditions, Equation 18.10 takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation: $$ \ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right) $$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2},\) respectively (b) Given that at \(25^{\circ} \mathrm{C} K_{c}\) is \(4.63 \times 10^{-3}\) for the reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\).

(a) Over the years, there have been numerous claims about "perpetual motion machines," machines that will produce useful work with no input of energy. Explain why the first law of thermodynamics prohibits the possibility of such a machine existing. (b) Another kind of machine, sometimes called a "perpetual motion of the second kind," operates as follows. Suppose an ocean liner sails by scooping up water from the ocean and then extracting heat from the water, converting the heat to electric power to run the ship, and dumping the water back into the ocean. This process does not violate the first law of thermodynamics, for no energy is created energy from the ocean is just converted to electric energy. Show that the second law of thermodynamics prohibits the existence of such a machine.

Hydrogenation reactions (e.g., the process of converting \(\mathrm{C}=\mathrm{C}\) bonds to \(\mathrm{C}-\mathrm{C}\) bonds in the food industry) are facilitated by the use of a transition metal catalyst, such as \(\mathrm{Ni}\) or \(\mathrm{Pt}\). The initial step is the adsorption, or binding, of hydrogen gas onto the metal surface. Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) when hydrogen gas is adsorbed onto the surface of Ni metal.
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