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What values of \(\Delta H, \Delta S,\) and \(\Delta G\) would you expect for the formation of an ideal solution of liquid components? (Is each value positive, negative, or zero?)

Short Answer

Expert verified
\(\Delta H = 0\) (neither absorbs nor releases heat), \(\Delta S > 0\) (entropy increases upon mixing), \(\Delta G < 0\) (process is spontaneous) for an ideal solution.

Step by step solution

01

Determine \(\Delta H\)

For an ideal solution, the change in enthalpy (\(\Delta H\)) should be zero. This is because, in an ideal solution, the interactions between the two types of particles (solvent and solute) are assumed to be equal before and after the mixture. If the intermolecular forces do not change upon mixing, no energy is absorbed or released, hence \(\Delta H = 0\).
02

Determine \(\Delta S\)

When two substances mix to form a solution, the entropy (\(\Delta S\)) typically increases because the mix represents a more disordered state compared to the separate pure components. Therefore, \(\Delta S\) for formation of an ideal solution should be positive.
03

Determine \(\Delta G\)

Free energy change (\(\Delta G\)) can be calculated by the formula \(\Delta G = \Delta H - T\Delta S\). From steps 1 and 2, we know that \(\Delta H = 0\) and \(\Delta S > 0\). For a spontaneous process, \(\Delta G\) must be negative or zero. Since the \(\Delta S\) value is positive, and multiplying it by any temperature will be positive, subtracting a positive number from zero will give a negative \(\Delta G\), indicating a spontaneous process. Therefore, the \(\Delta G\) for the formation of an ideal solution should be negative.

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Key Concepts

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

Enthalpy Change
The concept of enthalpy change (\( \Delta H \)) is critical in understanding thermodynamics. Enthalpy change refers to the heat absorbed or released during a process at constant pressure. For an ideal solution, like when two liquids form a perfect mixture, there's no change in the total energy due to their interactions.

In an ideal solution, the intermolecular forces between similar and dissimilar molecules before and after mixing are assumed to be the same. For this reason:
  • Energy absorption or release is negligible.
  • \( \Delta H = 0 \), as the energy state remains constant.
This might seem abstract, but imagine you're mixing two similar substances, like ethanol and methanol. The bonds and interactions between them don't "care" who's next to whom, which means mixing doesn't require or release energy. This illustrates the zero enthalpy change for an ideal solution.
Entropy Change
Entropy is a measure of disorder or randomness in a system. For any spontaneous process, entropy tends to increase, leading to a higher degree of disorder. When two separate pure components mix, they enter a state of more chaos, and therefore, entropy generally rises.

Let's break down the change in entropy (\( \Delta S \)):
  • A system of separated pure components is ordered, with each molecule enjoying its own space.
  • Upon mixing, different molecules are found in the same space, dramatically increasing disorder.
  • This results in a positive \( \Delta S \).
If you picture pouring together colored beads, arranged initially by color, into a single container, the level of mix and disorder confirms that entropy has increased. Hence, the formation of an ideal solution leads to a positive entropy change.
Gibbs Free Energy
Understanding Gibbs free energy (\( \Delta G \)) helps predict whether a process will occur spontaneously. It combines both enthalpy and entropy changes, reflecting the balance between energy conservation and disorder.
  • Formula: \( \Delta G = \Delta H - T \Delta S \)
  • Where \( T \) is the absolute temperature.
For an ideal solution:
  • \( \Delta H = 0 \) (no heat change)
  • \( \Delta S > 0 \) (disorder increases)
  • Thus, \( \Delta G \) is generally negative when solved as \( 0 - T \Delta S \).
This negative value indicates that the process of forming an ideal solution is spontaneous. Just as a roller coaster naturally moves downhill, the formation of an ideal solution happens without additional input, thanks to this energy balance. Gibbs free energy makes it clear why some chemical processes occur on their own, emphasizing the natural tendency towards greater disorder and energy efficiency.

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Most popular questions from this chapter

Arrange the entropy changes of the following processes, all at \(25^{\circ} \mathrm{C},\) in the expected order of increasing \(\Delta S,\) and explain your reasoning: (a) \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{atm}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{atm})\) (b) \(\mathrm{CO}_{2}(\mathrm{s}, 1 \mathrm{atm}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}, 10 \mathrm{mm} \mathrm{Hg})\) (c) \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{atm}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 10 \mathrm{mmHg})\)

The following standard Gibbs energy changes are given for \(25^{\circ} \mathrm{C}\) (1) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\) \(\Delta G^{\circ}=-33.0 \mathrm{kJ}\) (2) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(1)\) \(\Delta G^{\circ}=-1010.5 \mathrm{kJ}\) (3) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})\) \(\Delta G^{\circ}=+173.1 \mathrm{kJ}\) (4) \(\mathrm{N}_{2}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=+102.6 \mathrm{kJ}\) (5) \(2 \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta G^{\circ}=+208.4 \mathrm{kJ}\) Combine the preceding equations, as necessary, to obtain \(\Delta G^{\circ}\) values for each of the following reactions. (a) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \quad \Delta G^{\circ}=?\) (b) \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(1) \quad \Delta G^{\circ}=?\) (c) \(2 \mathrm{NH}_{3}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(1)\) \(\Delta G^{\circ}=?\) Of reactions (a), (b), and (c), which would tend to go to completion at \(25^{\circ} \mathrm{C}\), and which would reach an equilibrium condition with significant amounts of all reactants and products present?

Use the following data to estimate the standard molar entropy of gaseous benzene at \(298.15 \mathrm{K} ;\) that is, \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \mathrm{atm})\right] .\) For \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{s}, 1 \mathrm{atm})\) at its melting point of \(5.53^{\circ} \mathrm{C}, S^{\circ}\) is \(128.82 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(9.866 \mathrm{kJ} \mathrm{mol}^{-1} .\) From the melting point to 298.15 K, the average heat capacity of liquid benzene is \(134.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The enthalpy of vaporization of \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at \(298.15 \mathrm{K}\) is \(33.85 \mathrm{kJ} \mathrm{mol}^{-1},\) and in the vapor- ization, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) is produced at a pressure of 95.13 Torr. Imagine that this vapor could be compressed to 1 atm pressure without condensing and while behaving as an ideal gas. Calculate \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \text { atm) }] .[ \text { Hint: Refer to }\right.\) Exercise \(88,\) and note the following: For infinitesimal quantities, \(d S=d q / d T ;\) for the compression of an ideal gas, \(d q=-d w ;\) and for pressure-volume work, \(d w=-P d V\).

Sodium carbonate, an important chemical used in the production of glass, is made from sodium hydrogen carbonate by the reaction \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Data for the temperature variation of \(K_{\mathrm{p}}\) for this reaction are \(K_{\mathrm{p}}=1.66 \times 10^{-5}\) at \(30^{\circ} \mathrm{C} ; 3.90 \times 10^{-4} \mathrm{at}\) \(50^{\circ} \mathrm{C} ; 6.27 \times 10^{-3}\) at \(70^{\circ} \mathrm{C} ;\) and \(2.31 \times 10^{-1}\) at \(100^{\circ} \mathrm{C}\) (a) Plot a graph similar to Figure \(19-12,\) and determine \(\Delta H^{\circ}\) for the reaction. (b) Calculate the temperature at which the total gas pressure above a mixture of \(\mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is \(2.00 \mathrm{atm}\).

Indicate whether each of the following changes represents an increase or a decrease in entropy in a system, and explain your reasoning: (a) the freezing of ethanol; (b) the sublimation of dry ice; (c) the burning of a rocket fuel.

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