Standard Enthalpy of Formation
The standard enthalpy of formation is like a building block for chemistry—it tells us how much energy is needed to create one mole of a compound from its basic elements in their standard states. Imagine constructing a molecule, brick by brick, and measuring the energy changes as you build it up from scratch. Whether it’s forming water from hydrogen and oxygen gas or creating carbon dioxide from carbon and oxygen. Let’s understand in detail.
1.0What is standard enthalpy
Standard enthalpy of formation (ΔH°f) is the amount of heat absorbed or released when one mole of a compound is formed from its elements in their standard states under standard conditions (1 atmosphere pressure and 298 K or 25°C temperature). The reference state of an element is its most stable form at these conditions. For example:
- Dihydrogen: H2(gas)
- Dioxygen: O2 (gas)
- Carbon: C (graphite)
- Sulfur: S (rhombic solid)
It is denoted by the symbol ΔfH∘.
- Δ: Represents a change in enthalpy.
- °: Indicates standard conditions (298 K, 1 bar).
- f: Refers to the formation of the substance from its elements.
2.0Examples of Enthalpy of Formation (ΔfH∘)
Some examples of reactions with their corresponding standard molar enthalpies of formation are as follows:
- H2(g) + 12O2(g) → H2O(l) ΔfH∘ = −285.8 kJ mol−1
- C(graphite,s) + 2H2(g) → CH4(g) ΔfH∘ = −74.81 kJ mol−1
- 2C(graphite,s) + 3H2(g) + 12O2(g) → C2H5OH (l) ΔfH∘ = −277.7 kJ mol−1
These equations represent the formation of 1 mole of water, methane, and ethanol from their constituent elements, which are at their standard states.
Not All Enthalpy Changes Are Standard Enthalpy of Formation:
It’s important to note that not all enthalpy changes represent the standard enthalpy of formation. For instance:
CaO(s) + CO2(g) → CaCO3(s) ΔfH∘=−178.3 kJ mol−1
This reaction is not the standard enthalpy of the formation of calcium carbonate because calcium carbonate is formed from other compounds (CaO and CO₂), not from its constituent elements.
Similarly: H2(g) + Br2(l) → 2HBr(g) ΔfH∘=−72.8 kJ mol−1
This is not the standard enthalpy of formation of HBr because it forms 2 moles of HBr instead of 1 mole. Therefore, dividing the coefficients by 2, the standard enthalpy of formation for 1 mole of HBr is: H2(g) + ½ Br2 (l) → HBr(g) ΔfH∘ =−36.4 kJ mol−1
3.0Key Points
- ΔfH∘ = 0 for all elements in their reference states
e.g., H2(g), O2(g),Cgraphite(s),Srhombic(s)H2(g), O2(g.
- Standard Enthalpy of Reaction from Standard Enthalpy of Formation:
The standard enthalpy of any reaction can be calculated from the standard enthalpy of formation of the products and reactants. The formula is:
ΔrH∘ = ΣΔfH∘(products) − ΣΔfH∘(reactants)
This equation states that the standard enthalpy change of a reaction is the sum of the standard enthalpies of the formation of the products minus the sum of the standard enthalpies of formation of the reactants.
For a general reaction: aA + bB → cC + dD
The enthalpy change is: ΔfH∘ = [cΔfH∘(C) + dΔfH∘(D)] − [aΔfH∘(A) + bΔfH∘(B)]
4.0Solved Examples of Standard Enthalpy of Formation
Q. Standard enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, –393, –81 and 10 kJ/mole, respectively. Find the ΔrH° for the reaction : N2O4(g) + 3CO → N2O(g) + 3CO2(g)
Solution:
ΔrH° = ΣΔfH°(products) – ΣΔfH° (reactants)
= [ΔfH°N2O(g) + 3×ΔfH°CO2(g)]−[ΔfH°N2O4(g)∘ + 3×ΔfH°CO(g)∘]
= [81 + 3 × (–3393)] – [10 + 3 × (–110)] = –778 kJ/mol.
Q. Between Br2(l) and Br2(g) at 298.15 K, which substance has a nonzero standard enthalpy of formation?
Solution:
At 298.15 K, Br2(l) is the more stable form, meaning it has a lower enthalpy, so its standard enthalpy of formation is ΔfH°= 0. Consequently, Br2(g) has a nonzero standard enthalpy of formation because it is less stable. Standard enthalpy of formation values for various compounds at standard conditions (usually 298 K and 1 bar)
Standard Enthalpy of Formation (ΔfH°) values: