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When a bottle of perfume is opened, odorous molecules mix with air and slowly diffuse throughout the entire room. Is \(\Delta G\) for the diffusion process positive, negative, or zero? What about \(\Delta H\) and \(\Delta S\) for the diffusion?

Short Answer

Expert verified
ΔG is negative, ΔH is approximately zero, and ΔS is positive.

Step by step solution

01

Understanding the Concept of Diffusion

When perfume molecules diffuse in a room after opening a bottle, they spread from an area of higher concentration (inside the bottle) to lower concentration (the room). This is a spontaneous process, which tends to increase the dispersal of molecules and energy.
02

Analyzing Gibbs Free Energy Change (ΔG)

In any spontaneous process, the change in Gibbs Free Energy (ΔG) is negative. Since diffusion is a spontaneous process, \[\Delta G < 0.\]
03

Analyzing Enthalpy Change (ΔH)

The process of diffusion is typically not associated with a significant heat exchange with the surroundings, hence the change in enthalpy (ΔH) is approximately zero. \[\Delta H \approx 0.\]
04

Analyzing Entropy Change (ΔS)

Entropy (ΔS) measures the randomness or disorder of a system. During diffusion, the molecules spread throughout the available space, increasing disorder. Therefore, \[\Delta S > 0.\]

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Key Concepts

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

Diffusion
Diffusion is all about molecules moving from where they are bunched up to where they are more spread out. Think of it like a perfume spreading its scent throughout a room after you open the bottle. It's a natural process, and it happens because molecules are always in motion. They bump into each other, and slowly but surely, they spread out to fill the space evenly. It’s a bit like pouring a drop of ink into a glass of water and watching it spread until the color is equal everywhere. This happens without us having to do anything.
It's a crucial concept in thermodynamics because it is a spontaneous process—meaning it happens on its own due to the innate energy and motion of molecules. This movement from high concentration (like inside the perfume bottle) to low concentration (like in the room) is driven by the molecules' kinetic energy.
Gibbs Free Energy
Gibbs Free Energy (\(\Delta G\)) tells us about the "useful" energy in a system that can do work. In a spontaneous process like diffusion, this energy decreases, meaning \(\Delta G\) is negative.
Why negative? Because in thermodynamics, spontaneous processes are ones that naturally occur without needing extra energy added in. It reflects that the system's energy is becoming more stable as it spreads out and lowers its potential for doing work. So, when considering diffusion, which occurs naturally and spreads odor molecules throughout a room, \(\Delta G < 0\). This indicates that no extra work is necessary to make diffusion happen; it just happens by itself.
Entropy
Entropy (\(\Delta S\)) is often called a measure of disorder or randomness in a system. It's all about how spread out and mixed up molecules are.
During diffusion, entropy increases (\(\Delta S > 0\)) because molecules move from a state of more order (concentrated in the bottle) to less order (evenly spread in the air). It’s like dumping a puzzle out on the floor—there's more disorder when pieces are scattered everywhere.
  • More spreading out equals higher entropy.
  • Higher entropy means the system is more "disordered" and less organized.
In thermodynamics, processes naturally proceed towards increasing entropy, which means things naturally progress to states of higher disorder.
Enthalpy
Enthalpy (\(\Delta H\)) is a measure of total heat content in a system. For diffusion, changes in enthalpy are often minimal or zero because no big shifts in heat energy occur. This is why we say \(\Delta H \approx 0\) during diffusion processes.
Diffusion doesn’t typically involve adding or releasing heat; it's more about the distribution of molecules. Hence, energy change isn't driven by heat exchange.
  • If \(\Delta H\) is zero, it suggests that the process isn't driven or hindered by heat changes.
  • Because diffusion doesn't alter the system's heat significantly, it maintains nearly constant enthalpy.
This characteristic supports the understanding that diffusion is mainly guided by entropy change rather than enthalpy.

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Most popular questions from this chapter

Tell whether the entropy changes, \(\Delta S\), for the following processes are likely to be positive or negative: (a) The conversion of liquid water to water vapor at \(100^{\circ} \mathrm{C}\) (b) The freezing of liquid water to ice at \(0{ }^{\circ} \mathrm{C}\) (c) The eroding of a mountain by a glacier

The boiling point of a substance is defined as the temperature at which liquid and vapor coexist in equilibrium. Use the heat of vaporization \(\left(\Delta H_{\text {vap }}=30.91 \mathrm{~kJ} / \mathrm{mol}\right)\) and the entropy of vaporization \(\left[\Delta S_{\text {vap }}=93.2 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})\right]\) to calculate the boiling point \(\left({ }^{\circ} \mathrm{C}\right)\) of liquid bromine.

When \(2.00 \mathrm{~g}\) of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is burned in a bomb calorimeter, the temperature of the water and calorimeter rises by \(18.6^{\circ} \mathrm{C}\). Assuming that the bath contains \(250.0 \mathrm{~g}\) of water and that the heat capacity for the calorimeter is \(623 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), calculate combustion energies \((\Delta E)\) for glucose in both \(\mathrm{kJ} / \mathrm{g}\) and \(\mathrm{Cal} / \mathrm{g}\).

Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) in \(\mathrm{kJ} / \mathrm{mol}\) for benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), from the following data: $$\begin{aligned}2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-6534 \mathrm{~kJ} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right)=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H^{\circ} \mathrm{f}\left(\mathrm{H}_{2} \mathrm{O}\right)=-285.8 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$$

How much heat in kilojoules is evolved or absorbed in the reaction of \(2.50 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) with enough carbon monoxide to produce iron metal? Is the process exothermic or endothermic? $$\begin{aligned}\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) \\ \Delta H^{\circ}=-24.8 \mathrm{~kJ}\end{aligned}$$

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