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:
- Balanced Chemical Equation: Indicates the stoichiometric ratio of reactants and products.
- Physical States: Denoted by symbols such as (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solutions.
- 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:
- Exothermic Reaction:
- A reaction that releases heat energy into the surroundings.
- Example: Combustion of Ethanol
C2H5OH(ℓ) + 3O2(g) → 2CO2(g) + 3H2O (ℓ) ΔrHO=−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.
- Endothermic Reaction:
- A reaction that absorbs heat energy from the surroundings.
- Example: Dissolution of Copper(II) Sulfate Pentahydrate
CuSO4⋅5H2O(s) + aq → CuSO4 (aq) ΔsolH=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⋅5H2O (s) + aq + 11.71kJ → CuSO4(aq)
- These representations show that 11.71 kJ of energy is absorbed during the process.
Important Points:
- Enthalpy Change (ΔH): Indicates whether the reaction is exothermic (negative ΔH) or endothermic (positive ΔH).
- 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:
- H2(g) + Cl2(g) → 2HCl(g);ΔH=−185 kJ
- 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:
- 2HgO(s) → 2Hg(l)+O2 (g) ; ΔH = +182 kJ
- 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:
- CO(g) + 12 O2(g) → CO2(g); ΔH2=−283.0 kJ
- C(s)+O2(g) → CO2(g); ΔH3=−393.5 kJ
Steps to Calculate ΔH1:
- Write the given reaction (#3):
C(s) + O2(g) → CO2(g); ΔH3 = −393.5 kJ - 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
ΔH1=−110.5kJ
Thus, the enthalpy change for the reaction C(s)+12O2(g) → CO(g) is ΔH1=−110.5 k