For example, when you're given the enthalpy $\Delta H$ of a reaction and you want to switch that reaction around in order to add it to something else, you switch the sign. Furthermore, when adding two reactions, you add their $\Delta H$ values.

But when you're working with equilibium problems and you have the $K$ of a reaction, if you want to switch the reaction you take the inverse of $K$ and when you add reactions you multiply $K$ values.

When working with cell potentials $E_{cell}$ of a redox reaction, you switch the sign if you switch the equation around, but when you're adding two equations you do not multiply $E_{cell}$ by whatever you multiply the coeffecients on the species by.

Are there any types of patterns other than the three I've described here? Any help is greatly appreciated.


2 Answers 2


Equilibrium constant

When a reaction is at equilibrium, the species will have concentrations called the equilibrium concentrations. They do not depend on whether the reaction is written in one direction or in the reverse direction, and whether all the stoichiometric coefficients are multiplied by the same value. However, your equilibrium expression will be different (switching concentrations from numerator to denominator if reversing the chemical equation, or changing exponents when changing stoichiometric coefficients). As a result, the value of the equilibrium constant will change while the equilibrium concentrations themselves will remain the same.

Enthalpy of reaction

For a given reaction, the enthalpy change associated with making - say - 1 mol of species A is a given. If you use 1 mol of species A in the reaction instead, the enthalpy change will be of the same magnitude but opposite sign. If I make a different amount of species A, the enthalpy change will change in a manner proportional to that amount. In a similar manner, if I change the chemical equation for $\ce{A -> B}$ to $\ce{2A -> 2B}$, the molar enthalpy of reaction will double as well. The same goes for Gibbs energy or for entropy of reaction.

Cell potential

For a given reaction with certain concentrations of species in the two half cells of a voltaic cell, I will measure a certain cell potential. If I switch the role of reactants and products (in the equation), I will still measure the same potential (the direction of the actual reaction does not change). If I multiply all coefficients in the chemical equation by the same number, it still correctly describes the reaction, but this will not change the potential.

Mathematical relationships

$$\Delta G^\circ = - R T \ln(K)$$

The way logarithms work, if you multiply $\Delta G^\circ$ by a number, you have to raise K to the power of that number. If you add Gibbs energies, you have to multiply equilibrium constants.

$$\Delta G^\circ = - z F E^\circ $$

If you multiply $\Delta G^\circ$ by a number, z (the number of electrons transferred) will also increase by that number. Therefore, $E^\circ $ does not change.

Multiplying all coefficient in a chemical equation by (-1) changes the role of reactants and products. Thus, once you know the rules for changing coefficients by a common factor, you don't need to memorize another rule for reversing the equation.


Energies are additive.

Concentrations, partial pressures, and activities which are components of equilibrium constants are strongly related to probabilities. The probability of two things happening at the same time is computed by multiplying the probabilities, and likewise, the equilibrium constant is computed from products of concentrations/activities.

This is explicitly expressed in:

$$\Delta G^{\varnothing} = -RT \ln K$$

The equilibrium constant on the right is related to the free energy on the left via a log, since

$$\ln (ab) = \ln a + \ln b$$

So summing on the left hand side is related to multiplying inside the equilibrium constant on the right.

The potential $\mathscr{E}$ is related to the free energy, but it is not an energy. Potential is energy scaled per electron, so moving around extra electrons does not change the per-electron value.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.