Effect of a change in Concentration

If an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction

$$H_2\;(g) + I_2\;(g) ⇌ 2HI\;(g)\qquad K_c=50.0\;at\;400°C$$

The rate laws for the forward and reverse reactions are:

$\text{forward}\quad H_2\;(g)+I_2\;(g)→2HI\;(g)$$\text{rate}_f=k_f[H_2]^m[I_2]^n$
$\text{reverse}\quad 2HI\;(g)→H_2\;(g)+I_2\;(g)$$\text{rate}_r=k_r[HI]^x$

When this system is at equilibrium, the forward and reverse reaction rates are equal.


If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction:

$$\text{rate}_f\gt \text{rate}_r$$

The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.

The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction

$$\text{rate}_f\lt \text{rate}_r$$

and a temporary net reaction in the reverse direction to re-establish equilibrium.

As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,


If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then $Q_c\lt; K_c$ and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If $H_2$ is added, the right shift will consume $I_2$ and produce $HI$ as equilibrium is re-established, yielding a mixture with a greater concentrations of $H_2$ and $HI$ and a lesser concentration of $I_2$ than was present before. If $I_2$ is added, the new equilibrium mixture will have greater concentrations of $I_2$ and $HI$ and a lesser concentration of $H_2$. Finally, if $HI$ is removed, the concentrations of all three species will be lower when equilibrium is reestablished. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case $Q_c\lt;K_c$ and the equilibrium will shift left.

For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure $P$ of an ideal gas is proportional to its molar concentration $M$,


and so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered.

At equilibrium, the reaction $H_2\;(g) + I_2\;(g) ⇌ 2HI\;(g)$ is described by the reaction quotient


If the volume occupied by an equilibrium mixture of these species is decreased by a factor of 3, the partial pressures of all three species will be increased by a factor of 3:

$$Q’_{p}=\frac{{3P_{HI}}^2}{3P_{H_2}3P_{I_2}}\\ Q’_{p}=\frac{9P{HI^2}}{9P_{H_2}P_{I_2}}\\ Q’_{p}=\frac{P{HI^2}}{P_{H_2}P_{I_2}}\\ Q’_{p}=Q_P=K_P\\ Q'{p}=Q_P=K_P$$

And so, changing the volume of this gas-phase equilibrium mixture does not results in a shift of the equilibrium.

A similar treatment of a different system, $2NO_2\;(g) ⇌ 2NO\;(g) + O_2\;(g)$, however, yields a different result:

$$Q_p=\frac{P_{NO}^2P_{O_2}}{P_{NO_2}^2} \\
Q’_{P}=\frac{(3P_{NO})^2 3P_{O_2}}{(3P_{NO_2})^2}\\
Q’_{P}=\frac{(9P_{NO}^2)( 3P_{O_2})}{9P_{NO_2}^2}\\
Q’_{P}=\frac{27P_{NO}^2 P_{O_2}}{9P_{NO_2}^2}=3Q_P \\
Q’_{P}=3Q_P\gt K_P$$

In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.

These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.

Equilibrium and Soft Drinks

The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804) developed a method of infusing water with carbon dioxide to make carbonated water. Priestley’s approach involved production of carbon dioxide by reacting oil of vitriol (sulfuric acid) with chalk (calcium carbonate).

The carbon dioxide was then dissolved in water, reacting to produce hydrogen carbonate, a weak acid that subsequently ionized to yield bicarbonate and hydrogen ions:

dissolution $CO_2\;(g) ⇌ CO_2\;(aq)$
hydrolysis $CO_2\;(aq) + H_2O\;(l) ⇌ H_2CO_3\;(aq)$
ionization $H_2CO_3\;(aq)⇌HCO_3^−\;(aq) + H^+\;(aq)$

These same equilibrium reactions are the basis of today’s soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace.

When a carbonated beverage container is opened, a hissing sound is heard as pressurized CO2 escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved CO2 and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the CO2(aq) concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as “flat.”

Opening a soft-drink bottle lowers the CO2 pressure above the beverage, shifting the dissolution equilibrium and releasing dissolved CO2 from the beverage. (credit: modification of work by “D Coetzee”/Flickr)