Objective 1: Given the balanced equation of a reversible reaction, write the equilibrium constant expression.

The Law of Mass Action and Equilibrium Constants

The Law of Mass Action allows one to write what is called and equilibrium constant expression for any chemical reaction. It is best explained as follows:

For a chemical reaction at equilibrium such as:

$\text{aA}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ bB(g)}\rightleftharpoons \text{cC}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ dD(g)}$

the following ratio has a constant value at a given temperature:

$\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$

This constant is called the equilibrium constant and the ratio is called the equilibrium constant expression. The square brackets ([ ]) indicate concentration in terms of molarity, so [A] is the molarity of A, [B] is the molarity of B, etc.

The equilibrium constant is symbolized by an upper case K. K (the equilibrium constant) is also called K_eqor K_cis (where c is for concentration). K, K_eq, and K_c can be used interchangeably. I will use K in these notes.

Therefore, we could say that for the reaction:

$\text{aA}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ bB(g)}\rightleftharpoons \text{cC}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ dD(g)}$

The equilibrium constant expression is:

$K=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$ .

More on the Equilibrium Constant (K)

As stated earlier, the square brackets ([ ]) indicate concentration in terms of molarity, so [A] is the molarity of A. The equilibrium constant is called a constant because it has a constant value for the equilibrium reaction at a given temperature. Therefore, if for the above reaction, if K = 5 at a particular temperature (say 300 K for example), that means that the ratio

$K=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}=5$ if the reaction is at equilibrium at 300K. If the ratio does not equal 5 at 300K, the reactio is not at equilibrium.

The numerical value of K varies with temperature. For our above example, if the temperature is different (something other than 300K), K will have a different numerical value (it may be higher, it may be lower – we’ll discuss that later).

Let’s look at the patterns between the reaction and K to see how the expression is written:

The products (C and D) are in the numerator, and the reactants (A and B) are in the denominator.
The coefficients (c for C, d for D, a for A and b for B) are the exponents
the product concentrations (raised to the powers of their exponents) are multiplied together in the numerator and the reactant concentrations (raised to the powers of their exponents) are multiplied together in the denominator.

K is often expressed as a numerical value without units. The reason is that the units will depend on the exponents in the expression, and the exponents are determined by the reaction coefficients. Since the units of K depend on the particular reaction coefficients, it is customary to ignore them.

Pressure-Based Equilibrium Constants for Gas-Phase Reactions (K_p)

In gas-phase reactions, the concentration of a gas in the reaction mixture is proportional to its partial pressure. The ideal gas law PV=nRT can be rearranged as:

$\frac{n}{V}=\frac{P}{RT}$

Since n/V = molarity, molarity is directly proportional to pressure.

The equilibrium constant K can also be expressed as the ratio of the partial pressures of the gases (raised to the powers of the coefficients)

For aA(g) + bB(g) ⇆ cC(g)+ dD(g) , the pressure based equilibrium constant (K_p) expression is:

${{K}_{p}}=\frac{{{P}_{C}}^{c}{{P}_{D}}^{d}}{{{P}_{A}}^{a}{{P}_{B}}^{b}}$

Where P_c is the partial pressure of the gas C, P_d is the partial pressure of the gas D, etc.

Equilibrium Constant Examples

For the reaction ${{\text{N}}_{2}}(\text{g)}+3\text{ }\!\!~\!\!\text{ }{{\text{H}}_{2}}(\text{g)}\rightleftharpoons \text{2 }\!\!~\!\!\text{ N}{{\text{H}}_{3}}(\text{g)}$ ,

the equilibrium constant expression is

$K=\frac{{{[\text{N}{{\text{H}}_{3}}]}^{2}}}{[{{\text{N}}_{2}}]{{[{{\text{H}}_{2}}]}^{3}}}$

For the reaction ${{\text{N}}_{2}}{{\text{O}}_{4}}(\text{g)}\rightleftharpoons \text{2 }\!\!~\!\!\text{ N}{{\text{O}}_{2}}(\text{g)}$ ,

the equilibrium constant K_p is:

${{K}_{p}}=\frac{{{P}_{N{{O}_{2}}}}^{2}}{{{P}_{{{N}_{2}}{{O}_{4}}}}}$

Objective 1 Practice:

The correct equilibrium constant expression for the reaction below is:

$4\text{ }\!\!~\!\!\text{ N}{{\text{H}}_{3}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ 7 }\!\!~\!\!\text{ }{{\text{O}}_{2}}(\text{g)}\rightleftharpoons 4\text{ }\!\!~\!\!\text{ N}{{\text{O}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ 6 }\!\!~\!\!\text{ }{{\text{H}}_{2}}\text{O }\!\!~\!\!\text{ (g)}$

The correct pressure-based equilibrium constant expression for the reaction below is:

$\text{2 }\!\!~\!\!\text{ }{{\text{O}}_{3}}(\text{g)}\rightleftharpoons 3\text{ }\!\!~\!\!\text{ }{{\text{O}}_{2}}(\text{g)}$

The correct equilibrium constant expression for the reaction below is:

$2\text{ }\!\!~\!\!\text{ N}{{\text{O}}_{{}}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ C}{{\text{l}}_{2}}(\text{g)}\rightleftharpoons 2\text{ }\!\!~\!\!\text{ NOC}{{\text{l}}_{2}}(\text{g)}$

Objective 2:Given the equilibrium constant for one reaction, determine the value of the equilibrium constant for a related reaction.

K for Related Reactions

We saw earlier that for a chemical reaction at equilibrium such as:

$\text{aA}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ bB(g)}\rightleftharpoons \text{cC}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ dD(g)}$

the expression for K would be

$\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$

Similarly, for the reaction $\text{cC}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ dD(g)}\rightleftharpoons \text{aA}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ bB(g)}$ , the equilibrium constant expression would be:

$K=\frac{{{[A]}^{a}}{{[B]}^{b}}}{{{[C]}^{c}}{{[D]}^{d}}}$

The second reaction is simply the first reaction reversed, and the second expression of K is the reciprocal or inverse of the first one. Therefore, the expression (and value) of K for an equilibrium reaction is the reciprocal of K for the reverse reaction.

Let’s consider the reaction $\text{A}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ B(g)}\rightleftharpoons \text{C}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ D(g)}$ , which has the expression: $K=\frac{[C][D]}{[A][B]}$

The reaction $2\text{A}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ 2B(g)}\rightleftharpoons 2\text{C}(\text{g})\text{ }\!\!~\!\!\text{ +2D(g)}$ is the above reaction, except all of the coefficients have been multiplied by 2. The equilibrium constant expression for this reaction is:

$K=\frac{{{[C]}^{2}}{{[D]}^{2}}}{{{[A]}^{2}}{{[B]}^{2}}}$ .

Notice this expression is the square of the previous expression:

$K=\frac{{{[C]}^{2}}{{[D]}^{2}}}{{{[A]}^{2}}{{[B]}^{2}}}={{\left( \frac{[C][D]}{[A][B]} \right)}^{2}}$

Therefore, the value of K for a reaction multiplied by a factor (2 in our example) is K of the original reaction raised to the power of the factor (second power or squared in our example).

Objective 2 Example

At equilibrium, [H₂] = 0.106 M, [I₂] = 0.0220 M, and [HI] = 1.29 M.

Calculate K for the reaction ${{\text{H}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{I}}_{2}}(\text{g)}\rightleftharpoons 2\text{ }\!\!~\!\!\text{ HI}(\text{g)}$

$K=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}=\frac{{{[1.29]}^{2}}}{[0.106][0.220]}=714$

Calculate K for the reaction $2\text{ }\!\!~\!\!\text{ HI }\!\!~\!\!\text{ (g) }\!\!~\!\!\text{ }\rightleftharpoons {{\text{H}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{I}}_{2}}(\text{g)}$

Note that this reaction is the reverse of the original reaction ${{\text{H}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{I}}_{2}}(\text{g)}\rightleftharpoons 2\text{ }\!\!~\!\!\text{ HI}(\text{g)}$ . Therefore its K expression (and value) will be the reciprocal of the original K:

$K=\frac{1}{{{K}_{original}}}=\frac{1}{714}=0.00140$

Calculate K for the reaction ${}^{1}\!\!\diagup\!\!{}_{3}\;{{\text{H}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{}^{1}\!\!\diagup\!\!{}_{3}\;{{\text{I}}_{2}}(\text{g)}\rightleftharpoons {}^{2}\!\!\diagup\!\!{}_{3}\;\text{ }\!\!~\!\!\text{ HI}(\text{g)}$

Note that this reaction is the original reaction ${{\text{H}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{I}}_{2}}(\text{g)}\rightleftharpoons 2\text{ }\!\!~\!\!\text{ HI}(\text{g)}$ , but multiplied through by a factor of 1/3. Therefore its K expression (and value) will be the the original K raised to the 1/3 power:

$K={{\left( {{K}_{original}} \right)}^{\frac{1}{3}}}={{\left( 714 \right)}^{\frac{1}{3}}}=8.94$

Objective 2 Practice

The equilibrium constant for the reaction below is K = 3.7 x 10⁸at 25 °C.

${{\text{N}}_{2}}(\text{g)}+3\text{ }\!\!~\!\!\text{ }{{\text{H}}_{2}}(\text{g)}\rightleftharpoons \text{2 }\!\!~\!\!\text{ N}{{\text{H}}_{3}}(\text{g)}$

Calculate the value of equilibrium constant at the same temperature for the following reactions:

$\text{2 }\!\!~\!\!\text{ N}{{\text{H}}_{3}}(\text{g)}\rightleftharpoons {{\text{N}}_{2}}(\text{g)}+3\text{ }\!\!~\!\!\text{ }{{\text{H}}_{2}}(\text{g)}$

$\frac{1}{2}{{\text{N}}_{2}}(\text{g)}+\frac{3}{2}\text{ }\!\!~\!\!\text{ }{{\text{H}}_{2}}(\text{g)}\rightleftharpoons \text{N}{{\text{H}}_{3}}(\text{g)}$ .

Objective 3:Relate equilibrium constant magnitude to the position of the equilibrium and the relative equilibrium concentrations of reactants and products.

In general, equilibrium constants are a ratio of product concentrations (or partial pressures for K_p) divided by reactant concentrations (or partial pressures). These concentrations (or pressures) will be raised to powers corresponding to their coefficients in the equation.

$K=\frac{{{[\text{products}]}^{p}}}{{{\text{ }\!\![\!\!\text{ reactants}]}^{^{r}}}}$

If the value of K is much larger than 1 (K>>1), the numerator in the expression is much larger than the denominator. Typically, this means the product concentrations are much higher than the reactant concentrations at equilibrium (at equilibrium, mostly products are present). We can conclude for K>>1, the reaction favors the products and the equilibrium lies to the right.

If the value of K is much smaller than 1 (K<<1), the numerator in the expression is much smaller than the denominator. Typically, this means the product concentrations are much lower than the reactant concentrations at equilibrium (at equilibrium, mostly reactants are present). We can conclude for K<<1, the reaction favors the reactants and the equilibrium lies to the left.

Objective 4: Write an equilibrium expression for reactions involving a solid or a liquid.

K and K_p expressions only include solutions (symbolized by (aq) in chemical equations) and gases symbolized by (g)). K and K_p expressions DO NOT include pure solids (s) or liquids (l).

Where the equilibrium lies (and therefore the value of K and the concentrations of solutions and gases at equilibrium) DOES NOT depend on the amount of pure solids or liquids present. Why? The concentration (mol/L) of a pure solid or liquid does not change, even when the amount changes.

Objective 4 example

Write the equilibrium constant expression and the K_p expression for the reaction:

$\text{CaC}{{\text{O}}_{3}}(\text{s)}\rightleftharpoons \text{CaO }\!\!~\!\!\text{ (s) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ C}{{\text{O}}_{2}}(\text{g)}$

When writing the expression for K, the calcium oxide (CaO) and the calcium carbonate (CaCO₃) will be ignored as they are both solids. Only the carbon dioxide gas will be included. Since it is a product, it will be in the numerator and the expression is:

$K=[\text{C}{{\text{O}}_{2}}]$

Similarly, the K_p expression is:

${{K}_{P}}={{P}_{C{{O}_{2}}}}$

Objective 4 Practice:

Consider the reaction:

$\text{C}{{\text{O}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ C(s)}\rightleftharpoons 2\text{ }\!\!~\!\!\text{ CO }\!\!~\!\!\text{ (g) }\!\!~\!\!\text{ }$

Answer the following question:

Consider the reaction:

$\text{Ti }\!\!~\!\!\text{ }(\text{s) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ C}{{\text{l}}_{2}}(\text{g)}\rightleftharpoons \text{TiC}{{\text{l}}_{4}}(\text{l)}$

Answer the following question:

Objective 5: Given the concentrations of all reacting species, calculate the reaction quotient (Q) and predict the direction of reaction.

Equilibrium Constant (K) vs. Equilibrium Position

The Equilibrium Constant (K) is a ratio of concentrations defined by an expression. Its numerical value is constant at a given temperature for a reaction at equilibrium. Its numerical value does vary with temperature.

An equilibrium position is a set of reactant and product concentrations that coexist at equilibrium. Substitution of these concentrations into the equilibrium expression gives the value of K. There is more than one equilibrium position for a reaction at equilibrium.

Equilibrium Constant (K) vs. Equilibrium Position – example

Consider the reaction: $A(g)+B(l)\rightleftharpoons 2C(g)$

The equilibrium constant expression for the reaction would be

$K=\frac{{{[C]}^{2}}}{[A]}$

If the value of K was 10 at a particular temperature, then:

$K=\frac{{{[C]}^{2}}}{[A]}=10$

An equilibrium position for the reaction is a set of reactant and product concentrations that coexist at equilibrium. [C] = 0.70 M and [A]=0.049 M is an equilibrium position since

$K=\frac{{{[C]}^{2}}}{[A]}=\frac{{{[0.70]}^{2}}}{[0.049]}=10$

That means that concentrations of [C] = 0.70 M and [A]=0.049 M would co-exist at equilibrium.

There are other equilibrium positions (an infinite number, in fact). For example, [C] = 0.60 M and [A]=0.036 M would also be an equilibrium position.

The Reaction Quotient (Q)

Expressions for Q can be written just like an expressions for K. The concentrations in Q are not necessarily equilibrium concentrations.

We can consider Q to be a “test” of whether or not a system is at equilibrium.

If we consider our previous example:

Reaction: $A(g)+B(l)\rightleftharpoons 2C(g)$

Equilibrium constant expression for the reaction:

$K=\frac{{{[C]}^{2}}}{[A]}$

If the value of K was 10 at a particular temperature, then:

$K=\frac{{{[C]}^{2}}}{[A]}=10$

Would the system be at equilbrium if [C] = 0.90 M and [A]=0.10 M? We would plug it into the expression to see:

$Q=\frac{{{[C]}^{2}}}{[A]}=\frac{{{[0.90]}^{2}}}{[0.10]}=8.1$

Since Q (our “test”) is not equal to K (K=10 in our example), the system is not at equilibrium. So what will happen? The system will shift to reach equilibrium. In general, Q=[products]/[reactants]. Since the Q of 8.1 is less than the K value of 10, it means there is not enough product and too much reactant to be at equilibrium. The reaction will therefore shift to the right (decreasing reactant A’s concentration and increasing product C’s concentration) until equilibrium is reached (when Q=K=10).

System or equilibrium reaction shifts to reach equilibrium

If Q>K, there is too much product and not enough reactant for equilibrium. The reaction will shift to left to reach equilibrium.
If Q<K, there is too much reactant and not enough product for equilibrium. The reaction will shift to right to reach equilibrium.
If Q=K, the reaction is at equilibrium

Equilibrium SHIFTS left or right vs equilibrium LIES left or right

Students often confuse whether an equilibrium reaction will shift left or right to reach equilibrium with whether an equilibrium lies left or right (favoring reactants or products). These are not the same thing.

To determine whether an equilibrium lies to the left (favoring reactants) or the right (favoring products), compare Q and 1. (see Objective 3).

To determine whether an equilibrium reaction will shift left or right to reach equilibrium, compare Q and K.

For example, a reaction has an equilibrium constant value of K=0.000060. It is under conditions where the reaction quotient Q =0.000050. Since Q<K, the reaction would shift right to reach equilibrium. Since K<1 the reaction equilibrium lies left (favoring the reactants) at equilibrium.

Objective 5 Practice:

A 3.0 L reaction vessel contains 0.0060 moles of N₂O₅, 3.4 moles of NO₂, and 3.7 moles of O₂. The mixture is approaching the following equilibrium:

$2\text{ }\!\!~\!\!\text{ }{{\text{N}}_{2}}{{\text{O}}_{5}}(\text{g)}\rightleftharpoons 4\text{ }\!\!~\!\!\text{ N}{{\text{O}}_{2}}\text{ }\!\!~\!\!\text{ (g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{O}}_{2}}(\text{g)}$

Answer the following two questions:

Objective 6: Calculate Kc or Kp from each of the following:

a.Equilibrium concentrations or partial pressure of all species.

b.Initial concentrations and one equilibrium concentration.

Objective 7: Use the equilibrium constant to calculate equilibrium concentrations.

Solving Equilibrium Problems

There are two general types of equilibrium problems you will encounter. The first type are relatively straightforward. If you have a reaction such as

$\text{aA}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ bB(g)}\rightleftharpoons \text{cC}(\text{g})\text{ }\!\!~\!\!\text{ + }\!\!~\!\!\text{ dD(g)}$

with the equilibrium expression $K=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$ ,

there are five variables in the equation:

K, [A], [B], [C], and [D]. The molarities [A], [B], [C], and [D] are equilibrium concentrations, or concentrations at equilibrium. Given any four of these five, you should be able to solve for the remaining one.

In the second type, you are given some initial concentrations instead of equilibrium concentrations. In that case, you will have to employ a strategy known as an Initial, Concentration, Equilibrium Table (also called ICE Table). We will go over those problems later. First, you will have a chance to try some examples of the first, straightforward, type.

Objective 6 and 7 practice – first (easier) type

Consider the following reaction:

${{\text{H}}_{2}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{I}}_{2}}(\text{g)}\rightleftharpoons 2\text{ }\!\!~\!\!\text{ HI}(\text{g)}$

At equilibrium in a particular experiment, the concentrations of H₂, I₂, and HI were 0.15 M, 0.033 M, and 0.55 M respectively.

Dinitrogen tetroxide decomposes as follows:

${{\text{N}}_{2}}{{\text{O}}_{4}}(\text{g)}\rightleftharpoons \text{2 }\!\!~\!\!\text{ N}{{\text{O}}_{2}}(\text{g)}$

At a temperature where Kp = 0.133, N₂O₄gas was introduced to a flask. After the system was allowed to reach equilibrium, the pressure of N₂O₄ was found to be 2.71 atm.

Objective 6 and 7 practice – second (slightly more difficult) type

In the second type, you are given some initial concentrations instead of equilibrium concentrations. In that case, you will have to employ a strategy known as an Initial, Concentration, Equilibrium Table (also called ICE Table). Let’s look at an example that will require this type of problem:

Objective 6 and 7 practice – second (slightly more difficult) type – example

At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl₃(g) and 0.00870 mol PCl₅(g). After the system was allowed to reach equilibrium, 2.00 x 10^-3 mol Cl₂(g) was found in the flask. PCl₅(g) decomposes as follows:

$\text{PC}{{\text{l}}_{5}}(\text{g)}\rightleftharpoons \text{PC}{{\text{l}}_{3}}(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ C}{{\text{l}}_{2}}\text{(g)}$

Calculate the equilibrium concentrations of all three species and K.

Objective 6 and 7 practice – second (slightly more difficult) type – practice

When SO₂and O₂ , each at 1.00 atm are placed in a 1.00 L flask, at a certain temperature, they react according to the equation:

$2\text{ }\!\!~\!\!\text{ S}{{\text{O}}_{2}}\text{ }\!\!~\!\!\text{ }(\text{g) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ }{{\text{O}}_{2}}(\text{g)}\rightleftharpoons 2\text{ }\!\!~\!\!\text{ S}{{\text{O}}_{3}}(\text{g)}$

The system is allowed to come to equilibrium and 0.26 atm of SO₃are produced. Calculate the value of K_p and enter it below.

More Objective 6 and 7 practice – second (slightly more difficult) type – example

What do you do when K is given but no equilibrium concentrations are given

In this example, we will look at a problem in which K is given and we are solving for the equilibrium concentrations (no equilibrium concentrations are given).

Example 1

Consider the gas phase reaction in which molecular hydrogen and molecular fluorine combine to form hydrogen fluoride:

${{H}_{2}}(g)+{{F}_{2}}(g)\rightleftharpoons 2HF(g)$

3.00 moles of each of the three components were added to a 1.50 L flask at a temperature where K = 121. Calculate the equilibrium concentrations of all three components.

To start, we can write the equilibrium expression and realize that it will equal the given value of K, which is 121:

$K=\frac{{{[HF]}^{2}}}{[{{H}_{2}}][{{F}_{2}}]}=121$

Setting up an ICE table with the given information:

	H₂	F₂	↔	2 HF
I	2.00	2.00		2.00
C
E

In the previous examples you have looked at, you were given an equilibrium concentration, which allowed you to determine the change. In this problem, though, we are given the value of K, which we will have to use to solve for the change. We do know that the change will be somewhere between zero (no change) and 2.00 M (changes greater than 2.00 M would result in a negative equilibrium molarity, which is of course impossible.

Since we have to calculate the change, we will make the change a variable x to be solved for. Before we do that, though, we will have to determine whether the equilibrium will shift right (subtracting from reactants and adding to products) or shift left (adding to reactants and subtracting from products). As you learned earlier, the way we determine whether the initial concentrations will shift right or left to reach equilibrium is to compare Q and K.

Q is calculated as follows: $Q=\frac{{{[HF]}^{2}}}{[{{H}_{2}}][{{F}_{2}}]}=\frac{{{[2.00]}^{2}}}{[2.00][2.00]}=1$

Since the Q of 1 is less than the K of 121 the reaction will shift to the right. This means the change in the reactants will be a decrease (subtracted) in the reactants H₂ and F₂ and an increase (added) to the product HF:

	H₂	F₂	↔	2 HF
I	2.00	2.00		2.00
C	-x	-x		+2x
E	2.00-x	2.00-x		2.00 + 2x

Note the change added for HF is +2x, due to the coefficient of 2 for HF in the balanced equation (remember, the ratio of the changes corresponds to the ratio of the coefficients).

Substituting the equilibrium concentrations into the K expression yields:

$K=\frac{{{[HF]}^{2}}}{[{{H}_{2}}][{{F}_{2}}]}=\frac{{{[2.00+2x]}^{2}}}{[2.00-x][2.00-x]}=121$

Solving this for x will tell us the change and the equilibrium concentrations.

this can be rearranged to:

$\frac{{{[2.00+2x]}^{2}}}{{{[2.00-x]}^{2}}}=121$

The easiest way to solve for x is to recognize that this is a square over a square. Therefore the easiest way to solve for x is to take the square root of both sides:

$\sqrt{\frac{{{[2.00+2x]}^{2}}}{{{[2.00-x]}^{2}}}}=\sqrt{121}$

leaving $\frac{[2.00+2x]}{[2.00-x]}=11$

Solving this for x leaves x=1.54.

Remember though, x=1.54 is not the final answer to the problem (students often make that mistake as they are asked to solve for x so often in math classes. Now that we have x, we have to go back and put the value of 1.54 in for x in the ICE table to determine the equilibrium concentrations:

	H₂	F₂	↔	2 HF
I	2.00	2.00		2.00
C	-1.54	-1.54		+2(1.54)
E	0.46	0.46		5.08

The final equilibrium concentrations are therefore [H₂]=[F₂]=0.46 M and [HF]=5.08 M

Example 2

1.0 mol NOCl gas is placed into a 2.0 L flask and allowed to decompose into the gases NO and Cl₂:

$2NOCl(g)\rightleftharpoons 2NO(g)+C{{l}_{2}}(g)$

If the equilibrium constant is 1.6 X 10^-5 , what are the equilibrium concentrations of all three species ?

We can start with the expression for K:

$K=\frac{{{[NO]}^{2}}{{[C{{l}_{2}}]}^{{}}}}{{{[NOCl]}^{2}}}=1.6\times {{10}^{-5}}$

Since we are given only the initial concentration of NOCl, we will need an ICE table:

	2 NOCl	↔	2 NO	Cl₂
I	0.50		0	0
C
E

The initial molarity of NOCl is 1.0 mol/2.0 L = 0.50 L.

Since we are not given any equilibrium concentrations, we will put x’s in the change row and use the given value of K later to solve for x.

	2 NOCl	↔	2 NO	Cl₂
I	0.50		0	0
C	-2x		+2x	+x
E	0.050-2x		2x	x

Notice there is a -2x for NOCl and a 2x for NO in the change row due to the reaction coefficients for NOCl and NO.

Substituting the equilibrium values into the expression for K yields:

$K=\frac{{{[2x]}^{2}}[x]}{{{[0.50-2x]}^{2}}}=1.6\times {{10}^{-5}}$

Solving for x yields 0.0097. You can solve for x easily if you use a calculator with algebraic solving ability or an algebraic solver such as Wolfram Alpha.

There is an approximate solution strategy that makes this problem much easier to solve without the use of an algebraic solver. This strategy can be used any time K is very small (much less than one)

Since K is much less than one, the equilibrium favors the reactants, and very little of the NOCl reactant will decompose to go to products. Therefore the subtracted 2x in the change row will only be a very small fraction of the initial 0.50 M. Since the 2x will be very small compared to the 0.50, we can say that 0.50-2x equals about 0.50:

$0.50-2x\approx 0.50$

Therefore,

$K=\frac{{{[2x]}^{2}}[x]}{{{[0.50-2x]}^{2}}}=\frac{{{[2x]}^{2}}[x]}{{{[0.50]}^{2}}}$

solving,

$1.6\times {{10}^{-5}}=\frac{{{[2x]}^{2}}[x]}{{{[0.50]}^{2}}}$

$1.6\times {{10}^{-5}}=\frac{4{{x}^{3}}}{{{[0.50]}^{2}}}$

$1.6\times {{10}^{-5}}({{[0.50]}^{2}})=4{{x}^{3}}$

$4.0\times {{10}^{-6}}=4{{x}^{3}}$

and x=0.010.

As you can see, this is very close to the 0.0097 obtained by the algebraic solver.

Again, this strategy can be used any time K is much less than 1, and will be used extensively in our next topic (acids and bases).

Using our x of 0.010, we can solve for the equilibrium concentration of all three species:

[NOCl]=0.50-2x=0.50-2(0.010)=0.48 M

[NO]=2x=2(0.010)=0.020 M

[Cl₂]=x=0.010 M

Objective 8:Based on LeChatelier’s principle explain how relative equilibrium quantities of reactants and products are shifted by changes to the equilibrium conditions.

LeChatlier’s Principle will allow you to know how a reaction in equilibrium will respond when reaction conditions are changed. It can be summarized as “shift relieves stress”. What this means is that

A reaction is at equilibrium
A change imposed on the reaction system causes stress, throwing the reaction out of equilibrium
The equilibrium shifts left or right to reduce the stress, returning to equilibrium. The new concentrations will be different, but the equilibrium constant will be the same (unless you change the temperature)

What do I mean by a “change imposed on the reaction system”? It can mean several things, including

Addition or removal of a reactant or product
Increasing or decreasing pressure
Increasing or decreasing volume
addition of a catalyst
increasing or decreasing reaction temperature

We will look at each of these changes next and see how we can use LeChatlier’s Principle to see how the reaction would respond.

Addition or removal of a reactant or product

Let’s consider an example with the following equilibrium reaction:

${{H}_{2}}(g)+{{I}_{2}}(s)\rightleftharpoons 2HI(g)$

If we were to add H₂ gas to a system at equilibrium, the system will respond by shifting right to consume the extra added H₂ gas. Similarly, if we were to add HI gas, the system will respond by shifting left to consume the extra added HI gas. If a reactant or product is added, the system will shift away from that reactant or product that is added.

Conversely, If we were to remove H₂ gas from a system at equilibrium, the system will respond by shifting left to replenish the removed H₂ gas, returning the system to equilibrium. If a reactant or product is removed, the system will shift away toward the reactant or product that is removed.

What if the solid I₂ is added or removed? Since they do not appear in the K expression and do not affect equilibrium position, addition or removal of liquid (l) or solid (s) reactants or products will not affect equilibrium and no change or shift will occur.

Changes in concentration of a reactant or product

Increasing or decreasing the concentration has a similar effect to addition or removal of a reactant or product — increasing concentration is the same as addition and decreasing concentration is the same as removal. Liquids and solids have no effect (as pure, their concentrations cannot change. In our example,

${{H}_{2}}(g)+{{I}_{2}}(s)\rightleftharpoons 2HI(g)$ ,

decreasing the concentration of HI (g) would cause the system to shift right to return to equilibrium.

Pressure or volume effects

Changing pressure or volume only effects equilibrium reactions with gases. We can consider pressure and volume together since they are inversely proportional (see the notes on gases). Therefore, an increase in pressure will have the same effect as a decrease in volume (or vice versa)

Effect of an increase in pressure or decrease in volume

It’s easiest to think of this as a decrease in volume – whether pressure is increased or volume is decreased, the system will respond by shrinking its gas volume by shifting toward the side with fewer moles of gas.

Effect of an decrease in pressure or increase in volume

It’s easiest to think of this as a increase in volume – whether pressure is decreased or volume is increased, the system will respond by expanding to fill the larger gas volume by shifting toward the side with more moles of gas.

Example – pressure or volume changes

Our example reaction,

${{H}_{2}}(g)+{{I}_{2}}(s)\rightleftharpoons 2HI(g)$ ,

has one mole of gas on the reactant (left) side and two moles of gas on the product (right side).

Therefore, a decrease of pressure or an increase in volume will shift the reaction right (toward the higher number of moles of gas). An increase of pressure or a decrease in volume will shift the reaction left.

Addition of a catalyst

A catalyst is a substance that speeds up a chemical reaction without being consumed. They will be studied more in CHEM 152. In an equilibrium reaction, the forward rate is equal to the reverse rate. A catalyst speeds them both up but they are still equal. Therefore, addition of a catalyst does not affect equilibrium and no shift is expected.

Effect of temperature changes

Increasing or decreasing temperature will cause reaction shifts. Additionally, temperature changes are the only variable that changes the value of K.

To understand and correctly predict the reaction shift resulting from a temperature change, you must first consider whether the reaction is exothermic or endothermic:

Exothermic reactions produce heat, so heat can be considered a “product”
Endothermic reactions require heat, so heat can be considered a “reactant”

Increasing temperature can then be seen as adding heat, and decreasing temperature can be seen as removing heat. Let’s look at an example:

Example – Effect of temperature changes

Our example reaction,

${{H}_{2}}(g)+{{I}_{2}}(s)\rightleftharpoons 2HI(g)$ , is endothermic with ΔH=53 kJ

Therefore. we could think of heat as a reactant:

${{H}_{2}}(g)+{{I}_{2}}(s)+heat\rightleftharpoons 2HI(g)$

If we increased temperature, that would be like adding heat. Adding heat, like adding other reactants (see above) would shift the reaction to the right. This would also increase product concentration, increasing the value of K. Conversely, decreasing temperature would shift the reaction left and decrease K. Recall, temperature changes are the only changes that change the value of K.

Objective 8 Practice

The following endothermic reaction is at equilibrium:

$\text{CaC}{{\text{O}}_{3}}(\text{s)}\rightleftharpoons \text{CaO }\!\!~\!\!\text{ (s) }\!\!~\!\!\text{ + }\!\!~\!\!\text{ C}{{\text{O}}_{2}}(\text{g)}$

Answer the following questions:

OpenStax Section 13.3 gives additional examples of LeChatlier’s Principle for all of the variable changes discussed above.

Objective 1: Given the balanced equation of a reversible reaction, write the equilibrium constant expression.

The Law of Mass Action and Equilibrium Constants

More on the Equilibrium Constant (K)

Pressure-Based Equilibrium Constants for Gas-Phase Reactions (Kp)

Equilibrium Constant Examples

Objective 1 Practice:

Objective 2:Given the equilibrium constant for one reaction, determine the value of the equilibrium constant for a related reaction.

K for Related Reactions

Objective 2 Example

Objective 2 Practice

Objective 3:Relate equilibrium constant magnitude to the position of the equilibrium and the relative equilibrium concentrations of reactants and products.

Objective 4: Write an equilibrium expression for reactions involving a solid or a liquid.

Objective 4 example

Objective 4 Practice:

Objective 5: Given the concentrations of all reacting species, calculate the reaction quotient (Q) and predict the direction of reaction.

Equilibrium Constant (K) vs. Equilibrium Position

Equilibrium Constant (K) vs. Equilibrium Position – example

The Reaction Quotient (Q)

System or equilibrium reaction shifts to reach equilibrium

Equilibrium SHIFTS left or right vs equilibrium LIES left or right

Objective 5 Practice:

Objective 6: Calculate Kc or Kp from each of the following:

a.Equilibrium concentrations or partial pressure of all species.

b.Initial concentrations and one equilibrium concentration.

Objective 7: Use the equilibrium constant to calculate equilibrium concentrations.

Solving Equilibrium Problems

Objective 6 and 7 practice – first (easier) type

Objective 6 and 7 practice – second (slightly more difficult) type

Objective 6 and 7 practice – second (slightly more difficult) type – example

Objective 6 and 7 practice – second (slightly more difficult) type – practice

More Objective 6 and 7 practice – second (slightly more difficult) type – example

What do you do when K is given but no equilibrium concentrations are given

Example 1

Example 2

Objective 8:Based on LeChatelier’s principle explain how relative equilibrium quantities of reactants and products are shifted by changes to the equilibrium conditions.

Addition or removal of a reactant or product

Changes in concentration of a reactant or product

Pressure or volume effects

Effect of an increase in pressure or decrease in volume

Effect of an decrease in pressure or increase in volume

Example – pressure or volume changes

Addition of a catalyst

Effect of temperature changes

Example – Effect of temperature changes

Objective 8 Practice

Pressure-Based Equilibrium Constants for Gas-Phase Reactions (K_p)