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# Equilibrium

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## What is an equilibrium?

You are watching a soccer game on television. After the extra time, the score is 1: 1, a tie. Everything comes down to a penalty shoot-out the first player scores, then the next one, and the next one after him. Everybody scored! 5:5 is still a draw, basically, the score kept increasing but overall, nothing changed... Chemical equilibrium is just like that. Even though soccer players will eventually miss and the match will be decided, chemicals could and sometimes keep pushing shoulder to shoulder forever and ever.

A dynamic equilibrium is a state where things are happening both ways. Molecules are formed and decomposed simultaneously at exactly the same rate. If you look at a chemical equilibrium what you will usually see is nothing happening. Think back to our soccer game from before; if you could only see if it is a draw or an advantage for one team after every round, you would always see the draw. Basically, on the scale of big ( you, me, an elephant), nothing is happening in an equilibrium reaction, but on the scale of small (molecules, atoms, and ions), everything is in a constant motion.

But what is a dynamic equilibrium? Keep reading to find it out!

• Dynamic Chemical Equilibrium - here we go over the basic concepts and show how equilibrium relates to reaction reversibility.
• The Equilibrium Constant, Keq, and the Reaction Quotient - at this point, we formulate the equilibrium constant and reaction quotient for a couple of sample reactions.
• Reaction Rate Law - Justification for the form of the reaction rate law, is presented briefly.

### Dynamic Chemical Equilibrium:

• When the rate of the reaction leading to products, the forward reaction, is equal to the rate of the reaction leading back to reactants, reverse reaction, we have a system in chemical equilibrium.
• The rates of reaction correspond to how fast components are changing in a dynamic system.
• The rates of reaction for a system at equilibrium are not zero.
• When a chemical system reaches equilibrium, the rate of the forward reaction will be equal to the rate of the reverse reaction.

Ultimately, when the reaction system reaches chemical equilibrium, there is no change in the concentrations of products and reactants, however, one must keep in mind that chemical bonds in a system at equilibrium are dynamic and will be continuously forming, breaking, and reforming.

Figure 1: System at dynamic chemical equilibrium.

In the above figure, lowercase letters $a$, b, and c, correspond to the stoichiometric coefficients of the balanced equation, and the components of the chemical reaction are denoted by the uppercase/bold letters A, B, and AB.

Consider, the reaction for the formation of carbonic acid from water and aqueous carbon dioxide. When this reaction system reaches dynamic chemical equilibrium we would have:

Figure 2: Formation of carbonic acid in dynamic chemical equilibrium.

The formation of carbonic acid from carbon dioxide and water is a reversible reaction. When the rate of formation of carbonic acid (forward reaction) is equal to the decomposition of carbonic acid back to carbon dioxide and water (reverse reaction) then we have reached chemical equilibrium.

In what follows, we will refer to dynamic chemical equilibrium as chemical equilibrium, or sometimes simply as equilibrium. It is the net effect on a chemical system that chemical equilibrium measures. When a reaction is reversible and there is no change in the concentrations of reactants and products, we have reached the equilibrium state.

### Chemical Equilibrium Constant and the Reaction Quotient:

Let us again consider a hypothetical chemical system in equilibrium:

• Chemical equilibrium is the endpoint of a dynamic process in a chemical system undergoing a reaction.
• Chemical equilibrium occurs when the forward and reverse reaction rates are equal and when the concentrations of components of the reaction no longer change.
• The composition of a reaction mixture at chemical equilibrium does not depend on the route taken to arrive at the final concentration of components.

For example, let us call the ratio, of the concentration of products over reactants in a reaction mixture that has reached chemical equilibrium, the equilibrium constant, Keq, such that:

${K}_{eq}=\frac{{\left[\mathbit{A}\mathbit{B}\right]}^{c}}{{\left[\mathbit{A}\right]}^{a}{\left[\mathbit{B}\right]}^{b}}$

where $\left[\mathbit{A}\right]$, $\left[\mathbit{B}\right]$, are the concentration of the reactants, and$\left[\mathbit{A}\mathbit{B}\right]$, is the concentration of the final product. This expression is an example of the Reaction Quotient, also called the Law of Mass Action.

• The stoichiometric coefficients of the balanced equation are the exponents, $a$, b, and c.
• The value of the Equilibrium Constant, Keq, does not change whether we approach the equilibrium mixture from the left-hand side (forward reaction) or the right-hand side (reverse reaction):

For instance, consider the reaction for the formation of water from hydrogen and oxygen gas at the equilibrium point:

$2{H}_{2}\left(g\right)+{O}_{2}\left(g\right)⇌2{H}_{2}O\left(l\right)$

where $a=2$, $\mathbit{A}={H}_{2}$, $b=1$, $\mathbit{B}={O}_{2}$, and $c=2$, $\mathbit{A}\mathbit{B}={H}_{2}O$. The Equilibrium Constant for this system is then:

${K}_{eq}=\frac{{\left[{H}_{2}O\right]}^{2}}{\left[{H}_{2}\right]\left[{O}_{2}\right]}$

such that the rate for the forward reaction is equal to the rate for the reverse reaction and the equilibrium constants for forward and reverse are the same:

This example gives us an indication of the notion that for a system at chemical equilibrium the concentration of components does not change and that the rate of the forward reaction must equal the rate of the reverse reaction.

Note, that the chemical reaction, $2{H}_{2}\left(g\right)+{O}_{2}⇌2{H}_{2}O$, is just one instance of many different types of reaction systems.

For example, we could have considered the reaction, ${C}_{3}{H}_{8}\left(g\right)+5{O}_{2}\left(g\right)⇌3C{O}_{2}\left(g\right)+4{H}_{2}O\left(g\right)$. Then the corresponding Equilibrium Coefficient would be:

${K}_{eq}=\frac{{\left[C{O}_{2}\right]}^{3}{\left[{H}_{2}O\right]}^{4}}{\left[{C}_{3}{H}_{8}\right]{\left[{O}_{2}\right]}^{5}}$

or we could have considered the reaction, $PC{l}_{5}\left(g\right)⇌PC{l}_{3}\left(g\right)+C{l}_{2}\left(g\right)$, for which the Equilibrium Coefficient would be:

${K}_{eq}=\frac{\left[PC{l}_{3}\right]\left[C{l}_{2}\right]}{\left[PC{l}_{5}\right]}$

I hope you are seeing a trend here; for the general hypothetical chemical system:

$a\mathbit{A}+b\mathbit{B}+...+z\mathbit{Z}⇌a\text{'}\mathbit{A}\text{'}+b\text{'}\mathbit{B}\text{'}+...+z\text{'}\mathbit{Z}\text{'}$

the Equilibrium Constant would be:

${K}_{eq}=\frac{{\left[\mathbit{A}\text{'}\right]}^{a\text{'}}{\left[\mathbit{B}\text{'}\right]}^{b\text{'}}...{\left[\mathbit{Z}\text{'}\right]}^{z\text{'}}}{{\left[\mathbit{A}\right]}^{a}{\left[\mathbit{B}\right]}^{b}...{\left[\mathbit{Z}\right]}^{z}}$

where the stoichiometric coefficients for the products are $a\text{'}$, b', ... z', the concentrations of products are [A'], [B '], ... , [Z '], and the stoichiometric coefficients of the reactants are $a$, b,..., z, and the concentrations of the reactants are [A], [B], ..., [Z].

${K}_{eq}=\frac{{\left[\mathbit{C}\right]}^{c}{\left[\mathbit{D}\right]}^{d}}{{\left[\mathbit{A}\right]}^{a}{\left[\mathbit{B}\right]}^{b}}$

As we have already noted, the notion of chemical equilibrium is based on component concentrations and reaction rates. We will now move on to a simple discussion of reaction rates, or the speed at which chemical reactions take place.

### Reaction Rate:

• The reaction rate, or speed at which a chemical reaction occurs, differs substantially for different reaction conditions, types of reactants, and stability of products.
• Some reactions occur almost instantly, while others may take a very long time, even years, to reach a state in which the concentrations of products and reactants do not change.
• The change in the concentration of products or reactants per unit time is referred to as the reaction rate.

In particular, for a given chemical reaction the rate of the formation of products is called the forward reaction rate, and the rate of formation of reactants is called the reverse reaction rate. Consider, the formation of products AB, from a chemical system consisting of reactants A and B. In the forward reaction, reactants A and B, are consumed in the reaction while the formation of the product AB, increases:

Figure 2: Hypothetical forward chemical reaction.

Graphically, the above hypothetical forward chemical reaction can be depicted as:

Figure 3: Hypothetical reaction profile where the product AB, increases whilst the reactants A and B, decrease.

Let's consider the hypothetical forward reaction in greater detail,

$a\mathbit{A}+b\mathbit{B}\mathbf{}\mathbf{\to }\mathbf{}c\mathbit{A}\mathbit{B}$

where $a$, b, and c, correspond to the stoichiometric coefficients of the balanced equation. For any point in time, we can write an expression for the rate of the forward reaction as:

We can then write an expression for the rate of the forward reaction as a rate law:

$Rate=\frac{-1}{a}\frac{d\left[\mathbit{A}\right]}{dt}=\frac{-1}{b}\frac{d\left[\mathbit{B}\right]}{dt}=\frac{1}{c}\frac{d\left[\mathbit{A}\mathbit{B}\right]}{dt}$

where the negative sign denotes a decrease in the concentrations of the reactants [A] and [B], and changes in the concentrations with time, of all components, are denoted by the symbol,$\frac{d\left[...\right]}{dt}$.

We can translate the rate law into words as, 'the decrease per second in the concentration of reactant A, is equal to the decrease in the concentration per second in the concentration of reactant B, which are both equal to the increase per second in the concentration of the product AB '. In particular, for any point in time, we can determine the reaction rate by determining any one of the equations of the rate law because all of the changes in the concentrations of the components are proportional to one another. Thus, we can monitor the rate of change of any of the reaction components to arrive at the rate law:

$Rate=\frac{-1}{a}\frac{d\left[\mathbit{A}\right]}{dt},or,Rate=\frac{-1}{b}\frac{d\left[\mathbit{B}\right]}{dt},or,Rate=\frac{1}{c}\frac{d\left[\mathbit{A}\mathbit{B}\right]}{dt}$

The rate law for the formation of water: For example, consider the forward reaction for the formation of water from hydrogen and oxygen gas:

$2{H}_{2}\left(g\right)+{O}_{2}\left(g\right)\to 2{H}_{2}O\left(l\right)$

where $a=2$, $\mathbit{A}={H}_{2}$, $b=1$, $\mathbit{B}={O}_{2}$, and $c=2$, $\mathbit{A}\mathbit{B}={H}_{2}O$. Then, at any point in time, the rate law for the forward chemical reaction corresponding to the formation of water would be:

$Rate=\frac{-1}{2}\frac{d\left[{H}_{2}\right]}{dt},or,Rate=\frac{-1}{1}\frac{d\left[{O}_{2}\right]}{dt},or,Rate=\frac{1}{2}\frac{d\left[{H}_{2}O\right]}{dt}$

## Equilibrium - Key takeaways

• At equilibrium, for a reversible reaction, the rate of the forward reaction is equal to the rate of the reverse reaction.
• At equilibrium, the concentration of the components of the chemical reaction will not change.
• At equilibrium, the ratio of the concentration of products over the concentration of reactants, this is called the Equilibrium Constant:

${K}_{eq}=\frac{{\left[A\text{'}\right]}^{a\text{'}}{\left[B\text{'}\right]}^{b\text{'}}...{\left[Z\text{'}\right]}^{z\text{'}}}{{\left[A\right]}^{a}{\left[B\right]}^{b}...{\left[Z\right]}^{z}}$

We have only touched the surface of a complex subject whose ultimate goal is to picture the reaction mechanism at the molecular level. If you are interested you may learn more in our other explanations for each of the topics mentioned here.

## Final Equilibrium Quiz

Question

Does the concentration of reactants change when equilibrium is reached?

No, at chemical equilibrium the concentrations of reactants does not change.

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Question

Does the concentration of products change when a chemical system has reached the equilibrium point?

No, the concentration of products does not change at equilibrium.

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Question

Does the equilibrium constant change depend on the route taken to arrive at the final concentrations of components?

No, the route that was taken to arrive at the final concentration of components does not change the equilibrium constant.

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Question

What is the formula for the equilibrium constant for the reaction, ?

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Question

What is the equilibrium constant for the reaction, ?

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Question

What is the equilibrium constant for the reaction,

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Question

What is the equilibrium constant for the reaction, ?

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Question

Is the following false? "At equilibrium, for a reversible reaction, the rate of the forward reaction is equal to the rate of the reverse reaction".

No, it is true.

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Question

Is the following True or False? "At equilibrium, the concentrations of components is always changing"?

The statement is false because at equilibrium the concentrations of components remain constant.

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Question

Is the following statement True or False? "When the system is not at equilibrium the equilibrium constant will always give you the concentrations of components in the reaction mixture."

False, the equilibrium constant only applies to reactions that show no changes in the concentrations of components.

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Question

True or False: "The rates of reaction for a system at equilibrium are zero."

False

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Question

True or False: "Is it true that when the rate of the reverse reaction equals the rate of the forward reaction the system is far from equilibrium?"

False, the system is at equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.

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Question

The denaturation of proteins is an equilibrium reaction.

False

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Question

Which of the following are equilibrium reactions?

Weak acid + weak base

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Question

Equilibrium reactions can happen only in the gas phase.

False

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