Reversible chemical reactions tend to equilibrium, a state in which all chemical conversions take place at the same rate. This situation is characterised by a constant called the equilibrium constant, which relates the concentrations of the reactants and products of the reaction.
This value is very useful to assess the expected results of any given reversible reaction and to determine the concentration of any reactant or product once it has reached equilibrium.
Since equilibrium constants for many reactions are available in literature, they can be easily applied to the design of chemical reactors or any device that relies on the formation of any expected chemical compound.
Let’s see what chemical equilibrium is about and how to easily calculate the equilibrium constant of any reversible reaction.
How to calculate the equilibrium constant
- Determine if the chemical reaction has reached equilibrium, meaning, if the concentrations of both products and reactants are constant.
- Measure the molarities of products and reactants which are not in solid or pure liquid state.
- Examine the reaction’s chemical equation and find the stoichiometric coefficients of the substances measured in step 2.
- Plug into the following equation the values of the concentrations and the stoichiometric coefficients:
| Kc=[P1]p1 × [P2]p2 × …[R1]r1 × [R2]r2 × … | (1) |
Where [P1], [P2], … and p1, p2, … are the molarities and stoichiometric coefficients of the products; and [R1], [R2], … and r1, r2, … are the molarities and stoichiometric coefficients of the reactants, respectively.
What is the equilibrium constant
Imagine you are in a ball pit filled with plastic balls and a friend of yours is sitting in an empty ball pit a few meters away. You start throwing balls at your friend’s pit and they start accumulating as time pases. Now, imagine your friend has had enough and starts throwing balls back at you. A war has started!
Let’s say you initially have 100 balls in your pit, and you start throwing 1 ball per second at your friend. As time passes, the number of balls on your side decreases and the number on your friend’s side increases. At the beginning, your friend has a hard time finding a ball to throw back, since there are not many of them in his pit. So, let’s say that after half a minute your friend is only able to throw back 1 ball every two seconds (0,5 balls/s). In this scenario, the total number of balls on your side will still decrease, and the number on your friend’s side will increase, but at a lower rate.
As the amount of balls in your friend’s pit increases, it becomes easier for him to grab them and throw them back to you. So, his velocity starts to increase. Let’s say that after one minute he has enough balls on his side to be able to throw 1 of them per second. At this point, balls are flying back and forth at the same rate, so the final number of balls in each pit will not change. In this situation we have reached equilibrium. The following sequence summarizes our thought experiment.
In chemistry, equilibrium is reached by reversible reactions, meaning those that can take place in both directions: reactants form products, and products form reactants at the same time. Equilibrium is the state at which, similar to our ball-throwing example, the rates of both reactions (forward and reverse) are equal. Now, keep in mind this does not necessarily mean we will find the same amount of products and reactants in a chemical reaction that has reached equilibrium or, in the case of our ball-throwing game, the same number of balls in each pit. This only means that the velocity at which one is being converted into the other is the same.
The actual final amount of products and reactants once equilibrium is reached depends on the initial amounts. Think of the ball pits. The ball-throwing rate depends on each person’s ability (how fast they are) and on the number of balls present in his or her pit. Let’s think about this in detail. First, the greater the amount of balls in your pit —which is equivalent to the amount of chemical substance available to react—, the easier it gets to find one and throw it at your opponent.
So, if the game had started with you and your friend having the same number of balls in each pit, equilibrium would have probably been reached a lot faster than before, without your friend having to first accumulate enough balls in his pit to make it easier for him to find them and throw them back at the same rate as you. In this scenario, the total number of balls on each side at equilibrium would have been different than in our initial example.
On the other hand, each person’s ability to throw a ball to their opponent (how fast they can do it) is equivalent to the reaction kinetics, which is an intrinsic quality of the reactants and products.
Now, if the amount of products and reactants to be found in equilibrium depends on their initial amounts and on the kinetics of the chemical reaction, is there anything constant at all? Actually, there is! We can get an answer if we examine the rates of the reactions taking place.
Let’s consider the general reversible reaction aA + bB ↔cC+dD, where A and B are the reactants; a, b, their respective stoichiometric coefficients; C and D are the products; and c, d, their respective coefficients. First, for the reaction to take place, substance A has to collide with substance B with enough energy to initiate it. How often these collisions take place determines the reaction rate, which is defined as the amount of substance that reacts per unit time. Since our reactants are very often part of a solution, the reaction rate is usually expressed in units of molarity (mol/L) per unit time, so mol/L/s.
If you think about it, if you are walking on a crowded street in the middle of the day, it is more likely you will bump into someone than if you do so in the middle of the night, when there are almost no people around. The same thing happens in a reaction: the collision of the reactants, and thus how many of them react together, depends on their concentration inside the reaction recipient. In consequence, we may say that the reaction rate is directly proportional to the reactants’ concentrations. If the reaction takes place in a liquid phase, the concentration is measured in molarity, and if it takes place in a gas phase, it is measured using the substances’ partial pressure.
In the case of a reversible reaction, two reaction rates are present: one for the forward reaction and one for the reverse reaction. The first one can be then written as:
| Vf=kf[A]a[B]b | (2) |
Where [A] and [B] indicate the molarities of A and B, respectively; and a and b the orders of reaction for A and B. These are usually equal to their stoichiometric coefficients. The proportionality constant in the equation, kf, is called the reaction rate constant. Since the stoichiometric coefficients can vary a lot from reaction to reaction, this constant has different units depending on them. This way, the reaction rate keeps its usual units. On the other hand, the rate of the reverse reaction is:
| Vr=kr[C]c[D]d | (3) |
Where [C] and [D] are the molarities of C and D; c and d their respective stoichiometric coefficients; and kr the reverse reaction rate constant. We have established that, in equilibrium, the reaction rates of both the forward and the reverse reactions are equal, meaning the same amount of products and reactants is formed per unit time. In consequence:
| kf[A]a[B]b=kr[C]c[D]d | (4) |
If we separate on one side the reaction rate constants, we get:
| kfkr=[C]c[D]d[A]a[B]b | (5) |
Since a constant divided by another constant forms a third constant, we can conclude that the term to the right of equation 5 must remain constant, provided the reaction has reached equilibrium. We call this the equilibrium constant, which is usually denoted as Kc. Here “c” stands for concentration, since equation 5 is usually implemented using the molarities of the reactants and the products.
If you have trouble understanding how concentrations and reaction rates behave during a reaction, go ahead and read the section called “Time behaviour of concentrations and reaction rates”.
The equilibrium constant formula
For a general reversible reaction with P1, P2,… products and R1, R2,… reactants, which stoichiometric coefficients are p1, p2,… and r1, r2, …, respectively, the equilibrium constant is given by:
| Kc=[P1]p1 × [P2]p2 × …[R1]r1 × [R2]r2 × … | (6) |
If the reaction is taking place in the gaseous phase, partial pressures are used for each reactant and product instead of molarity. Since the pressure of a gas depends on its temperature, an additional term is added to equation 6:
| Kc=pP1p1 × pP2p2 ×… pR1r1 × pR2r2 ×…(RT)Δn | (7) |
Where pX1x1 is the partial pressure of product/reactant number 1 to the power of its stoichiometric coefficient; R is the gas constant, T the absolute temperature (measured in Kelvin), and Δn the difference in the moles of product gas and reactant gas once equilibrium is reached. The quotient of the partial pressures is usually denoted as Kp, where “p” stands for pressure.
The value of the equilibrium constant can let you assess the reaction taking place, according to these widely accepted ranges:
- Kc>1000: mostly products are formed in equilibrium
- 0,001<Kc<1000: both products and reactants are found in significant concentrations in equilibrium
- Kc<0,001: mostly reactants are formed in equilibrium
How to calculate the equilibrium constant
Given that the strict definition of the equilibrium constant actually uses the activities instead of the concentrations of the substances participating in the reaction, equation 6 is only a good approximation for those reactions where products and reactants are diluted. This is because the activities of these substances approach their molarities when they are dissolved. Therefore, no substances in solid state or pure liquids, like solvents, should be considered for the calculation of the equilibrium constant using equation 6.
Let’s consider the following reaction as an example:
| NH3(g)+HCl(g)NH4Cl(s) | (8) |
First, both reactants (ammonia and hydrochloric acid) are in gaseous state. The product, on the other hand, is found in solid state. Since only those substances in gaseous and liquid (non pure) states should be considered, the equilibrium constant of this reaction would be calculated like this:
| Kc=1[NH3]1 × [HCl]1 | (9) |
Let’s consider another example: the dissociation of hydrogen fluoride in water:
| HF(aq)+H2O(l)H3O(aq)++F(aq)- | (10) |
The reactant HF is dissolved in pure water and forms two ionic species. In this case, the calculation of the equilibrium constant should not consider the concentration of the pure liquid: water. Therefore, it would look like this:
| Kc=[H3O+][F-][HF] | (11) |
Time behaviour of concentrations and reaction rates
Consider a simple reversible reaction of the form aA ↔cC. Let’s suppose this reaction starts with an initial concentration of A and no C. As the reaction proceeds, more C is formed, increasing its concentration, while the amount of A decreases. Simultaneously, since there are less and less molecules of A to react with time, the forward reaction rate decreases. On the other hand, since more and more molecules of C are formed, it becomes more likely that they will react to form A, and thus the reverse reaction rate increases.
This entire situation is summarized by the following graphs. Go ahead and analyze them thoroughly:
