Thermochemistry Equations & Formulas - Lecture Review & Practice Problems
Thermochemistry Overview
Focus on thermochemistry and important equations relevant to the subject.
Internal Energy
Internal Energy Change Equation:
[ \Delta E = Q + W ]
Q: Heat energy entering (+) or leaving (-) the system.
W: Work done on (+) or by (-) the system.
Example of Q
System at 100°C and surroundings at 50°C:
Heat flows from the system to surroundings.
Therefore, Q is negative for the system (exothermic) and positive for surroundings (endothermic).
Measurement Units:
1 kJ = 1,000 J
1 Cal = 4.184 J
1 kcal = 1,000 Cal
Work (W)
Equation for Work:
[ W = P \Delta V ]
Positive W occurs when work is done on the system and negative W when done by the system.
Expansion or Compression:
Expansion: Gas does work, leads to negative W.
Compression: Force applied to gas, leads to positive W.
Example Calculation
300 J of heat absorbed (Q = +300), gas expands from 2 L to 3 L with P = 5 atm:
[ \Delta V = 3 \text{L} - 2 \text{L} = 1 \text{L} ]
Convert pressure-volume work into Joules:
[ W = P \Delta V = 5 \text{atm} \times 1 \text{L} = 5 \text{atm} \cdot \text{L} = 5 \times 101.3 \text{J} = 506.5 \text{J} ]
Thus, W = -506.5 J (work done by gas).
Calculate change in internal energy:
[ \Delta E = Q + W = 300 - 506.5 = -206.5 \text{J} ]
Energy lost = -206.5 J.
Calculating Heat (Q)
Heat equation:
[ Q = mC \Delta T ]
m: Mass (g), C: Specific heat capacity (J/g°C), ΔT: Change in temperature (°C).
Example: Heating 50 g of water from 25°C to 75°C:
[ \Delta T = 75 - 25 = 50 \text{°C} ]
Calculate energy:
[ Q = 50 \text{g} \times 4.184 \text{J/g°C} \times 50 °C = 10,460 ext{J} ]
Phase Changes
Heat absorbed or released during phase changes:
[ Q = m \Delta H ] (where ΔH refers to heat of fusion/vaporization).
Example: Melting 54 g of ice at 0°C:
Molar mass of water = 18 g/mol, heat of fusion for water = 6 kJ/mol.
Convert grams into moles and then calculate kJ:
[ \text{Moles of ice} = \frac{54\text{g}}{18\text{g/mol}} = 3 \text{mol} ]
[ Q = 3 \text{mol} \times 6 \text{kJ/mol} = 18 \text{kJ} ]
Thermochemical Equations
Example: Combustion of propane:
Balanced reaction releases 12,200 kJ of heat.
Exothermic if heat is a product.
Question: How many kJ released by 64 g of O2?
Molar mass O2 = 32 g/mol.
Use mole ratio to find kJ per moles:
12,200 kJ for every 5 moles of O2.
Calculate:
[ ext{Moles of O2} = \frac{64}{32} = 2 \text{mol} ]
[ \text{Heat} = \frac{2 \times 12,200}{5} = 4,880 \text{kJ} ]
Hess's Law
Used to find enthalpy from enthalpies of other reactions:
Sum of products – sum of reactants.
Example Reaction: 2 A + 2 B → D + E
Two additional reactions provided to find required reaction enthalpy.
Steps:
Adjust coefficients as needed and reverse reactions ensuring enthalpy sign changes accordingly.
Gather information, add, and simplify to find enthalpy of desired reaction.
Conclusion
Key equations and principles of thermochemistry important for mastering concepts in energy exchanges, work, and phase changes.