Problem 124 - Chemistry The Molecular Science

Question

Without consulting tables of \(\Delta_{\mathrm{f}} H^{\circ}, S^{\circ},\) or \(\Delta_{\mathrm{f}} G^{\circ}\) values, predict which of these reactions is (i) always product-favored. (ii) product-favored at low temperatures, but not productfavored at high temperatures. (iii) not product-favored at low temperatures, but productfavored at high temperatures. (iv) never product-favored. (a) \(2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) (b) \(\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{~g})+8 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 5 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{P}_{4}(\mathrm{~g})+10 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PF}_{5}(\mathrm{~g})\)

Step-by-Step Solution

Answer

(i) b; (ii) a and c; (iii) none; (iv) none.

Step 1: Analyze Reaction (a)

The reaction \( 2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \) is a formation of a dimer from two mono-molecules. This reaction reduces the number of gas moles, suggesting that \( \Delta S < 0 \) (entropy decreases). The reaction is exothermic because forming bonds releases energy, so \( \Delta H < 0 \). According to \( \Delta G = \Delta H - T \Delta S \), \( \Delta G \) will be negative at low temperatures because both \( \Delta H \) is negative and \( T \Delta S \) will not dominate if \( T \) is low.

Step 2: Analyze Reaction (b)

The combustion reaction \( \mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{~g})+8 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 5 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) is highly exothermic, with \( \Delta H < 0 \) and usually \( \Delta S > 0 \) due to an increase in the number of molecules of gas and resulting higher disorder. Therefore, \( \Delta G \) is negative at all temperatures, implying the reaction is always product-favored.

Step 3: Analyze Reaction (c)

For the reaction \( \mathrm{P}_{4}(\mathrm{~g})+10 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PF}_{5}(\mathrm{~g}) \), the entropy \( \Delta S < 0 \) because the number of moles of gas decreases. The reaction is also exothermic \( \Delta H < 0 \), as forming stable \( \mathrm{PF}_5 \) molecules releases energy. Lower temperatures will make \( \Delta G \) negative because the \(-T\Delta S\) term is less effective at lower temperatures compared to \( \Delta H \), suggesting product-favoredness at lower temperatures.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) plays a crucial role in determining whether a chemical reaction will be spontaneous or not. This thermodynamic potential measures the maximum reversible work that can be performed by a system at constant temperature and pressure. The formula for Gibbs Free Energy change is given by:
\[ \Delta G = \Delta H - T \Delta S \]
where:

• \( \Delta H \) is the change in enthalpy or heat content of the system
• \( T \) is the temperature in Kelvin
• \( \Delta S \) is the change in entropy or disorder

A negative \( \Delta G \) indicates a reaction that is favorable and spontaneous under the given conditions. Conversely, a positive \( \Delta G \) suggests the reaction is not spontaneous. Thus, Gibbs Free Energy helps us predict the extent to which a reaction might proceed.

Entropy

Entropy (\( \Delta S \)) is a measure of disorder or randomness in a system. In chemical reactions, entropy tends to increase as the number of possible arrangements or states of the system increases. When a reaction results in more gas molecules being formed, it usually implies an increase in entropy.
In the exercise's reaction (a), \( \Delta S \lt 0 \) as two \( NO_2 \) molecules combine to form a single \( N_2O_4 \) molecule. This decrease in the number of gas particles results in lower entropy. A negative entropy change suggests that the reaction could be less favorable as temperature increases, particularly in cases where enthalpy doesn't outweigh the entropy loss.

Enthalpy

Enthalpy (\( \Delta H \)) is the heat content of a system and is an indicator of the energy transferred as heat during a chemical reaction. It helps determine whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). In the reactions discussed:
- Reaction (a) and (c) are exothermic (\( \Delta H \lt 0 \)), as bond formation releases energy.- Reaction (b), being a combustion process, is also exothermic, indicating the release of considerable heat energy.
When a reaction is significantly exothermic, it can drive the reaction to be product-favored even if there is a decrease in entropy. Thus, the negative \( \Delta H \) can often compensate for a negative entropy change in making Gibbs Free Energy negative.

Product-favored Reactions

Product-favored reactions are those where products are thermodynamically more stable than the reactants, leading to a negative Gibbs Free Energy change. They can be subdivided based on the conditions under which they occur:

• **Always Product-Favored:**
  Reactions that are product-favored regardless of temperature have both a negative \( \Delta H \) and a positive \( \Delta S \), as seen in reaction (b). They remain spontaneous throughout.
• **Temperature-dependent Product-Favored:**
  Some reactions depend on temperature for product favorability. For example, reaction (a) is favored at lower temperatures due to a negative \( \Delta H \) but becomes less favorable as temperature increases because \( \Delta S \lt 0 \). Conversely, reaction (c) is more product-favored at lower temperatures for similar reasons.
• **Never Product-Favored:**
  A reaction is never product-favored if both \( \Delta H \) is positive and \( \Delta S \) is negative, leading to positive \( \Delta G \) at all temperatures.

Understanding these factors helps predict and manipulate reaction conditions to favor desired product outcomes.