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Hess's Law

Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.


Hess's Law is named after Russian Chemist and Doctor Germain Hess. Hess helped formulate the early principles of thermochemistry. His most famous paper, which was published in 1840, included his law on thermochemistry. Hess's law is due to enthalpy being a state function, which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out. The principle underlying Hess's law does not just apply to Enthalpy and can be used to calculate other state functions like changes in Gibb's Energy and Entropy.


  • The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps.
  • If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations.

Why it works

The enthalpy of a reaction does not depend on the elementary steps, but on the final state of the products and initial state of the reactants. Enthalpy is an extensive property and hence changes when the size of the sample changes. This means that the enthalpy of the reaction scales proportionally to the moles used in the reaction. For instance, in the following reaction, one can see that doubling the molar amounts simply doubles the enthalpy of the reaction.

H2 (g) + 1/2O2 (g) → H2O (g)             ΔH° = -572 kJ

2H2 (g) + O2 (g) → 2H2O (g)             ΔH° = -1144kJ

The sign of the reaction enthalpy changes when a process is reversed.

H2 (g) + 1/2O2 (g) → H2O (g)             ΔH° = -572 kJ

When switched:

H2O (g) → H2 (g) + 1/2O2 (g)             ΔH° = +572 kJ

Since enthalpy is a state function, it is path independent. Therefore, it does not matter what reactions one uses to obtain the final reaction.

Steps involved in solving enthalpy of combustion problems

  1. Balance the individual equations
  2. If necessary look up standard enthalpies
  3. Flip equations around if necessary to cancel out terms on opposite sides
  4. Changing the equation around requires a sign change of the H of that individual step
  5. Sum up the individual steps 


Example 1: Graphite to Diamond

Calculate the standard enthalpy of combustion of the transition of C(s, graphite) C(s, diamond), given

C(s, graphite) + O2 → CO2 ΔHo = -393.5 kJ/mol

CO2 → C(s, diamond) + O2 ΔHo = + 395.41 kJ/mol


First we see that both equations are balanced. The enthalpies were given and there is no need to flip an equation around because it is possible to cancel out a couple terms as is. What is left is canceling out the O2 and the CO2 species, writing the overall reaction and then summing the two enthalpies together.

C(s, graphite) + O2CO2

CO2 → C(s, diamond) + O2

Overall Equation becomes: C(s, graphite) C(s, diamond)

  • adding the enthalpies gives (-393.5 kJ/mol + 395.41 kJ/mol) = + 1.91 kJ/mol

Since the Ho is positive, the reaction is endothermic.

Standard Enthalpy of Formation

The following equation is used to calculate the Standard Enthalpy of Formation:

\(\Delta_{r} H^{\ominus} = \sum \upsilon H^{\ominus} (products) - \sum \upsilon H^{\ominus} (reactants)\)

  • If \(\Delta H^o\) is positive, the reaction is endothermic 
  • If \(\Delta H^o\) is negative, the reaction is exothermic

Standard Gibb's Energy of Formation

As mentioned above, the Standard Gibb's Free Energy of Formation is calculated using the principles of Hess's law. The following equation is used for those calculations:

\(\Delta{_fG^\ominus} = \sum v\Delta{_fG^\ominus}(products) - \sum v\Delta{_fG^\ominus}(reactants)\)

Standard Entropy

The Standard Entropy of a reaction is calculated using the following equation:

\(\Delta_{r} S^{\ominus} = \sum \upsilon S^{\ominus} (products) - \sum \upsilon S^{\ominus} (reactants)\)


Example 2

Using Hess’ Law, the enthalpy of reaction of the major process of steam reforming can be determined.

CH4(g) + H2O(l) → CO(g) + 3H2(g)                 ΔH° = ??

Using the two postulates, given enough information, we are able to solve the enthalpy of reaction of an untabulated equation. Using the reaction of Carbon Dioxide and Hydrogen gas and reaction of methane decomposition;

1) CO(g) + H2(g) → C(graphite) + H2O(g)                  ΔH° = -131.3 kJ

2) C(graphite) + 2H2(g) → CH4(g)                      ΔH° = -74.8 kJ


  1. Flip Equation #2 to and change the sign of the enthalpy reaction.
    1. Ex. CH4(g) C(graphite) + 2H2(g)                 ΔH° = +74.8
  2. Flip Equation #1 as well in order to get the unnecessary equation parts to cancel out.
    1. C(graphite) + H2O CO(g) + H2(g)               ΔH° = +131.3 kJ
  3. With the two equations packed together, notice the equations will cancel out and come out with the final equation. C(graphite’s) cancel out between the two equations.

C(graphite) + H2O → CO(g) + H2(g)                 ΔH° = +131.3 kJ

CH4(g) → C(graphite) + 2H2(g)                  ΔH° = +74.8

Final Equation:

CH4(g) + H2O → CO + 3H2                ΔH° = +74.8 kJ + 131.3 kJ

  1. Add the final enthalpies to the two fixed up equations and receive the enthalpy of the unknown equation: ΔH° = +201.1 kJ

CH4(g) + H2O → CO + 3H2                   ΔH° = 201.1 kJ



  •  Shelly Cohen (UCD)
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Last Modified
08:40, 11 Mar 2014

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