Problem 20
Question
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?)
Step-by-Step Solution
Verified Answer
\(\Delta H = 0\) (neither absorbs nor releases heat), \(\Delta S > 0\) (entropy increases upon mixing), \(\Delta G < 0\) (process is spontaneous) for an ideal solution.
1Step 1: 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 \(\Delta H = 0\).
2Step 2: 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.
3Step 3: 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.
Key Concepts
Enthalpy ChangeEntropy ChangeGibbs Free Energy
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:
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.
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 (\( \Delta S \)):
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 \( \Delta S \).
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.
- \( \Delta H = 0 \) (no heat change)
- \( \Delta S > 0 \) (disorder increases)
- Thus, \( \Delta G \) is generally negative when solved as \( 0 - T \Delta S \).
Other exercises in this chapter
Problem 15
Which of the following changes in a thermodynamic property would you expect to find for the reaction \(\mathrm{Br}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Br}
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If a reaction can be carried out only by electrolysis, which of the following changes in a thermodynamic property must apply: (a) \(\Delta H>0 ;\) (b) \(\Delta
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Explain why (a) some exothermic reactions do not occur spontaneously, and (b) some reactions in which the entropy of the system increases do not occur spontaneo
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Explain why you would expect a reaction of the type \(\mathrm{AB}(\mathrm{g}) \longrightarrow \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})\) always to be spont
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