For the reaction $2Cl(g) \rightarrow Cl_2(g)$, what are the signs of $\Delta H$ and $\Delta S$?

> [!TLDR] > Bond formation releases energy ($\Delta H < 0$) and two particles combining into one decreases disorder ($\Delta S < 0$); both are **negative**. ## Understanding the Process When two chlorine atoms in the gas phase combine to form a chlorine molecule, we're witnessing bond formation. To predict the signs of $\Delta H$ and $\Delta S$, we need to think about what happens energetically and entropically during this transformation. The key insight: this reaction converts two separate, independent particles into one bonded molecule. That simple fact tells us almost everything. --- ## Analyzing $\Delta H$ (Enthalpy Change) 1. **Bond formation is exothermic.** When two chlorine atoms approach each other, their unpaired electrons pair up to form a covalent Cl–Cl bond. This bond has a well-defined bond energy (approximately 242 kJ/mol). 2. **Energy is released to the surroundings.** The system goes from a higher-energy state (two isolated atoms with unpaired electrons) to a lower-energy state (one stable molecule with a shared electron pair). The difference in energy leaves the system as heat. 3. **Therefore $\Delta H [!FORMULA] > $$\Delta H = H_{\text{products}} - H_{\text{reactants}} = H_{Cl_2} - 2H_{Cl} < 0$$ --- ## Analyzing $\Delta S$ (Entropy Change) 1. **Count the particles.** We start with 2 moles of gaseous atoms and end with 1 mole of gaseous molecules. The number of independent particles has decreased. 2. **Fewer particles means less disorder.** Entropy measures the number of ways energy can be distributed among particles and their microstates. Two separate atoms can move independently in three-dimensional space, each with its own translational, rotational, and electronic degrees of freedom. When they bond, they become a single entity with shared motion. 3. **The system becomes more ordered.

For the reaction 2Cl(g) arrow Cl₂(g), what are the signs of Δ H and Δ S?

Exercises · 5.18

2Cl(g) \rightarrow Cl_2(g)

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Concept understanding — Gibbs Free Energy

Gibbs Free Energy: The "Why" Behind Spontaneous Reactions

Imagine you're pushing a boulder downhill. It's going to happen naturally — you don't need to keep pushing. But if you want to roll it uphill, you have to work against gravity the whole way. Chemistry works the same way: some reactions happen on their own (spontaneous), and others need a constant push of energy.

The question is: what decides which is which? That's exactly what Gibbs Free Energy answers.

The Two Competing Forces

Two things drive every chemical change:

Enthalpy ( $H$ ) — the total heat content. Nature tends to move toward lower energy. A fire releases heat; that's enthalpy driving the reaction forward. Reactions that release heat (exothermic, $\Delta H < 0$ ) are favoured.
Entropy ( $S$ ) — the measure of disorder. Nature also tends toward more chaos. A messy room doesn't tidy itself; a gas spreads to fill its container. Reactions that increase disorder (positive $\Delta S$ ) are favoured.

But here's the catch: these two can pull in opposite directions. An endothermic reaction (absorbs heat, $\Delta H > 0$ ) might still happen if it creates enough disorder. Ice melting is a perfect example — it absorbs heat, but the liquid water is far more disordered than the crystal.

The Resolution: Gibbs Free Energy

Josiah Willard Gibbs combined these two forces into one number that tells you the net direction:

\Delta G = \Delta H - T \Delta S

Where:

$\Delta G$ = change in Gibbs free energy (kJ/mol)
$\Delta H$ = change in enthalpy (kJ/mol)
$T$ = absolute temperature (Kelvin)
$\Delta S$ = change in entropy (J/K·mol — careful with units!)

The sign of $\Delta G$ is the final verdict:

$\Delta G$ sign	What it means
Negative ( $\Delta G < 0$ )	Spontaneous — reaction happens on its own
Positive ( $\Delta G > 0$ )	Non-spontaneous — needs constant energy input
Zero ( $\Delta G = 0$ )	Equilibrium — no net change

Why Temperature Matters

Notice the $T$ in front of $\Delta S$ . Temperature decides which force wins. At low temperatures, the enthalpy term ( $\Delta H$ ) dominates. At high temperatures, the entropy term ( $T\Delta S$ ) takes over.

This explains everyday observations:

Ice melts spontaneously above 0°C (entropy wins at higher $T$ )
Water freezes spontaneously below 0°C (enthalpy wins at lower $T$ )
At exactly 0°C, $\Delta G = 0$ — ice and water coexist in equilibrium

The Precise Statement

Gibbs Free Energy is the maximum useful work obtainable from a closed system at constant temperature and pressure. When a reaction proceeds, the system loses free energy ( $\Delta G < 0$ ), and that energy is available to do work — like running a muscle or powering a battery. …

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Bond formation releases energy ( $\Delta H < 0$ ) and two particles combining into one decreases disorder ( $\Delta S < 0$ ); both are negative.

Understanding the Process

When two chlorine atoms in the gas phase combine to form a chlorine molecule, we're witnessing bond formation. To predict the signs of $\Delta H$ and $\Delta S$ , we need to think about what happens energetically and entropically during this transformation.

The key insight: this reaction converts two separate, independent particles into one bonded molecule. That simple fact tells us almost everything.

Analyzing $\Delta H$ (Enthalpy Change)

Bond formation is exothermic. When two chlorine atoms approach each other, their unpaired electrons pair up to form a covalent Cl–Cl bond. This bond has a well-defined bond energy (approximately 242 kJ/mol).
Energy is released to the surroundings. The system goes from a higher-energy state (two isolated atoms with unpaired electrons) to a lower-energy state (one stable molecule with a shared electron pair). The difference in energy leaves the system as heat.
Therefore $\Delta H < 0$ . The reaction is exothermic. The enthalpy of the products (one $Cl_2$ molecule) is lower than the enthalpy of the reactants (two $Cl$ atoms).

\Delta H = H_{\text{products}} - H_{\text{reactants}} = H_{Cl_2} - 2H_{Cl} < 0

Analyzing $\Delta S$ (Entropy Change)

Count the particles. We start with 2 moles of gaseous atoms and end with 1 mole of gaseous molecules. The number of independent particles has decreased.
Fewer particles means less disorder. Entropy measures the number of ways energy can be distributed among particles and their microstates. Two separate atoms can move independently in three-dimensional space, each with its own translational, rotational, and electronic degrees of freedom. When they bond, they become a single entity with shared motion.
The system becomes more ordered. The two atoms are now constrained relative to each other by the chemical bond. While the $Cl_2$ molecule still has translational and rotational freedom, the overall number of accessible microstates has decreased compared to two independent atoms.
Therefore $\Delta S < 0$ . The entropy of the system decreases. Going from 2 particles to 1 particle always reduces entropy in the gas phase. …

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Q.For the reaction 2Cl(g)→Cl2(g)2Cl(g) \rightarrow Cl_2(g)2Cl(g)→Cl2​(g), what are the signs of ΔH\Delta HΔH and ΔS\Delta SΔS?

Understanding the Process

Analyzing ΔH\Delta HΔH (Enthalpy Change)

Analyzing ΔS\Delta SΔS (Entropy Change)

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Q.For the reaction $2Cl(g) \rightarrow Cl_2(g)$ , what are the signs of $\Delta H$ and $\Delta S$ ?

Analyzing $\Delta H$ (Enthalpy Change)

Analyzing $\Delta S$ (Entropy Change)

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