Types of chemical Equilibrium with Examples Definition

Types of chemical equilibrium
In the chemical industry, knowledge of chemical equilibrium is essential to obtain synthesis with better yields. Source: Pxhere.

Once the chemical equilibrium is established, no further changes or responses can be obtained unless it is disrupted or affected by external actions. Thus, in the synthesis of a product, several parameters such as pressure, volume or temperature are modified , so that the maximum amount of product is generated in equilibrium.

Otherwise, in equilibrium the quantities of products will be unsatisfactory; that is, it will have a poor reaction yield. That is why chemical balance is vital for the chemical industry, and in general for any synthesis regardless of its scale.

In the chemical equilibrium there may be more quantity of products, or else more quantity of reactants. It all depends on where the balance is shifted. Taking into account several factors, an equilibrium can be shifted in either direction of the double arrow in the reversible reaction.

Explanation of chemical equilibrium (Types of chemical equilibrium)

Before balance

To understand what chemical equilibrium is, consider the following reversible reaction:

2 O 4 (g) ⇌ 2 NO 2 (g)

2 O 4 gas is colorless, while NO 2 gas is brown or brown in color. If a certain amount of N 2 O 4 is injected into a vial or small container , it will be colorless until equilibrium is established.

On the left of the upper image, it can be seen that almost all the molecules are N 2 O 4 and that the NO 2 concentration is zero.

Two graphs are also shown above that represent the trend in equilibrium, with the dotted line indicating the position at the moment when only N 2 O 4 is injected . One graph corresponds to Concentration vs. Time (bottom left hand) and the other corresponds to the graph at Speed ​​vs. Time (lower right hand).

The concentration of N 2 O 4 , [N 2 O 4 ] (blue line) will gradually decrease, since part of it will dissociate to produce NO 2 molecules . Thus, with the concentration of NO 2 , [NO 2 ] (red line) equal to zero at first, it will soon increase as the N 2 O 4 dissociates.

However, it is a reversible reaction: part of the NO 2 molecules will bind to form N 2 O 4 again . Thus, there will be two reactions, the direct and the inverse, each with its own speeds.

Reaction rates

At first, the rate of consumption of N 2 O 4 is higher than the rate of consumption of NO 2 . Obviously, since there is only N 2 O 4 , the few molecules formed from NO 2 will hardly be able to meet each other to react. At this point, the vial will start to turn orange, because you have a mixture of N 2 O 4 and NO 2 .

Gradually, as there are more NO 2 molecules in the vial and the N 2 O 4 molecules dissociate, the rates of the two reactions will become equal, even as the concentrations differ from each other more and more. That is, [NO 2 ] tends to be greater than [N 2 O 4 ], which is why the red line is above the bluish line.

Note that the speed becomes dC / dt, that is, the change in concentration with respect to time. That is why the two C vs. t and V vs. t are not identical.

On balance

Once the pair N 2 O 4 -NO 2 establishes the equilibrium, the speeds of both reactions will equalize and the following mathematical expressions will be obtained:

Direct V = k 1 [N 2 O 4 ]

Inverse V = k -1 [NO 2 ] 2

direct = inverse

1 [N 2 O 4 ] = k -1 [NO 2 ] 2

1 / k -1 = [NO 2 ] 2 / [N 2 O 4 ]

eq = [NO 2 ] 2 / [N 2 O 4 ]

At this point, the vial will stain even more brown, as the equilibrium is further shifted towards the formation of [NO 2 ]. That is, K eq , the equilibrium constant, must be greater than 1 taking into account the higher ratio, [NO 2 ] 2 / [N 2 O 4 ].

At equilibrium, the concentrations [N 2 O 4 ] and [NO 2 ] remain constant, with both reactions happening at the same speed: as soon as a certain amount of N 2 O 4 is dissociated , the same amount will be produced again immediately because of the reaction of a certain amount of NO 2 .

For this reason, it is said that the chemical equilibrium is dynamic: the molecules N 2 O 4 and NO 2 continue to participate in the reactions even when there is no change in their concentrations.

And since the reaction rates are the same for both directions, the red and blue lines in the V vs. t touch on a horizontal line.

Equilibrium constant

The equilibrium constant for the above reaction will always be the same, at a certain temperature, no matter how much N 2 O 4 is first injected into the vial. This will be the case even if a mixture of N 2 O 4 and NO 2 is injected directly , then left to rest until equilibrium is reached.

When equilibrium is reached and the [N 2 O 4 ] and [NO 2 ] concentrations are measured , the [NO 2 ] 2 / [N 2 O 4 ] ratio will equal K eq for this reaction. The larger (Keq >> 1), the more products there will be in equilibrium. And the smaller it is (Keq << 1), the more reactants there will be in equilibrium.

The concentrations [N 2 O 4 ] and [NO 2 ] will not always be the same. For example, if small volumes are injected into the vial, or if torrents of these gases are instead injected into a reactor, the amounts will vary considerably. However, K eq will remain the same as long as the temperature is the same in both processes.

eq is susceptible to changes in temperature: the higher the temperature, the higher its value. That is, the higher the temperature, in equilibrium there will be a tendency to the formation of more products. This is so unless the reaction is exothermic, as will be explained in the next section.

Factors Affecting Chemical Balance – (Types of chemical equilibrium)

The factors that affect the chemical equilibrium in a reversible reaction are: concentration, pressure and temperature.


In a reversible reaction, according to Le Chatelier’s Principle, increasing the concentration of a reactant will cause a shift of the chemical equilibrium to the right, in order to increase the concentration of the reaction products.

On the contrary, if a product is added to the reaction, the chemical equilibrium will shift to the left, in order to increase the concentration of the reactants. If the following reaction is taken as an example:

C (s) + O 2 (g) ⇌ 2 CO (g)

As the concentration of the reactants (C and O 2 ) increases, the equilibrium will shift to the right, that is, towards the formation of the carbon monoxide (CO) product. But an increase in the concentration of CO will produce a shift of the equilibrium to the left, to increase the concentration of C and O 2 .


In a reaction carried out in the gas phase, a variation in the volume or in the pressure exerted on the gases that intervene in it, will produce an alteration in the equilibrium, depending on the number of moles of the reactants and the products (Types of chemical equilibrium).

If the components of a reaction (reactants and products) have differences in the number of moles involved in a reaction, for example, the dimerization of nitrogen dioxide (NO 2 ):

2 NO 2 (g) ⇌ N 2 O 4 (g)

An increase in the pressure exerted on the gas phase reaction, consequence of a decrease or not in the volume of the reaction, will cause a displacement of the chemical equilibrium towards the side that has fewer gaseous moles. In this example: the product N 2 O 4 .

On the contrary, a decrease in pressure will cause an equilibrium shift towards the side of the reaction that has the largest number of moles of gas. In the given example, it will move towards the reactants (NO 2 ), in order to counteract the decrease in pressure.

If in a gas phase reaction the moles of the reactants are equal to the moles of the products, the changes in pressure will not affect the chemical equilibrium (Types of chemical equilibrium).


Endothermic reaction

In an endothermic reaction, heat can be considered as a reactant, since its supply is necessary for the reaction to occur:

A + Q ⇌ B

Q = heat supplied

Therefore, in an endothermic reaction with increasing temperature, the equilibrium will shift to the right. Meanwhile, as the temperature decreases, the equilibrium shifts to the left.

Since heat is part of the reactants, A will consume it to become B.

Exothermic reaction

In an exothermic reaction, heat is generated, this being a product of the reaction:

A ⇌ B + Q

In an exothermic reaction with increasing temperature, the equilibrium will shift to the left (reactants), as heat production increases. Meanwhile, as the temperature decreases, the equilibrium will shift to the right (products).

Since the heat is part of the products, as the temperature increases there will be no need to add more heat to the medium. And therefore, the equilibrium will seek to take advantage of the extra heat to produce more reagents; in this case, more than A.

Types of chemical equilibrium

Depending on the physical state of the reaction components, the Types of chemical equilibrium can be homogeneous or heterogeneous.

Homogeneous equilibrium

In this type of equilibrium, all reactants and products have the same phase or liquid state . For instance:

2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g)

Here both N 2 , H 2 and NH 3 are gaseous substances.

Heterogeneous equilibrium

It exists when not all reactants and products have the same phase or physical state. For instance:

2 NaHCO 3 (s) ⇌ Na 2 CO 3 (s) + CO 2 (g) + H 2 O (g)

Here we have NaHCO 3 and Na 2 CO 3 as solids, and CO 2 and H 2 O as gases or vapors.

Examples of chemical equilibrium

Dissociation of acetic acid in water

Acetic acid, CH 3 COOH, dissociates in water establishing an equilibrium:

CH 3 COOH (aq) + H 2 O (l) ⇌ CH 3 COO  (aq) + H 3 O + (aq)

A reaction whose equilibrium constant is called the acidity constant.

Isomerizaci or n butane

The n- butane gas can be isomerized according to the following reversible reaction to produce the isobutane isomer:

CH 3 CH 2 CH 2 CH 3 (g) ⇌ CH 3 CH (CH 3 ) 2 (g)

Nitric oxide formation in the atmosphere

Nitric oxide can be formed in the atmosphere , although very sparingly, from nitrogen and oxygen in the air:

2 (g) + O 2 (g) ⇌ 2NO (g)

The K eq of this reaction is 2 · 10 -31 at 25 ° C, so only a negligible amount of NO will be formed.

REACTION or n complexation of silver with amon í aco

Silver ions, Ag + , can complex with ammonia according to the following reversible reaction:

Ag + ( sol ) + 2NH 3 (l) ⇌ [Ag (NH 3 ) 2 ] + (sol)

Solved exercises

Exercise 1

What type of chemical equilibrium corresponds to the following reaction?

2NOCl (g) ⇌ 2NO (g) + Cl 2 (g)

This reaction corresponds to a homogeneous equilibrium, since all the substances involved, NOCl, NO and Cl 2 , are gases.

Exercise 2

For the same reaction above, what change in equilibrium would be expected if the pressure suddenly increased?

Let’s count the moles of the reactants:


We have 2 moles of NOCl. Now let’s count the moles of the products:

2NO + Cl 2

We have three moles: two of NO, and one of Cl 2 . There are more gaseous moles on the product side. Therefore, an increase in pressure will cause the equilibrium to shift to the left side, towards the formation of NOCl. In this way, the system seeks to soften the effects of pressure favoring the formation of NOCl and not NO and Cl 2 .

Exercise 3

For the same reaction above, suppose a considerable volume of Cl 2 is suddenly injected . What will happen to the balance?

As Cl 2 is added to the reactor or vessel, its concentration is increasing as a reaction product. And therefore the equilibrium will shift to the left again, thus forming more NOCl.

Exercise 4

If the dissociation of NOCl at 227 ºC has an equilibrium constant of 4.5 · 10 -4 , in what direction is equilibrium displaced?

Again, since K eq is less than 1 (4.5 · 10 -4 <1), in equilibrium it is expected that there will be more NOCl than NO or Cl 2 , since it is shifted to the left.

Exercise 5

According to the following equilibrium reaction:

[Co (OH 2 ) 6 ] 2+ (aq) + 4Cl  (aq) + Q ⇌ [CoCl 4 ] 2- (aq) + 6H 2 O (l)

And knowing that the [Co (OH 2 ) 6 ] 2+ complex is pink in color, and that the complex is [CoCl 4 ] 2- it is bluish, what changes would you expect to observe if you heat a container with [Co (OH 2 ) 6 ] 2+ ? What would you expect to see if after heating the container, you placed it in an ice bath?

The reaction is endothermic, absorbing heat Q as a reactant. Therefore, when heating the container with [Co (OH 2 ) 6 ] 2+ , the equilibrium will shift to the right, towards the formation of [CoCl 4 ] 2- . You will see a color change from pink to bluish.

Then, if the bluish container with [CoCl 4 ] 2- is placed in an ice bath, the reaction will now shift to the left, towards the formation of Co (OH 2 ) 6 ] 2+ :

[CoCl 4 ] 2- (aq) + 6H 2 O (l) ⇌ [Co (OH 2 ) 6 ] 2+ (aq) + 4Cl  (aq) + Q

This is because the reverse reaction is exothermic, having heat Q as a product. Therefore, on cooling the container with [CoCl 4 ] 2- of blue color, the complex [Co (OH 2 ) 6 ] 2+ will form again , and the pink color will reappear.

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