Introductory Chemistry Online/Principles of Chemical Equilibrium

  Chapter 10. Principles of Chemical Equilibrium

As we have studied chemical reactions in this course, we have used a “reaction arrow” to indicate the process of reactants being converted into products. The implication here is that the reaction is “irreversible”, proceeding in the direction of the arrow. Many simple reactions that we encounter in chemistry, however, are not irreversible, but proceed in both directions with products readily be converted back into reactants. When a set of reactions, such as this, proceed so that the rate of conversion in one direction equals the rate of conversion in the other, we say the reactions are in equilibrium. An equilibrium system is shown by using a set of double arrows, proceeding in opposite directions. An understanding of equilibrium is essential to an appreciation of the concepts behind acid-base behavior, solubility phenomena, etc.

  10.1 The Concept of Equilibrium Reactions

Pure dinitrogen tetroxide (N₂O₄) is a colorless gas that is widely used as a rocket fuel. Although N₂O₄ is colorless, when a container is filled with pure N₂O₄, the gas rapidly begins to turn a dark brown (Figure 10.1). A chemical reaction is clearly occurring, and indeed, chemical analysis tells us that the gas in the container is no longer pure N₂O₄, but has become a mixture of dinitrogen tetroxide and nitrogen dioxide; N₂O₄ is undergoing a decomposition reaction to form NO₂. If the gaseous mixture is cooled, it again turns colorless and analysis tells us that it is again, almost pure N₂O₄; this means that the NO₂ in the mixture can also undergo a synthesis reaction to re-form N₂O₄. A more detailed analysis of the relative concentrations of N₂O₄ and NO₂ at various times gives us the data that are shown in the plot in Figure 10.2. Initially, only N₂O₄ is present. As the reaction proceeds, the concentration of N₂O₄ decreases and the concentration of NO₂ increases. However, if you examine the figure, after some time, the concentrations of N₂O₄ and NO₂ have stabilized and, as long as the temperature is not changed, the relative concentrations of the two gasses remain constant.

The reversible reaction of one mole of N₂O₄, forming two moles of NO₂, is a classic example of a chemical equilibrium. We encountered the concept of equilibrium in Chapter 9 when we dealt with the autolysis of water to form the hydronium and hydroxide ions, and with the dissociation of weak acids in aqueous solution.

2 H₂O ⇄ H₃O⁺ + HO^–

The autolysis of water; see Section 8.5

When we wrote these chemical equations, we used a double arrow to signify that the reaction proceeded in both directions. Using this convention, the dissociation of dinitrogen tetroxide to form two molecules of nitrogen dioxide can be shown as:

N₂O₄ ⇄ 2 NO₂

If the temperature of our gas mixture is again held constant and the total pressure of the gas in the container is varied, analysis shows that the partial pressure of N₂O₄ varies as the square of the partial pressure of NO₂ (Figure 10.3; remember, the Ideal Gas Laws tell us that the partial pressure of a gas, P_gas, is directly proportional to the concentration of that gas in the container). Mathematically, the relationship between the partial pressures of the two gasses can be expressed by Equation 1.

${\frac {\left(P_{NO_{2}}\right)^{2}}{P_{N_{2}O_{4}}}}=K$

Equation 1.

  10.2 The Equilibrium Constant

The constant in Equation 1 is called the equilibrium constant for the reaction. What the equilibrium constant for this reaction tells us is that, regardless of pressures (or concentrations), a mixture of the two gasses will undergo reaction such that the ratios of the partial pressures reach a constant value, given by the equilibrium constant, K. Once this constant ratio has been reached does this mean the reactions stop? Of course not; examine Figure 10.2. In the region of the plot where the concentrations of N₂O₄ and NO₂ are constant (the lines are level) N₂O₄ is still decomposing to form two molecules of NO₂ and two molecules of NO₂ are still reacting to synthesize a molecule of N₂O₄, but the lines are level because the rates of the two chemical reactions have become constant; N₂O₄ is decomposing at the same rate as two molecules of NO₂ are reacting to form N₂O₄. (Figure 10.4)

In theory, all chemical reactions are equilibria. In practice, however, most reactions are so slow in the reverse direction that they are considered “irreversible”. When a reaction evolves a gas, forms a precipitate or proceeds with the generation of a large amount of heat or light (for example, combustion) the reaction is essentially irreversible. Many chemical reactions are, however, readily reversible and for these reactions the mathematical expression for the equilibrium constant can be written using a simple set of rules.

1. Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator.
2. If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
3. The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
4. Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.

Thus, for the reaction of nitrogen with hydrogen gas to form ammonia:

N₂ (g) + 3 H₂ (g) ⇄ 2 NH₃ (g)

The expression for the equilibrium constant will have the partial pressure of ammonia in the numerator, and it will be squared, corresponding to the coefficient “2” in the balanced equation; $\left(P_{NH_{3}}\right)^{2}$. Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the “first power” (which is not shown) and the partial pressure of hydrogen will be cubed, corresponding to the coefficient “3”; $\left(P_{N_{2}}\right)\left(P_{H_{2}}\right)^{3}$. The final expression for the equilibrium constant is given in Equation 2.

${\frac {\left(P_{NH_{3}}\right)^{2}}{P_{N_{2}}\left(P_{H_{2}}\right)^{3}}}=K$

Equation 2.

Example 10.1 Writing Equilibrium Expressions

For the chemical reactions shown below, write an expression for the equilibrium constant in terms of the partial pressures of the reactants and products.

PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)
2 NOCl (g) ⇄ 2 NO (g) + Cl2 (g)

Exercise 10.1 Writing Equilibrium Expressions

For the chemical reaction shown below, write an expression for the equilibrium constant in terms of the partial pressures of the reactants and products.

PCl3 (g) + 3 NH3 (g) ⇄ P(NH2)3 (g) + 3 HCl (g)

  10.3 Calculating Equilibrium Values

The numeric value of the equilibrium constant tells us something about the ratio of the reactants and products in the final equilibrium mixture. Likewise, the magnitude of the equilibrium constant tells us about the actual composition of that mixture.

In the three equilibrium systems shown in Figure 10.5, the first depicts a reaction in which the ratio of products to reactants is very small. Because the expression for the equilibrium constant is given by the pressure (or concentration) of products divided by the pressure (or concentration) of reactants, the equilibrium constant, K, for this system is also small. In the second example, the concentrations of reactants and products are shown to be equal, making the ratio (the equilibrium constant) equal to “1”. In the last example, the products are shown to dominate the equilibrium mixture, making the ratio ${}^{\left(P_{Products}\right)}\!\!\diagup \!\!{}_{\left(P_{Reactants}\right)}\;$very large. In these examples, the stoichiometric ratios of the reactants and products are one and there is only one reactant and only one product; if multiple reactants or products are involved, the relationship between their concentrations would be more complex, but that ratio is always given by the expression for K. This fact allows us to take data for an equilibrium reaction and, if K is known, calculate concentrations for reactants and products. Likewise, if all of the equilibrium concentrations are known, we can use these to calculate a value for the equilibrium constant.

In these types of problems, an ICE table is often useful. This table has entries for Initial concentrations (or pressures), Equilibrium concentrations and any Change between the initial and equilibrium states. For example, consider the reaction between carbon monoxide and chlorine to form phosgene, a deadly compound that was used as a gas warfare agent in World War I. (Figure 10.6)

CO (g) + Cl₂ (g) ⇄ COCl₂ (g)

A typical equilibrium problem might read as follows:

A mixture of CO and Cl₂ has initial partial pressures of 0.60 atm for CO and 1.10 atm for Cl₂. After the mixture reaches equilibrium, the partial pressure of COCl₂ is 0.10 atm. Determine the value of K.

The initial pressures for carbon monoxide and chlorine are placed in the first row and the equilibrium pressure for phosgene is placed in the last row. Initially, the pressure of phosgene was zero, so that goes in the first row; the change for phosgene is therefore “+ 0.10 atm”.

	$P_{CO}$	$P_{Cl_{2}}$	$P_{COCl_{2}}$
=== Initial ===	0.60 atm	1.10 atm	0 atm
Change			+ 0.10 atm
Equilibrium			0.10 atm

Because one mole of CO is required to make one mole of COCl₂ the partial pressure of CO must have dropped by 0.10 atm (the Change) in order to make COCl₂ with a partial pressure of 0.10 atm, giving a final (Equilibrium) pressure of 0.50 atm for carbon monoxide. Likewise, one mole of chlorine is required to make one mole of COCl₂ making the Change for chlorine 0.10 atm and the Equilibrium partial pressure 1.00 atm. The completed table is shown below.

	$P_{CO}$	$P_{Cl_{2}}$	$P_{COCl_{2}}$
=== Initial ===	0.60 atm	1.10 atm	0 atm
Change	-0.10 atm	-0.10 atm	+ 0.10 atm
Equilibrium	0.50 atm	1.00 atm	0.10 atm

The equilibrium expression for the phosgene-forming reaction is given by Equation 3.

${\frac {P_{COCl_{2}}}{P_{CO}P_{Cl_{2}}}}=K$

Equation 3.

Substituting the values from the Table into this equation:

$K={\frac {P_{COCl_{2}}}{P_{CO}P_{Cl_{2}}}}={\frac {\left(0.10\right)}{\left(0.50\right)\left(1.00\right)}}=0.20$

Notice that equilibrium constants for gas phase reactions are not typically written with units, although units are sometimes used in equilibrium constants calculated from molar concentrations. Many textbooks differentiate between equilibrium constants calculated from partial pressures and molar concentrations by affixing subscripts; K_P and K_c. In this book, we will simply use K and K_c to represent the two; a value for K will always denote a constant calculated from partial pressure data.

Example 10.2 Determining Equilibrium Values

For the reaction shown below, all four gasses are introduced into a vessel, each with an initial partial pressure of 0.500 atm, and allowed to come to equilibrium; at equilibrium, the partial pressure of SO₃ is found to be 0.750 atm. Determine the value of K.

SO₂ (g) + NO₂ (g) ⇄ SO₃ (g) + NO (g)

Exercise 10.2 Determining Equilibrium Values

For the reaction shown above, the initial partial pressures of SO₃ and NO are 0.500 atm under conditions where the equilibrium constant is, K = 9.00. The equilibrium partial pressure for SO₂ is found to be 0.125 atm. Calculate the equilibrium partial pressure for SO₃.

 

  10.4 Using Molarity in Equilibrium Calculations

As we have pointed out several times in the preceding sections, the Ideal Gas Laws (Chapter 10) tell us that the partial pressure of a gas and the molar concentration of that gas are directly proportional. We can show this simply by beginning with the combined gas law:

$P_{gas}V=nRT$

If we divide both sides by the volume, V, and state that V must be expressed in liters, the right side of the equation now contains the term $\left({}^{n}\!\!\diagup \!\!{}_{V_{liters}}\;\right)$. Realizing that the number of moles of gas (n) divided by the volume in liters is equal to molarity, M, this expression can be re-written as:

$P_{gas}=MRT$

Using this expression, molar concentrations can easily be substituted for partial pressures, and visa versa.

Example 10.3 Using Molarity in Equilibrium Calculations

For the reaction shown below, if the molar concentrations of SO₃, NO and SO₂ are all 0.100 M, what is the equilibrium concentration of NO₂?

Exercise 10.3 arity in Equilibrium Calculations

For the reaction between carbon monoxide and chlorine to form phosgene, the equilibrium constant calculated from partial pressures is K = 0.20. How does this value relate to the equilibrium constant, K_C, under the same conditions, calculated from molar concentrations?

CO (g) + Cl₂ (g) ⇄ COCl₂ (g)

  10.5 Le Chatelier's Principle: Stress and Equilibria

Consider a simple chemical system that is at equilibrium, such as dinitrogen tetroxide: nitrogen dioxide. The Law of Mass Action states that when this system reaches equilibrium, the ratio of the products and reactants (at a given temperature) will be defined by the equilibrium constant, K. Now imagine that, after equilibrium has been reached, more dinitrogen tetroxide is introduced into the container. In order for the ratio to remain constant (as defined by K) some of the N₂O₄ that you added must be converted to NO₂. The addition of reactants or products to a system at equilibrium is commonly referred to as a “stress”. The response of the system to this stress is dictated by Le Chatelier's Principle, which states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. Again, stress refers to a change in concentration, a change in pressure or a change in temperature, depending on the system being examined. If pressure or temperature are changed, the numeric value K will change; if only concentration changes are involved, K does not change. We will consider temperature and pressure effects in General Chemistry, but for now, remember; in a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.

  10.6 Equilibria involving Acids and Bases

In Chapter 8, we learned that a “weak acid” was only partially dissociated in solution, while a “strong acid” was fully dissociated. Now that we better understand the concept of equilibrium, these two classes of Brønsted acids can simply be differentiated based on their equilibrium constants. For an acid, BH, that dissociates in water to form B^– and hydronium ion, we can write a simple equilibrium expression, as follows.

BH(aq) + H₂O(l) ⇄ B^–(aq) + H₃O⁺(aq)

$K_{C}={\frac {\left[H_{3}O^{+}\right]\left[B^{-}\right]}{\left[BH\right]}}=''K_{a}''$

Equation 4.

You should note two things in this equation; the term for water does not appear (remember, solids and liquids are never written in equilibrium expressions) and the equilibrium constant for K_C is written as K_a to denote that this is an acid dissociation equilibrium. Now, as we learned in Chapter 8, a strong acid is “fully dissociated”, which simply means that [BH] is very, very small, thus K_a for a strong acid is very, very large. A weak acid is only “partially dissociated” which means that there are significant concentrations of both BH and B^– in solution, thus K_a for a weak acid is “small”. For most common weak acids, the values for K_a will be in the range of 10^-3 to 10^-6. As an example, consider acetic acid (the acidic component of vinegar) where K_a = 1.8 × 10^-5.

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO^–(aq) + H₃O⁺(aq)

$K_{a}={\frac {\left[H_{3}O^{+}\right]\left[CH_{3}COO^{-}\right]}{\left[CH_{3}COOH\right]}}=1.8\times 10^{-5}$

Equation 5.

Example 10.4 Equilibria involving Acids and Bases

A series of acids have the following Ka values: rank these in descending order from the strongest acid to the weakest acid.

A. 6.6 × 10^–4B. 4.6 × 10^–4C. 9.1 × 10^–8D. 3.0 × 10²

Exercise 10.4 Equilibria involving Acids and Bases

At 25.0 oC, the concentrations of H3O+ and OH– in pure water are both 
1.00 × 10-7 M, making Kc = 1.00 × 10-14 (recall that this equilibrium constant is generally referred 
to as Kw). At 60.0o C, Kw increases 
to 1.00 × 10-13. What is the pH of a sample of pure water at 60.0o C?

  10.7 The pH of Weak Acid Solutions

For a solution of a strong acid, calculating the [H₃O⁺] concentration is simple; because the acid is 100% dissociated, the concentration of hydronium ions is equal to the molar concentration of the strong acid (this is, of course, only true for a monoprotic acid such as HCl or HNO₃; for H₂SO₄, [H₃O⁺] = 2 × [H₂SO₄], etc.). For a weak acid, however, the hydronium ion concentration will be much, much less than the molar concentration of the acid and [H₃O⁺] must be calculated using the value of K_a. We can approach this using an ICE table, like we did for previous equilibrium problems. If we prepared a solution of acetic acid (Figure 10.7)that was exactly 0.50 M, then initially [CH₃COOH] is 0.50 M and both [CH₃COO^–] and [H₃O⁺] are zero. A small amount of CH₃COOH will ionize; let’s call this x, making the change for [CH₃COOH] “-x”, increasing both [CH₃COO^–] and [H₃O⁺] by the amount “+x”. Finally, the equilibrium concentration of [CH₃COOH] will be (0.50 M – x) and both [CH₃COO^–] and [H₃O⁺] will be x. The completed table is shown below.

	[CH3COOH]	[CH3COO–]	[H3O]+
=== Initial ===	0.50 M	0	0
Change	- x	+ x	+ x
Equilibrium	0.50 M - x	x	x

The expression for K_a for acetic acid is given in Equation 5. Substituting for our equilibrium values:

$K_{a}=1.8\times 10^{-5}={\frac {\left[H_{3}O^{+}\right]\left[CH_{3}COO^{-}\right]}{\left[CH_{3}COOH\right]}}={\frac {x^{2}}{0.50-x}}$

$orx^{2}+9.0\times 10^{-6}x-1.8\times 10^{-5}=0$

Equation 6.

Equation 6 is a quadratic equation and we could solve it using the standard quadratic formula. This is not necessary, however, because acetic acid is a weak acid and by definition, very little of the dissociated form will exist in solution, making the quantity x very, very small. If x is much, much less than 0.50 M (our initial concentration of acetic acid), then (0.50 M – x)  0.50 M and Equation 6 simplifies to:

$K_{a}=1.8\times 10^{-5}={\frac {\left[H_{3}O^{+}\right]\left[CH_{3}COO^{-}\right]}{\left[CH_{3}COOH\right]}}={\frac {x^{2}}{0.50}}$

$orx=\left[H_{3}O^{+}\right]={\sqrt {\left(1.8\times 10^{-5}\right)\times 0.50}}=3.0\times 10^{-3}M$

We can test our assumption by substituting for x; (0.50 – 0.0030) = 0.497, which rounds to 0.50 to two significant figures. Because the concentration of hydronium ion is very small for a weak acid, for most typical solutions, the concentration of hydronium ion can be estimated simply as:

$\left[H_{3}O^{+}\right]={\sqrt {K_{a}\times C_{0}}}$

Equation 7.

where C₀ is the initial molar concentration of the weak acid.

Example 10.5 Weak Acid Solutions

a) Nitrous acid (HNO2) is a weak acid with a Ka of 4.3 × 10-4. 
Estimate the hydronium ion concentration and the pH for a 0.50 M solution 
of nitrous acid in distilled water. 

b) Acetic acid is a weak acid with Ka = 1.8 × 10-5. For a solution of 
acetic acid in water, the [H3O+] is found to be 
4.2 × 10-3 M. What is the concentration of unionized acetic acid in this solution? 

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO^–(aq) + H₃O⁺(aq)

c) A solution is prepared in which acetic acid is 0.700 M and its conjugate base, 
acetate anion is 0.600 M. As shown above, the Ka of acetic acid 
is 1.8 x 10-5; what will the pH of this solution be?

Exercise 10.5 Writing Equilibrium Expressions

What concentration of the weak acid, acetic acid (K_a = 1.8 × 10^-5)
must you have in pure water in order for the final pH to be 2.38?

  10.8 Solubility Equilibria

In Chapter 5 we learned about a class of reactions that involved the formation of a solid that was “insoluble” in water, and precipitated from the solution. In these “precipitation reactions”, one ionic salt was described as “insoluble”, driving the reaction towards the formation of products. Silver chloride is a classic example of this. If you mix silver nitrate (almost all nitrate salts are “soluble” in water) with sodium chloride, a copious white precipitate of silver chloride formed and the silver nitrate was deemed “insoluble” (Figure 10.8).

Nonetheless, if you took the clear solution from above the silver chloride precipitate and did a chemical analysis, there will be sodium ions, nitrate ions, and traces of chloride ions and silver ions. The concentrations of silver and chloride ions would be about 1.67 × 10^-5 M, far below the concentrations we typically work with, hence we say that silver chloride is “insoluble in water”. That, of course, is not true. Solubility is an equilibrium in which ions leave the solid surface and go into solution at the same time that ions are re-deposited on the solid surface. For silver chloride, we could write the equilibrium expression as:

AgCl(s) ⇄ Ag⁺(aq) + Cl^-(aq)

In order to write the expression for the equilibrium constant for this solubility reaction, we need to recall the rules stated in Section 10.2 of this chapter; Rule #4 states, “Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.” Because silver chloride is a solid, the expression for the equilibrium constant is simply,

$K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]$

Note that we have denoted the equilibrium constant as K_sp, where “sp” refers to solubility equilibrium, or “solubility product” (the product of the concentrations of the ions). We can calculate the value of K_sp for silver chloride from the analytical data that we cited above; an aqueous solution above solid silver chloride has a concentration of silver and chloride ions of 1.67 × 10^-5 M, at 25˚ C. Because the concentrations of silver and chloride ions are both 1.67 × 10^-5 M, the value of K_sp under these conditions must be:

$K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]=\left(1.67\times 10^{-5}\right)^{2}=2.79\times 10^{-10}$

This is very small, considering that K_sp for sodium chloride is about 29!

For a salt such as PbI₂ (Figure 10.9), chemical analysis tells us that the lead concentration in a saturated solution (the maximum equilibrium solubility under a specified set of conditions, such as temperature, pressure, etc.) is about 1.30 × 10^-3 M. In order to calculate K_sp for lead (II) iodide, you must first write the chemical equation and then the equilibrium expression for K_sp and then simply substitute for the ionic concentrations. As you do this, remember that there are two iodide ions for every lead ion, therefore the concentrations for lead (II) and iodide are 1.30 × 10^-3 M and 2.60 × 10^-3 M, respectively.

PbI₂(s) ⇄ Pb²⁺(aq) + 2 I^-(aq)

$K_{sp}=\left[Pb^{2+}\right]\left[I^{-}\right]^{2}=\left(1.30\times 10^{-3}\right)\left(2.60\times 10^{-3}\right)^{2}=8.79\times 10^{-9}$

  Study Points

• If two opposing chemical reactions proceed simultaneously at the same rate, the processes are said to be in equilibrium. The two opposing reactions are shown linked with a double arrow (⇄). An example of opposing chemical reactions and their equilibrium expressions are:

N₂(g) + 3 H₂(g) → 2 NH₃(g) and 2 NH₃(g) → N₂(g) + 3 H₂(g):

The equilibrium would be written as: N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

• An equilibrium constant (K) is a numerical value that relates the concentrations of the products and reactants for a chemical reaction that is at equilibrium. The numeric value of an equilibrium constant is independent of the initial concentrations of reactants, but is dependent on the temperature. Equilibrium constants are generally not written with units (although they may be).
• Because equilibrium constants are written as the concentrations (or partial pressures) of products divided by the concentrations (or partial pressures) of reactants, a large value of K means that there are more products in the equilibrium mixture than there are products. Likewise, a small value of K means that, at equilibrium, there are more reactants than products.
• Equilibrium constants that are based on partial pressures are often written as K_P, while equilibrium constants based on molar concentrations are written as K_c.
• An expression for an equilibrium constant can be written from a balanced chemical equation for the reaction. The Law of Mass Action states the following regarding equilibrium expressions:
  1. Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator.
  2. If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
  3. The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
  4. Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.

• As an example of an equilibrium expression, consider the reaction of nitrogen and hydrogen to form ammonia. The partial pressure of ammonia will be in the numerator, and it will be squared. Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the “first power” (which is not shown) and the partial pressure of hydrogen will be cubed.

N₂ (g) + 3 H₂ (g) ⇄ 2 NH₃ (g):

${\frac {\left(P_{NH_{3}}\right)^{2}}{P_{N_{2}}\left(P_{H_{2}}\right)^{3}}}=K$

• If equilibrium values for a given reaction are known, the equilibrium constant can be calculated simply by substituting those values in the equilibrium expression. Quite often, however, initial and equilibrium values are only given for selected reactants and products. In these cases, initial and equilibrium values are arranged in an ICE Table, and the changes between initial and equilibrium states are calculated based on reaction stoichiometry.
• Because the partial pressure of a gas and the molar concentration of that gas are directly proportional, the ideal gas law can be rearranged as follows, to give an expression relating molarity and partial pressure of a gas.

$P_{gas}V=nRT$; dividing by the volume in liters gives the term $\left({}^{n}\!\!\diagup \!\!{}_{V_{liters}}\;\right)$ which is equivalent to molarity, M; or, $P_{gas}=MRT$

• Le Chatelier's Principle states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. In this context, stress refers to a change in concentration, a change in pressure or a change in temperature, although only concentration is considered here. If pressure or temperature are changed, the numeric value K will change; if only concentration changes are involved, K does not change. In a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, and the introduction of more reactants will lead to the formation of more products, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.

• For a weak acid, dissociating in water to form it's conjugate base and hydronium ion, the equilibrium constant is referred to as K_a. Because a weak acid is only "partially dissociated", the concentration of BH in solution is large, thus K_a for a weak acid is "small" (in the range of 10^-3 to 10^-6). For example, acetic acid (the acidic component of vinegar), has an acid dissociation constant of K_a = 1.8 x 10^-5.

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO^-(aq) + H₃O⁺(aq)

• For a solution of a weak acid in water, the concentration of hydronium ion will be very small. If the concentration of the weak acid is fairly large (typically > 0.01 M) the concentration of the undissociated acid will be much larger than [H₃O⁺]. Because of this, the hydronium ion concentration (hence, the pH) can be fairly accurately estimated from the K_a of the weak acid and the initial concentration of the acid (C_o), by the equation:

$\left[H_{3}O^{+}\right]={\sqrt {K_{a}\times C_{0}}}$

• The equilibrium constant defining the solubility of an ionic compound with low solubility is defined as K_sp, where “sp” refers to “solubility product”. Because reactants or products that are present as solids or liquids do not appear in equilibrium expressions, for silver chloride, the expression K_sp will be written as:

AgCl(s) ⇄ Ag⁺(aq) + Cl^-(aq) ;

$K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]$

• For silver chloride, the solubility at 25 ^oC is 1.67 × 10^-5 M. That means the concentrations of silver and chloride ions in solution are each 1.67 × 10^-5 M, making the value of K_sp under these conditions:

$K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]=\left(1.67\times 10^{-5}\right)^{2}=2.79\times 10^{-10}$

  Supplementary Problems

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