Thermochemical Equations

Thermochemical equations and their rules are fundamental for understanding energy changes in chemical reactions. By using these equations, we can determine whether a reaction absorbs or releases heat. Hess's Law further aids in calculating the enthalpy changes for complex reactions by breaking them down into simpler, more manageable steps, making it an invaluable tool in thermochemistry.

1.0What is a Thermochemical Equation

A thermochemical equation represents a balanced chemical reaction along with the physical states of all reactants and products, as well as the heat change involved in the reaction. This equation provides information about the amount of energy absorbed or released during a chemical reaction.

2.0Components of a Thermochemical Equation

1. Balanced Chemical Equation: Indicates the stoichiometric ratio of reactants and products.
2. Physical States: Denoted by symbols such as (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solutions.
3. Enthalpy Change (ΔH): Shows whether the reaction is exothermic (releases heat) or endothermic (absorbs heat). It is expressed in kilojoules per mole (kJ/mol).

Here is an example of a Thermochemical Equation:

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l) + 890.4 kJ

3.0Types of Reactions

1. Exothermic Reaction:

A reaction that releases heat energy into the surroundings.

Example: Combustion of Ethanol               

 C₂H₅OH(ℓ) + 3O₂(g) → 2CO₂(g) + 3H₂O (ℓ)    Δ_rH^O=−1367 kJ/mol

The negative sign indicates that 1367 kJ of heat is released per mole of ethanol combusted.

Alternate Forms of the Thermochemical Equation: 

C₂H₅OH(ℓ) + 3O₂(g) → 2CO₂(g) + 3H₂O(ℓ) + 1367 kJ

C₂H₅OH (ℓ) + 3O₂(g) −1367 kJ → 2CO₂(g) + 3H₂O (ℓ)

These different representations all convey that the reaction releases 1367 kJ of energy.

2. Endothermic Reaction:

A reaction that absorbs heat energy from the surroundings.

Example: Dissolution of Copper(II) Sulfate Pentahydrate 

CuSO₄⋅5H₂O(s) + aq → CuSO₄ (aq) Δ_solH=11.71 kJ

The positive sign indicates that 11.71 kJ of heat is absorbed during the dissolution.

Alternate Forms of the Thermochemical Equation: 

CuSO₄⋅5H₂O (s) + aq → CuSO₄(aq) − 11.71 kJ

CuSO4​⋅5H₂​O (s) + aq + 11.71kJ → CuSO₄​(aq) 

These representations show that 11.71 kJ of energy is absorbed during the process.

Important Points:

1. Enthalpy Change (ΔH): Indicates whether the reaction is exothermic (negative ΔH) or endothermic (positive ΔH).
2. Coefficients: The numbers in front of the chemical species refer to the number of moles of each substance involved in the reaction. For example, in the combustion of ethanol:

• 1 mole of C₂H₅OH(ℓ)
• 3 moles of O₂(g)
• 2 moles of CO₂(g)
• 3 moles of H₂O(ℓ)

These coefficients define the proportion in which reactants combine and products are formed, and they are crucial for calculating the heat change per mole of reaction.

4.0Thermochemical Equations and Rules

Thermochemical equations provide information about the heat changes associated with chemical reactions. Understanding these equations and their rules is essential for predicting the energy requirements or releases during reactions.

Rule #1: Proportionality of Enthalpy Change (ΔH)

• The enthalpy change (ΔH) is directly proportional to the amount of reactants and products involved in the reaction.
• For example:
• • When one mole of ice melts, ΔH=+6.00 kJ.
  • When one gram of ice melts, ΔH=+0.333 kJ/g.

Example Reactions:

1. H₂(g) + Cl₂(g) → 2HCl(g);ΔH=−185 kJ
2.  12 H₂(g) + 12 Cl₂(g) → HCl(g);ΔH=−92.5 kJ

Rule #2: Reversibility of Enthalpy Change

• The enthalpy change for a reaction is equal in magnitude but opposite in sign to the enthalpy change of the reverse reaction.

Example Reactions:

1. 2HgO(s) → 2Hg(l)+O₂ (g) ; ΔH = +182 kJ
2. 2Hg(l) + O₂(g) →2HgO (s) ; ΔH = −182 kJ

Rule #3: Independence of Reaction Path

• The enthalpy change (ΔH) is independent of the pathway taken. It depends only on the initial and final states of the reaction, not on the steps taken to get there. This is known as Hess's Law.

5.0Hess’s Law and Calculating Enthalpy Change

Hess’s Law is useful for calculating ΔH for reactions that are difficult to carry out directly. It states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction pathway.

Example:

Target Reaction: C(s) + 12O₂(g) → CO(g);  ΔH₁=?

Given Reactions:

1. CO(g) + 12 O₂(g) → CO₂(g); ΔH₂=−283.0 kJ
2. C(s)+O₂(g) → CO₂(g); ΔH₃=−393.5 kJ

Steps to Calculate ΔH₁​:

1. Write the given reaction (#3):

C(s) + O₂(g) → CO₂(g); ΔH₃ = −393.5 kJ

2. Reverse the given reaction (#2):

CO₂(g) → CO(g) + 12O₂(g); −ΔH₂ = +283.0 kJ

3. Add the two reactions:

C(s) + 12O₂(g) → CO(g); ΔH₁=−393.5 kJ + 283.0 kJ

ΔH_1​=−110.5kJ

Thus, the enthalpy change for the reaction C(s)+12O₂(g) → CO(g) is ΔH₁=−110.5 kJ