Thermochemistry and Hess's Law Study Guide

Introduction to Thermochemistry

• Thermochemistry (Chemical Energetics): The branch of chemistry that studies the heat changes accompanying chemical reactions.

• Enthalpy Changes: Reactions can be classified based on whether heat is moving out of or into the system:

  • Exothermic: Heat energy is released to the surroundings.

  • Endothermic: Heat energy is absorbed from the surroundings.

Factors Affecting Enthalpy Change ($\Delta H$)

• Amount of Reactants or Products: The magnitude of heat change is directly proportional to the quantity of the substances involved.

  • Example: The heat released by the combustion of 2 moles of hydrogen is exactly double the heat released by the combustion of 1 mole of hydrogen.

• Physical States of Reactants and Products: The physical state (solid, liquid, or gas) of the substances involved significantly impacts the enthalpy change.

  • Example 1 (Formation of liquid water): $H_2 + \frac{1}{2}O_2 \rightarrow H_2O(l)$ where $\Delta H = -286\,kJ/mol$.

  • Example 2 (Formation of gaseous water): $H_2 + \frac{1}{2}O_2 \rightarrow H_2O(g)$ where $\Delta H = -242\,kJ/mol$.

• Temperature: Reactions occur at specific temperatures. In thermochemical measurements, the standard temperature is usually $25^{\circ}C$ ($298\,K$).

• Pressure: Standard pressure for these reactions is typically $1\,atm$ ($760\,mmHg$).

Standard Enthalpy Changes and Definitions

• Standard State: The most stable physical form of a substance under standard conditions ($298\,K$ and $1\,atm$ pressure).

• Standard Enthalpy Change of Reaction ($\Delta H_r^{\circ}$): The enthalpy change that occurs when reactants in their standard states react to form products in their standard states under standard conditions.

• Standard Enthalpy of Combustion ($\Delta H_c^{\circ}$):

  • Definition: The heat released when exactly one mole of a substance is completely burned in excess oxygen under standard conditions.

  • Sign: Since combustion reactions are always exothermic, the $\Delta H_c^{\circ}$ value is always negative.

Experimental Determination using a Bomb Calorimeter

• Setup:

  • A known mass ($m_{substance}$) of the substance is placed in a crucible inside a sealed steel bomb.

  • The bomb is filled with high-pressure oxygen and submerged in a known volume/mass of water ($m_{water}$) within an insulated calorimeter.

• Procedure:

  • Initial Temperature ($T_1$): The initial temperature of the water is recorded.

  • Ignition: The substance is ignited electronically.

  • Maximum Temperature ($T_2$): The temperature of the water rises as it absorbs the heat from the combustion. The peak temperature is recorded.

• Calculation Assumptions: It is assumed that the heat released by the combustion ($q_{combustion}$) is equal to the heat gained by the water ($q_{water}$), meaning heat absorbed by the bomb apparatus itself is considered negligible.

• Formulas:

  • $q_{water} = m \cdot c \cdot \Delta T$

  • Where $m$ is the mass of the water, $c$ is the specific heat capacity of water, and $\Delta T = T_2 - T_1$.

  • To find the molar enthalpy of combustion:

    • $\Delta H_c^{\circ} = -\frac{q_{water}}{n}$

    • The number of moles ($n$) is calculated as $\frac{m_{substance}}{M_w}$.

    • Final calculation form: $\Delta H_c^{\circ} = -\frac{m_{water} \cdot c \cdot (T_2 - T_1)}{m_{substance}/M_w}$.

  • The negative sign is added because the process is exothermic (heat is released).

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

• Definition: The enthalpy change when exactly one mole of a compound is formed from its constituent elements in their standard states under standard conditions.

• Reference Point: By definition, the standard enthalpy of formation for any element in its standard state is zero ($\Delta H_f^{\circ} = 0$).

• Implications of the Sign:

  • Negative $\Delta H_f^{\circ}$: Indicates an exothermic formation. The resulting compound is physically stable because it is at a lower energy level than its elements.

  • Positive $\Delta H_f^{\circ}$: Indicates an endothermic formation. The compound is less stable and its formation may not be spontaneous at room temperature.

• Note: Many $\Delta H_f^{\circ}$ values cannot be measured directly and must be determined via indirect methods.

Laws of Thermochemistry

• La Chatelier's Law (applied to Enthalpy): If a reaction is reversed, the magnitude of the enthalpy change remains the same, but the sign is reversed.

  • Example: If $A \rightarrow B$ has $\Delta H = -X\,kJ/mol$, then $B \rightarrow A$ has $\Delta H = +X\,kJ/mol$.

• Hess's Law of Constant Heat Summation:

  • Statement: If a reaction can be carried out in a series of steps, the total enthalpy change for the overall reaction is equal to the sum of the enthalpy changes for each individual step (provided conditions like temperature and pressure are constant).

  • Implication: The enthalpy change of a reaction is a state function: it depends only on the initial and final states and is independent of the chemical path taken.

Example Calculation: Methane Formation via Hess's Law

• Goal: Calculate the enthalpy of formation of methane ($CH_4$): $C(s) + 2H_2(g) \rightarrow CH_4(g)$.

• Given Data:

  1. Combustion of Carbon: $C(s) + O_2(g) \rightarrow CO_2(g)$ where $\Delta H_1 = -393\,kJ/mol$

  2. Combustion of Hydrogen: $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$ where $\Delta H = -286\,kJ/mol$

  3. Combustion of Methane: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ where $\Delta H = -890\,kJ/mol$

• Procedure:

  • Equation 1: Keep as is to provide 1 mole of carbon on the reactant side.

  • Equation 2: Multiply by 2 into $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$. The change is $\Delta H_2 = 2 \times (-286) = -572\,kJ/mol$.

  • Equation 3: Reverse the equation to make $CH_4$ a product: $CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)$. The change is $\Delta H_3 = +890\,kJ/mol$.

• Summation:

  • By summing the equations, $O_2$, $CO_2$, and $H_2O$ cancel from both sides.

  • Result: $C(s) + 2H_2(g) \rightarrow CH_4(g)$.

  • $\Delta H_f^{\circ}(CH_4) = \Delta H_1 + \Delta H_2 + \Delta H_3 = (-393) + (-572) + (+890) = -75\,kJ/mol$.

Other Enthalpy Change Classifications

• Standard Enthalpy of Atomization ($\Delta H_{atom}^{\circ}$): The energy required to convert one mole of a substance in its normal state into one mole of gaseous atoms. This is always an endothermic process (\Delta H > 0).

• Bond Dissociation Energy (BDE): The energy required to break one mole of a specific covalent bond in a diatomic molecule to form gaseous atoms. Process: $AB(g) \rightarrow A(g) + B(g)$.

• Bond Energy: The average energy required to break one mole of a specific type of bond across various gaseous molecules.

  • Bond Breaking: Endothermic (requires energy).

  • Bond Formation: Exothermic (releases energy).

• Relating Atomization and BDE: For a diatomic molecule, $\Delta H_{atom}^{\circ} = \frac{1}{2} \times BDE$. This is because BDE yields 2 moles of atoms, while atomization refers to the production of only 1 mole of atoms.

• Standard Enthalpy of Sublimation ($\Delta H_{sub}^{\circ}$): The energy required to convert one mole of a solid element directly into gaseous atoms under standard conditions. For solid elements, $\Delta H_{sub}^{\circ} = \Delta H_{atom}^{\circ}$.

Average Bond Energies and Cycle Diagrams

• Calculating Si-Cl Bond Energy Example:

  • Given:

    1. $\Delta H_f^{\circ}(SiCl_4(l)) = -610\,kJ/mol$

    2. $\Delta H_{atom}^{\circ}(Si(s)) = +338\,kJ/mol$

    3. $\Delta H_{atom}^{\circ}(Cl_2(g)) = +242\,kJ/mol$ (this is often used as the $\Delta H$ for $Cl_2(g) \rightarrow 2Cl(g)$).

  • Indirect Path Analysis:

    • $\Delta H_1 (Si(s) \rightarrow Si(g)) = +338\,kJ/mol$

    • $\Delta H_2 (2Cl_2(g) \rightarrow 4Cl(g)) = 2 \times (+242\,kJ/mol) = +484\,kJ/mol$

    • $\Delta H_3 (Si(g) + 4Cl(g) \rightarrow SiCl_4(l)) = ?$

  • Using Hess's Law:

    • $\Delta H_f^{\circ} = \Delta H_1 + \Delta H_2 + \Delta H_3$

    • $-610 = 338 + 484 + \Delta H_3$

    • $-610 = 822 + \Delta H_3 \rightarrow \Delta H_3 = -1432\,kJ/mol$

  • Bond Energy Calculation:

    • $\Delta H_3$ represents the formation of 4 moles of Si-Cl bonds.

    • Average Bond Energy = $\frac{-1432}{4} = -358\,kJ/mol$.

Predicting Reaction Enthalpy from Formation Data

• Formula:

  • $\Delta H_{rxn}^{\circ} = \sum (n \times \Delta H_f^{\circ} \text{ products}) - \sum (m \times \Delta H_f^{\circ} \text{ reactants})$

• Example: Hydrogenation of Ethane:

  • Reaction: $C_2H_6(g) + H_2(g) \rightarrow 2CH_4(g)$.

  • Given Values:

    • $\Delta H_f^{\circ}(CH_4(g)) = -74.8\,kJ/mol$

    • $\Delta H_f^{\circ}(C_2H_6(g)) = -84.7\,kJ/mol$

    • $\Delta H_f^{\circ}(H_2(g)) = 0\,kJ/mol$ (element in standard state).

  • Calculation:

    • $\Delta H_{rxn}^{\circ} = [2 \times (-74.8)] - [1 \times (-84.7) + 1 \times (0)]$

    • $\Delta H_{rxn}^{\circ} = (-149.6) - (-84.7)$

    • $\Delta H_{rxn}^{\circ} = -149.6 + 84.7 = -64.9\,kJ/mol$.", "title": "Thermochemistry and Hess's Law Study Guide"}