Thermodynamics: Hess's Law, Enthalpies of Formation, and Chemical Proofs

Introductory Remarks and Preliminary Concepts

• The lecture begins with a brief reference to a specific value: $H_1$ is equal to $100\,millivolts$.
• A generic reaction sequence is mentioned: $z + 2q \rightarrow y$.

Manipulating Thermochemical Equations (Hess's Law Context)

• The concept of a "proof" in thermodynamics involves manipulating known chemical reactions to match a target reaction and then summing their enthalpy changes ($\Delta H$).
• The professor distinguishes between the work required for a test versus the work required for homework:
  • On a test: Focus on identifying whether to flip a reaction or multiply it by a coefficient to find the final enthalpy value quickly.
  • For homework: A full "proof" is required, showing the manipulation of the full chemical equations and the subsequent cancellation of intermediate species.

Example Case Study 1: Sulfur Oxides

• Target Reaction: The goal is to derive the reaction associated with $2SO_3 \rightarrow 2SO_2 + O_2$.
• Step 1: Sulfur Dioxide Formation
  • Initial reaction: $S + O_2 \rightarrow SO_2$ with a value of $-296.8\,kJ$.
  • Manipulation: The product side must match the target. Because the target has a coefficient of 2 for $SO_2$, the initial reaction is multiplied by 2.
  • New enthalpy value: $-296.8\,times 2 = -593.6\,kJ$.
• Step 2: Sulfur Trioxide Formation
  • Initial reaction: $S + \frac{3}{2}O_2 \rightarrow SO_3$ with a value of $-395.6\,kJ$.
  • Manipulation: Since $SO_3$ is on the product side in the initial reaction but needed on the reactant side for the target, the reaction must be flipped (sign change). It must also be multiplied by 2 to match the coefficient.
  • New enthalpy value: Change $-395.6\,kJ$ to positive and multiply by 2, resulting in $+791.2\,kJ$.
• Final Summation:
  • Adding the adjusted values: $-593.6\,kJ + 791.2\,kJ = +197.6\,kJ$.
• The Proof:
  • Reaction 1 updated: $2S + 2O_2 \rightarrow 2SO_2$
  • Reaction 2 updated: $2SO_3 \rightarrow 2S + 3O_2$
  • Cancellation process: The $2S$ on both sides cancels out. The $2O_2$ on the reactant side cancels against portion of the $3O_2$ on the product side, leaving $1O_2$ on the product side.
  • Resulting equation: $2SO_3 \rightarrow 2SO_2 + O_2$. This matches the target, proving the validity of the numerical manipulations.

Example Case Study 2: Formation of Heptane ($C7H16$)

• Target Reaction: Synthesis of heptane from its base elements: $7C + 8H_2 \rightarrow C7H16$.
• Manipulation Strategy:
  • Process 1: A reaction involving heptane is flipped to put heptane on the correct side ($C7816$ is used in the transcript). The enthalpy for this step is provided as a positive $+4817\,kJ$.
  • Process 2: Carbon combustion. The carbon is on the correct side, but the reaction needs to be multiplied by 7. Calculation: $-393.5 \times 7 = -2754.5\,kJ$.
  • Process 3: Hydrogen reaction. This needs to be multiplied by 4 to achieve the required amount of hydrogen. Calculation: $-488.6 \times 4 = -1934.4\,kJ$.
• Calculation of $\Delta H$ for the reaction:
  • Total Sum: $4817 - 2754.5 - 1934.4 = 128.1\,kJ/mol$.
• The Proof Mechanism:
  • Flip Reaction: $7CO_2 + 8H_2O \rightarrow C7H16 + 11O_2$
  • Carbon Reaction: $7C + 7O_2 \rightarrow 7CO_2$
  • Hydrogen Reaction: $8H_2 + 4O_2 \rightarrow 8H_2O$ (Note: The speaker corrects that there are 4 oxygens involved here, summing to 11 oxygens on one side to cancel the 11 on the other).
  • Final additions: Carbon dioxides cancel, waters cancel, and all 11 oxygens cancel out, leaving $7C + 8H_2 \rightarrow C7H16$.

Standard Enthalpy of Formation ($\Delta H_f^\circ$)

• Defining "Standard":
  • Refers to the "standard state" of a substance.
  • Conditions: Room temperature ($25\,^\circ C$ or $298\,K$) and one atmosphere ($1\,atm$).
• Enthalpy of Formation Definition: The heat involved in forming exactly one mole of a compound from its constituent elements at standard temperature and pressure.
• Notation: The symbol is $\Delta H_f^\circ$, where the zero ($^\circ$) indicates standard conditions and the "f" indicates formation.
• Zero-Enthalpy Rule for Elements:
  • The standard enthalpy of formation for any element in its most stable form at standard conditions is defined as zero ($0$).
  • This is because elements already exist in nature; they are not "formed" from other substances.
  • Elements are typically omitted from enthalpy tables for this reason; students are expected to know they are zero.

Identifying Correct Formation Reactions

In determining the correct representation for the standard enthalpy of formation of a compound, several criteria must be met:

1. Diatomic Reality: Elements must be written in their natural standard states (e.g., nitrogen exists as $N_2$, not $N$; oxygen as $O_2$, not $O$).
2. Molar Constraint: The reaction must produce exactly one mole of the target compound. If the reaction produces two moles, it does not represent the standard enthalpy of formation for that compound.
3. Elemental Reactants: The reactants must be pure elements, not other compounds. If a compound is used to make another compound, it is a standard reaction but not an enthalpy of formation reaction.

Calculating $\Delta H_{rxn}$ Using Formation Tables

• Data is collected from experiments and organized into tables for various compounds.
• State Specificity: Values differ based on the state of matter. For example, liquid water and gaseous water have different formation enthalpies. It is critical to use the correct state from the table.
• The Fundamental Formula:
  • $\Delta H_{rxn} = \sum n \Delta H_f^\circ(\text{products}) - \sum m \Delta H_f^\circ(\text{reactants})$
  • This is effectively "Final minus Initial."

Example: Combustion of Propane ($C_3H_8$)

• Reaction: Propane reacts with oxygen to form carbon dioxide and liquid water ($C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$).
• Setup:
  • Sum of products: $3 \times \Delta H_f^\circ(CO_2) + 4 \times \Delta H_f^\circ(H_2O(l))$
  • Sum of reactants: $1 \times \Delta H_f^\circ(C_3H_8) + 5 \times \Delta H_f^\circ(O_2)$
  • Note: $\Delta H_f^\circ(O_2) = 0$
• Calculation Result: The result for this specific combustion is $-2,219.9\,kJ\,mol^{-1}$.
• Interpretation: This energy is released per mole of propane, per 5 moles of oxygen, etc., according to the stoichiometry of the reaction.

Questions & Discussion

• Q: If you have different states of matter, would they still be able to cancel out [in a proof]?
• A: No. States of matter are critically important. If you have liquid water on one side and gaseous water on the other, they do not cancel each other out. You must account for the specific state.
• General Advice for Tests: Students are required to memorize all thermo-related equations. Writing them out repeatedly during homework is recommended as a memorization strategy.
• Test Strategy Warning: Multiple-choice questions often include an option for "not enough information given." Students frequently mistakenly choose this when an element is missing from a provided table, forgetting that the value for elements is zero by definition.