Thermochemistry: Study Notes and Practice Problems

Thermochemistry

Introduction to Thermochemistry

Thermochemistry is the study of the energy and heat associated with chemical reactions and physical transformations. It focuses on the transfer of energy as heat and work, and the changes in enthalpy during chemical processes.

Specific Heat and Heat Calculations

• Specific Heat Capacity (C): The amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or Kelvin). Units: J/g·°C.

• Heat (q): The energy transferred due to a temperature difference. Calculated using:

$q = m \cdot C \cdot \Delta T$

• m: mass of the substance (g)

• C: specific heat capacity (J/g·°C)

• ΔT: change in temperature (°C or K)

Example: Calculate the heat needed to raise the temperature of 5.78 g of silver from 18°C to 35°C, given C = 0.235 J/g·°C.

$q = 5.78 \times 0.235 \times (35 - 18) = 23.1 \text{ J}$

Heat Transfer in Water

• Water has a relatively high specific heat capacity, making it effective for thermal regulation.

• To calculate the heat absorbed or released by water:

$q = m \cdot C \cdot \Delta T$

• Example: How much heat does 45.5 g of water absorb when heated from 22.5°C to 43.1°C? (C = 75.291 J/mol·K; convert to J/g·°C if needed.)

Enthalpy of Reaction and Combustion

• Enthalpy Change (ΔH): The heat change at constant pressure during a chemical reaction.

• Combustion Reaction: A reaction where a substance reacts with oxygen to produce heat and light.

• Standard enthalpy of combustion is usually given per mole of substance combusted.

Example: Calculate the heat evolved in the complete combustion of 25.0 g of ethanol (C2H5OH), given ΔHcomb = –1367 kJ/mol.

$\text{Moles of ethanol} = \frac{25.0}{46.07} = 0.543 \text{ mol}$ $q = 0.543 \times (-1367) = -743 \text{ kJ}$

Formation Reactions

• Formation Reaction: A reaction in which one mole of a compound is formed from its elements in their standard states.

• Only one product is formed, and all reactants are elements in their standard states.

Example: Which of the following is a formation reaction?

Option	Reaction	Formation Reaction?
A	C (diamond) → C (graphite)	No
B	H2(g) + O2(g) → H2O2(l)	Yes
C	C2H2(g) + H2(g) → C2H4(g)	No
D	Ca(s) + O2(g) + CO2(g) → CaCO3(s)	Yes

Additional info: Only reactions forming one mole of a compound from elements in their standard states are considered formation reactions.

Hess's Law and Enthalpy Calculations

• Hess's Law: The total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in.

• Enthalpy changes can be calculated using standard enthalpies of formation (ΔHf°).

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

Example: Calculate ΔH for the reaction: NH4NO3(s) → N2O(g) + 2 H2O(l), given ΔHf° values:

Substance	ΔHf° (kJ/mol)
NH4NO3(s)	–365.6
N2O(g)	82.05
H2O(l)	–285.83

$\Delta H_{rxn} = [82.05 + 2(-285.83)] - [-365.6] = [82.05 - 571.66] + 365.6 = -489.61 + 365.6 = -124.01 \text{ kJ}$

Enthalpy Change from Multiple Reactions

• When given multiple reactions and their enthalpy changes, you can combine them to find the enthalpy change for a target reaction.

• Reverse or multiply reactions as needed, adjusting ΔH accordingly.

Example: Given:

Reaction	ΔH (kJ)
2C(s) + 3H2(g) → C2H6(g)	–84.7
C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(l)	–1550.6
C(s) + O2(g) → CO2(g)	–393.5
H2(g) + 1/2 O2(g) → H2O(l)	–285.8

Calculate ΔH for the formation of C2H6(g) from its elements.

$\Delta H = -84.7 \text{ kJ}$

Additional info: This is an application of Hess's Law, where reactions are manipulated to yield the desired overall reaction.