Chemical Energetics

This section explores the changes in energy that occur during chemical reactions. We will cover enthalpy changes, bond energies, Hess’s Law, and calorimetry.

Enthalpy Changes

Definition

Enthalpy (H) is a state function that represents the total heat content of a system. A change in enthalpy (ΔH) during a chemical reaction is the heat absorbed or released at constant pressure.

$$ \Delta H = H_{products} - H_{reactants} $$

Exothermic and Endothermic Reactions

• Exothermic Reactions: Reactions that release heat to the surroundings. ΔH is negative.
• Endothermic Reactions: Reactions that absorb heat from the surroundings. ΔH is positive.

Enthalpy and Bond Energies

The enthalpy change of a reaction is related to the changes in bond energies. Breaking bonds requires energy (endothermic), and forming bonds releases energy (exothermic).

$$ \Delta H = \sum (\text{bond energies of reactants}) - \sum (\text{bond energies of products}) $$

The sign of ΔH indicates whether the overall process is exothermic or endothermic.

Hess’s Law

Statement

Hess’s Law states that the enthalpy change for a reaction is independent of the pathway taken. This means that the overall enthalpy change can be calculated using the enthalpy changes of a series of individual reactions that add up to the overall reaction.

Applying Hess’s Law

1. Write down the reactions that, when added together, give the overall reaction.
2. Multiply each reaction by a factor so that when added, the desired reactions cancel out.
3. Add the enthalpy changes of the modified reactions. The sum is the enthalpy change for the overall reaction.

Example

Consider the enthalpy change for the formation of carbon dioxide from carbon and oxygen:

$$ C(s) + O_2(g) \rightarrow CO_2(g) \qquad \Delta H_1 = ? $$

We can use Hess's Law to determine this by combining two other reactions:

1. $$ C(s) + O_2(g) \rightarrow CO_2(g) \qquad \Delta H_2 = -393.5 kJ/mol $$
2. $$ C(g) + O_2(g) \rightarrow CO(g) \qquad \Delta H_3 = -283.0 kJ/mol $$

To obtain the desired reaction, we need to reverse reaction 3 and multiply it by 2:

1. Reversed reaction 3: $$ CO(g) \rightarrow C(s) + O_2(g) \qquad \Delta H_4 = +283.0 kJ/mol $$
2. Multiply reaction 4 by 2: $$ 2CO(g) \rightarrow 2C(s) + 2O_2(g) \qquad \Delta H_5 = +566.0 kJ/mol $$

Adding reactions 2, 4 and 5 gives:

$$ C(s) + O_2(g) \rightarrow CO_2(g) \qquad \Delta H_1 = \Delta H_2 + \Delta H_4 + \Delta H_5 = -393.5 + 283.0 + 566.0 = 455.5 kJ/mol $$

Calorimetry

Definition

Calorimetry is the process of measuring the heat transferred during a chemical reaction. A calorimeter is an insulated container used for this purpose.

Types of Calorimeters

• Constant-Volume Calorimeter: Used for reactions where the volume doesn't change significantly. Also known as a bomb calorimeter.
• Constant-Pressure Calorimeter: Used for reactions where the pressure is kept constant. This is typically a coffee-cup calorimeter.

Heat Transfer and Specific Heat Capacity

The heat transferred (q) during a reaction can be calculated using the following equation:

$$ q = mc\Delta T $$

Where:

• q = heat transferred (in Joules or calories)
• m = mass of the solution (in grams)
• c = specific heat capacity of the solution (in J/g°C or cal/g°C)
• ΔT = change in temperature (in °C)

Calculations with Calorimetry

• Constant-Volume Calorimetry: The heat absorbed or released by the reaction is equal to the heat absorbed or released by the calorimeter. $$ q_{reaction} = -q_{calorimeter} $$
• Constant-Pressure Calorimetry: The heat absorbed or released by the reaction is equal to the heat absorbed or released by the calorimeter plus the heat absorbed or released by the solution. $$ q_{reaction} = -q_{calorimeter} - q_{solution} $$

Example

A 50.0 g solution of water with a specific heat capacity of 4.18 J/g°C is placed in a calorimeter. 25.0 g of a substance is added, and the temperature rises from 20.0 °C to 28.5 °C. The calorimeter itself absorbs 25.0 J of heat. Calculate the enthalpy change of the reaction.

Quantity	Value	Units
Mass of solution (m)	50.0	g
Specific heat capacity of water (c)	4.18	J/g°C
Initial temperature (Ti)	20.0	°C
Final temperature (Tf)	28.5	°C
Heat absorbed by solution (qsolution)	$q_{solution} = mc\Delta T = 50.0 \times 4.18 \times (28.5 - 20.0) = 50.0 \times 4.18 \times 8.5 = 1708.5$	J
Heat absorbed by calorimeter (qcalorimeter)	25.0	J
Heat absorbed by reaction (qreaction)	$q_{reaction} = -q_{solution} - q_{calorimeter} = -1708.5 - 25.0 = -1733.5$	J

The enthalpy change of the reaction is -1733.5 J/mol.