Hess's law

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A visual representation of the law in action.

Hess's law is a law of physical chemistry named for Germain Hess's expansion of the Hess Cycle and used to predict the enthalpy change and conservation of energy (denoted as state function ΔH) regardless of the path through which it is to be determined.

1 Explanation
2 Extension to entropy and free energy
3 Use
- 3.1 Typical use
- 3.2 Example
4 External links

[edit] Explanation

The law states that because enthalpy is a state function, the enthalpy change of a reaction is the same regardless of what pathway is taken to achieve the products. In other words, only the start and end states matter to the reaction, not the individual steps between.

"The total energy change for a chemical reaction is independent of the route by which the reaction takes place, provided initial and final conditions are the same." (Chemistry 1 book, Cambridge University Press)

This allows the change in enthalpy for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing arithmetic operations on chemical equations. Chemical equations may be multiplied (or divided) by a whole number. When an equation is multiplied by a constant, its ΔH must be multiplied by the same number as well. If an equation is reversed, ΔH for the reaction must also be reversed (i.e. -ΔH).

Addition of chemical equations can lead to a net equation. If enthalpy change is included for each equation and added, the result will be the enthalpy change for the net equation. If the net enthalpy change is negative (ΔH_net < 0), the reaction is more likely to be an exothermic reaction (positive ΔH yields endothermic reactions). (But entropy also plays an important role in determining spontaneity, so some reactions with positive enthalpy change are spontaneous anyway.)

Hess's Law says that enthalpy changes are additive. Thus the ΔH for a single reaction can be calculated from the difference between the heat of formation of the products minus the heat of formation of the reactants. In mathematical terms:

$\Delta H^\theta = \Sigma(\Delta H_{f~products}^\theta ) - \Sigma(\Delta H_{f~reactants}^\theta).$

For multiple reactions, the notation changes, but the concept remains the same:

$\Delta H^\theta = \Sigma(\Delta H_{products}^\theta ) - \Sigma(\Delta H_{reactants}^\theta).$

[edit] Extension to entropy and free energy

The concepts of Hess's law can be expanded to include changes in entropy and in free energy, which are also state functions. For example the Bordwell thermodynamic cycle is an example of such an extension which takes advantage of easily measured equilibriums and redox potentials to determine experimentally inaccessible Gibbs free energy values. Combining ΔG˚ values from Bordwell thermodynamic cycles and ΔH˚ values found with Hess's law can be helpful in determining entropy values which are not measured directly, and therefore must be calculated through alternative paths.

[edit] Use

If one knows the ΔH_f (Enthalpy change of formation) of both the reactants and the products one can deduce the enthalpy change of the reaction.

ΔH=ΔH_f(P)-ΔH _f(R)

Hess' Law of constant heat summation states that the overall enthalpy change for a process depends only on the initial and final states of the system, and is independent of the route or pathway between these states.

The same process can be used for information of enthalpy of combustions although the arrows in the illustration above would be pointing down. This has no effect except switching the signs of the enthalpies. This formula should be substituted:

ΔH=ΔH_cP-ΔH_cR

[edit] Typical use

Typical table for making a Hess cycle:

Substance	ΔH_f^ɵ /KJ.mol^-1
CH₄ (g)	-75
O₂ (g)	0
CO₂ (g)	-394
H₂O (l)	-286

Using this ΔH_f^ɵ data the ΔH_c^ɵ for the below reaction can be found:

CH₄_(g)+2O₂_(g) → CO₂_(g) + 2H₂O_(l)

ΔH_c^ɵ+-75+0=-394+2x-286

ΔH_c^ɵ-75=-966

ΔH_c^ɵ=-891KJ.mol^-1

[edit] Example

Given:

B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g) ΔH = +2035 kJ

H₂O(l) → H₂O(g) ΔH = +44 kJ

H₂(g) + (1/2)O₂(g) → H₂O(l) ΔH = -286 kJ

2B(s) + 3H*2B(s) + (3/2)O₂(g) → B₂O₃(s)