Enthalpies of Formation
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Core Concept
Enthalpy of Formation (ΔH∘f) is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (298 K and 1 atm). Standard enthalpies of formation are essential for calculating the enthalpy changes of reactions.
Practice Tips
Use Standard Enthalpies: Always use ΔH∘f values from standard tables to ensure accuracy.
Check Physical States: Ensure each substance’s physical state (solid, liquid, gas) matches its enthalpy of formation, as ΔH∘f values vary by state.
Remember Elements Have DeltaH∘f=0: Pure elements in their standard states have a formation enthalpy of zero.
Multiply by Stoichiometric Coefficients: Don’t forget to multiply each ΔH∘f∘ by its coefficient from the balanced chemical equation.
Calculating Enthalpy of a Reaction Using Enthalpies of Formation
The enthalpy change of a reaction can be calculated using the enthalpies of formation of the reactants and products:
ΔH∘reaction=∑ΔH∘f(products)−∑ΔH∘f(reactants)
Identify the Products and Reactants:
Write the balanced chemical equation for the reaction.
Find the Enthalpies of Formation:
Look up the ΔH∘f values for each reactant and product in a table.
Multiply by Coefficients:
Multiply each ΔH∘f value by its coefficient in the balanced equation.
Subtract the Sum of Reactants from Products:
Sum the enthalpies of formation for the products and reactants separately, then subtract the total for the reactants from the total for the products.
Key Concepts
Standard Enthalpy of Formation (ΔH∘f):
The enthalpy change when 1 mole of a compound is formed from its elements in their most stable forms at 1 atm pressure and 298 K.
Symbol: ΔH∘f (the degree symbol ∘∘ denotes standard conditions).
Measured in kilojoules per mole (kJ/mol).
Standard State:
Elements are in their most stable physical forms at 1 atm and 298 K (e.g., O2 for oxygen, N2 for nitrogen, and graphite for carbon).
Enthalpy of Formation of Elements:
The standard enthalpy of formation of any pure element in its standard state is zero.
Example: ΔH∘f for O2(g) and N2(g) is zero.
Applications:
Enthalpies of formation are used to calculate the enthalpy change for chemical reactions.
Hess’s Law and standard enthalpies of formation can be combined to find ΔH∘reaction.
Example Problem
Calculate the enthalpy change (ΔHrxn) for the combustion of methane (CH4) given the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Use the standard enthalpies of formation (ΔH∘f):
ΔH∘f(CH4(g))=−74.8kJ/mol
ΔH∘f(O2(g))=0kJ/mol (element in its standard state)
ΔH∘f(CO2(g))=−393.5kJ/mol
ΔH∘f(H2O(l))=−285.8kJ/mol
Step-by-Step Solution:
Write the formula for ΔHrxn∘:
ΔH∘rxn=∑(ΔH∘f of products)−∑(ΔH∘f of reactants)
Substitute the values for the products:
Products: CO2(g) and 2H2O(l)
∑(ΔHf∘ of products)=[1⋅ΔHf∘(CO2(g))]+[2⋅ΔHf∘(H2O(l))]\sum (\Delta H^\circ_f \text{ of products}) = [1 \cdot \Delta H^\circ_f (\text{CO}_2(g))] + [2 \cdot \Delta H^\circ_f (\text{H}_2\text{O}(l))]∑(ΔHf∘ of products)=[1⋅ΔHf∘(CO2(g))]+[2⋅ΔHf∘(H2O(l))] =[1⋅(−393.5)]+[2⋅(−285.8)]= [1 \cdot (-393.5)] + [2 \cdot (-285.8)]=[1⋅(−393.5)]+[2⋅(−285.8)] =−393.5+(−571.6)=−965.1 kJ= -393.5 + (-571.6) = -965.1 \, \text{kJ}=−393.5+(−571.6)=−965.1kJ
Substitute the values for the reactants:
Reactants: CH4(g)\text{CH}_4(g)CH4(g) and 2O2(g)2\text{O}_2(g)2O2(g)
\sum (\Delta H^\circ_f \text{ of reactants}) = [1 \cdot \Delta H^\circ_f (\text{CH}_4(g))] + [2 \cdot \Delta H^\circ_f (\text{O}_2(g))]
=[1⋅(−74.8)]+[2⋅(0)]
=−74.8+0=−74.8 kJ
Calculate ΔH∘rxn:
ΔH∘rxn=∑(ΔH∘f of products)−∑(ΔH∘f of reactants)
=−965.1−(−74.8)= -965.1 - (-74.8)
=−965.1+74.8=−890.3 kJ
