Hess's law

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.

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::$Delta\; H^\; heta\; =\; sum(Delta\; H\_\{f~products\}^\; heta\; )\; -\; sum(Delta\; H\_\{f~reactants\}^\; heta).$For multiple reactions, the notation changes, but the concept remains the same::$Delta\; H^\; heta\; =\; sum(Delta\; H\_\{products\}^\; heta\; )\; -\; sum(Delta\; H\_\{reactants\}^\; heta).$

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.

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 products}-ΔH_{f reactants}

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_{c products}-ΔH_{c reactants}

Typical use

Typical table for making a Hess cycle: Using this ΔH_f^ɵ data the ΔH_c^ɵ for the below reaction can be found:::CH_4(g)+2O_2(g) → CO_2(g) + 2H₂O_(l):ΔH^ɵ_c+-75+0=-394+2x-286:ΔH_c^ɵ-75=-966:ΔH_c^ɵ=-891KJ.mol^-1

=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) + 3H₂(g) → B₂H₆(g) ΔH = +36 kJ Find the ΔH_f of:
*2B(s) + (3/2)O₂(g) → B₂O₃(s)

After the multiplication and reversing of the equations (and their enthalpy changes), the result is:
*B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) ΔH = -2035 kJ
*3H₂O(g) → 3H₂O(l) ΔH = -132 kJ
*3H₂O(l) → 3H₂(g) + (3/2)O₂(g) ΔH = +858 kJ
*2B(s) + 3H₂(g) → B₂H₆(g) ΔH = +36 kJ Adding these equations and canceling out the common terms on both sides, we get
*2B(s) + (3/2)O₂(g) → B₂O₃(s) ΔH = -1273 kJ

Using extension of Hess's law to entropy and free energy: For entropy:
*ΔS^o = Σ(ΔS_f^o_products) - Σ(ΔS_f^o_reactants)
*ΔS = Σ(ΔS^o_products) - Σ(ΔS^o_reactants). For free energy:
*ΔG^o = Σ(ΔG_f^o_products) - Σ(ΔG_f^o_reactants)
*ΔG = Σ(ΔG^o_products) - Σ(ΔG^o_reactants).

Also see

*Thermochemistry

External links

* [http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html Hess's Law - The principle of conservation of energy - www.science.uwaterloo.ca]