Thermochemistry and Standard Enthalpy Review

Exam Review Session: Led by LAs on Sunday from 2:00 PM to 3:30 PM via Zoom.
Zoom link is provided on Canvas; another announcement will be made on Sunday afternoon regarding the recording.
Calculator Requirement: ensure your calculator is charged and ready to use, as many students showed up without calculators last time.
Identifications Needed: Bring an ID and pencil to the exam.
CD Chart: Same as for the previous exam, and assigned seating will be according to the chart.

Definition: Formation enthalpy is the energy change associated with making one mole of a substance from elements in their standard states.
Standard States: e.g., nitrogen at 1 atm and room temperature is a diatomic gas (N2), similarly for oxygen (O2).
Notation: Enthalpy of formation is denoted as ( \Delta H_f^\circ ).
Example: The formation enthalpy for nitrogen monoxide (NO) is ( \Delta H_f^\circ = 90.3 \text{ kJ/mol} ); thus, making one mole of NO from its elemental states requires 90.3 kJ.
Key Point: Enthalpies of formation for elements in their standard states are defined to be zero. Thus, for gaseous nitrogen or oxygen, ( \Delta H_f^\circ = 0 \text{ kJ/mol} ).

Formula: [ \Delta H = \Sigma (\Delta Hf^\circ \text{ of products}) - \Sigma (\Delta Hf^\circ \text{ of reactants}) ]
- The ( \Sigma ) denotes summation of energies involved.
- Final enthalpy change is calculated by taking the sum of the product energies and subtracting from it the sum of the reactant energies.

Problem Statement: Combust 1 mole of propane (C3H8) with oxygen to form CO2 and water.
Data from Table:
- ( \Delta H_f^\circ \text{ for C3H8} = -103.8 \text{ kJ/mol} )
- ( \Delta H_f^\circ \text{ for CO2} = -393.5 \text{ kJ/mol} ) (3 moles)
- ( \Delta H_f^\circ \text{ for H2O(l)} = -285.8 \text{ kJ/mol} ) (4 moles)
Calculation Steps:
1. Calculate total energy for products:
- For CO2: ( 3 \times (-393.5) = -1180.5 \text{ kJ} )
- For H2O: ( 4 \times (-285.8) = -1143.2 \text{ kJ} )
1. Total energy for products = -1180.5 + -1143.2 = -2323.7 kJ
2. Total energy for reactants:
- For C3H8: ( -103.8 \text{ kJ} )
- For O2: 0 kJ (zero for elemental state)
1. Calculate ( \Delta H ):
- ( \Delta H = -2323.7 - (-103.8) = -2219.9 \text{ kJ} )
Conclusions: The process is exothermic, releasing 2219.9 kJ of energy.

Summation, keeping track of coefficients, and ensuring proper signs lead to knowing standard values, including zero enthalpy for elemental states.
Problems require careful tracking of energy signs when summing or subtracting.

System Work: ( U = q + w ) where ( q ) is heat and ( w ) is work.
- Work done on the system: +1151 kJ
- Heat released from the system: -526 kJ
Calculating Internal Energy:
- Total change in internal energy = +1151 - 526 = +625 kJ.
- Key Insight: Even with heat released, if significant work is done on a system, the internal energy could still increase.

Heat exchange problem: Hot metal in cool water
- Given: 14.9 grams of metal (81.6 °C) placed in 88 g of water (initially at much lower temperature).
- Specific Heat Equation: ( q{ ext{metal}} = - q{ ext{water}} )
- Set up: (- (m{ ext{metal}} imes c{ ext{metal}} imes (T{ ext{final}} - T{ ext{initial, metal}}) = m{ ext{water}} imes c{ ext{water}} imes (T{ ext{final}} - T{ ext{initial, water}}))
- Solving these simultaneous equations results in a calculated final temperature.

When calculating for mass and energy exchanges, ensuring attention to units both in kJ and grams is crucial for correct calculations.
When setting up systems, identifying which is gaining versus losing heat or energy helps clarify the direction of the calculation.

Outline is given for solutions based on variations of Hess's Law calculation strategies.
Also covered: General techniques for thermochemical equations, including adjusting reactions and understanding each reaction's contribution to enthalpy changes.