Chemistry 101: Equilibrium

The law of mass action states that, at equilibrium, the composition of the reaction mixture can be expressed in terms of an ideal equilibrium constant, keq.

For the chemical reaction
aA (g) + bB (g) ⇔ cC (g)
if [C]eq is the concentration of C at equilibrium,

For the chemical reaction
aA (g) + bB (g) ⇔ cC (g) + dD (g),
if [C]eq is the concentration of C at equilibrium and [D]eq is the concentration of D at equilibrium, the equilibrium constant Keq can be calculated using the equation pictured above, where [A]eq and [B]eq are the concentrations of A and B at equilibrium, respectively.

The law implies that the rate of the forward reaction is equal to the rate of the reverse reaction at equilibrium, and the equilibrium constant depends only on the temperature of the system.

The expression for Q is very similar to the expression for keq. The only difference is that Q can apply to a chemical system at any concentrations, while keq specifically refers to the concentrations at equilibrium.

For this system,

keq = [B]eq/[A]eq

1. If keq >> 1, [B]eq >> [A]eq and, at equilibrium, B will be in excess.

2. If keq = 1, [B]eq = [A]eq and, at equilibrium, A and B will be at equal concentration.

3. If keq &laquo_space;1, [A]eq >> [B]eq and, at equilibrium, A will be in excess.

The value of the equilibrium constant is calculated from the actual concentrations of the products and reactants at equilibrium. The general formula for this reaction is:

keq=[CO2]

The value of the equilibrium constant (and reaction quotient) depends only on the concentration of reactants and products present in the aqueous or gaseous phases. Chemicals in the solid or liquid phase do not affect the equilibrium levels.

Q = Keq.

Equilibrium occurs when the reactants and products are at concentrations such that the reaction quotient, Q, is equal to the equilibrium constant keq.

When the system is at equilibrium, by definition, the rates of the forward and reverse reactions are equal.

The rates of the forward and reverse reactions are equal.

For every molecule of NO that comes together with a molecule of NO3 to make 2 molecules of NO2, 2 molecules of NO2 will also come together to make a molecule of NO and NO3.

Le Chatelier’s Principle states that when a system in equilibrium is placed under stress, the system adjusts to restore equilibrium.

There are three kinds of stress a chemical system can be placed under:
Concentration
Temperature
Pressure

More C will be created.

According to Le Chatelier’s Principle, the system will respond to relieve any stress placed on one side of the system.

In this case, the reactant side is the one placed under stress by the addition. As such, the system will respond by shifting to the right, favoring the forward reaction, and more products will be created.

Q decreases.

This circumstance favors the forward reaction, or the creation of more products.

As a general rule of reactions: when Q < keq, the forward reaction is favored and more products are created.

More A and B will be created.

In this case, the product side is the one placed under stress by the addition. As such, the system will respond by shifting to the left, favoring the reverse reaction, and more reactants will be created.

More C will be created.

For an endothermic reaction, heat must be added; hence it can be thought of as a reactant. Increasing the temperature is akin to adding more reactant, and therefore shifts the reaction to the right.

More A and B will be created.

For an exothermic reaction, heat is given off; hence it can be thought of as a product. Increasing the temperature is akin to adding more product, and therefore shifts the reaction to the left.

More C will be created.

Increasing the pressure will add stress to the side which has more moles of gas. In this case, that is the reactant side, with 3 moles of gas vs. 1 mole of gas on the product side.

So, when the pressure is increased, the system will shift right, creating more products.

More C will be created.

Decreasing the pressure lessens the inhibition for the side creating more moles of gas. In this case, the product side has 3 moles of gas vs. 2 total moles of gas on the reactant side.

So, when the pressure is decreased, the system will shift right, creating more products.

ΔGº = -RT ln(Keq)

Where:
R = the ideal gas constant in l-atm/K-mol
T = the absolute temperature in K

Kc is the value of Keq when all reactant and product concentrations are given in concentration units, such as mol/l.

When reactant and product concentrations are given in these units, Kc will be the calculated value of equilibrium.

Kp is the value of Keq when all reactant and product concentrations are given in pressure units, such as atm.

When reactant and product concentrations are given in these units, Kp will be the calculated value of equilibrium.

The equation relating Kp to Kc is: Kp = Kc (RT)Δn

where:
R = the ideal gas constant in l-atm/K-mol
T = the absolute temperature in K
Δn = the change in moles of gas (moles product gas - moles reactant gas)

Kc will be greater than Kp.

The equation relating Kp to Kc is: Kp = Kc (RT)Δn

Since at STP, R*T is greater than 1, and since Δn is negative if the number of moles of gas decreases, this gives :
Kp = Kc (1 / [number greater than 1]), hence Kp is a fraction of the size of Kc.

Kc will be equal to Kp.

The equation relating Kp to Kc is: Kp = Kc (RT)Δn

If the number of moles of gas is unchanged, Δn must equal zero, this gives:
Kp = Kc ([RT]0)=Kc * 1,
hence Kp = Kc.

The solubility product constant (Ksp) is the equilibrium constant for a solvation reaction.
Ksp = [anions]a[cations]c

Just like other equilibrium constants, Ksp is equal to the concentration of products over reactants raised to the power of their coefficients. However, since pure liquids and solids are omitted from equilibrium calculations, the equation simplifies to just the concentration of ions in solution, raised to the power of their coefficients.

Ksp increases.

A warmer solvent can dissolve more solutes. This leads to a higher concentration of ions in solution, thus Kspincreases.

Ksp = [Ag+][Cl-]

The coefficient in front of both silver and chloride is 1, so the exponents will be 1.

Ksp= [Ba++][F-]2

The coefficient in front of barium is 1 so the exponent will be 1 over [Ba++]. The coefficient in front of fluoride is 2, so the exponent will be 2 over [F-].

The common ion effect states that a salt will be less soluble in a solution containing product ions than it would be in a pure solvent. A solution containing ions that a salt would separate into when dissolved, is referred to as having common ions.

Ex: AgCl dissolved into chlorinated water (Cl- ions present) will be less soluble than AgCl in pure water. The added Cl- ion (product) shifts equilibrium to the side containing the solid salt (reactant).

1. NaCl will still dissolve.
   The presence of Cl- ions from AgCl will not prevent the highly soluble NaCl from dissociating.

2. AgCl will precipitate.
   The additional Cl- ions now present from NaCl exceed the saturation levels for AgCl in solution, and will force those ions back into solid AgCl salt form.

If a highly soluble chloride salt such as NaCl were added to the solution, the common ion effect of Cl- ions would lead to the selective precipitation of lead (II) chloride.

A complex is formed when a metallic atom or ion is bonded to a surrounding array of molecules or anions; usually written with brackets as: [complex]. Complexes cannot be solvated back into their original atoms/ions.

Ex: Cisplatin [PtCl2(NH3)2] is formed when the Pt++ cation is bonded to two Cl- anions and two NH3 groups.

1. Solubility decreases.
   Common ions prevent salts from fully dissociating, since the solution already contains some concentration of product ions.

2. Solubility increases.
   Complexes formed from product ions effectively remove those ions from solution, allowing more salt to dissociate and replace the lost product ions.

Solubility for both will be increased.

The solution will accept more K2[PtCl4] and NH3, since the [PtCl2(NH3)2] complex that forms effectively removes both PtCl4-- ions and NH3 molecules from solution.

According to LeChatelier's principle, the system then shifts towards the product ions, increasing the solubility of both species.

1. Base molecules already present in a basic solution will bond to the H+ ions, removing them from solution, shifting equilibrium to the right, and increasing solubility of HA.
   HA + H2O ⇔ H+ + A- 

2. Acid molecules already present in an acidic solution will bond to the OH- ions, removing them from solution, shifting equilibrium to the right, and increasing solubility of B-.
   B- + H2O ⇔ BH + HO-

1. Kw is the ion product for the water autoionization reaction,

2. 2 H2O ⇔ H3O+ + OH-

So Kw = [H3O+] [OH-]. In pure water at STP: [H3O+] = [OH-] = 10-7
Kw = 10-14

2 H2O ⇔ H3O+ + OH-

Since the creation of each H3O+ from a water molecule also requires the creation of an OH-, in pure water these concentrations will always be equal.

At STP, [H3O+] = [OH-] = 10-7

pH measures the acidity of a substance. A low pH means a high concentration of H+ ions. A high pH means there are few H+ ions.

pH can be calculated by the equation: pH= - log [H+]
In fact: p(anything) = - log (anything).
You only need to memorize that one general equation.

At STP, water has a pH of 7.0 and is a neutral substance.

Below 7

As a solution becomes more acidic, its pH decreases.

The conjugate acid XH+ will be stronger than the conjugate acid YH+.

In general: the weaker the base, the stronger its conjugate acid will be.

pH = -log[H+ ions]

You would need to be provided with the concentration of H+ ions in order to solve. Remember, for a strong acid, the acid concentration is equal to H+ ion concentration.

If [H+] = n x 10 -e
then pH = {e-1}.{10-n}

Ex: if [H+] = 6.2 x 10-4, then n = 6.2, and e = 4, so the pH can be approximated as equal to [4-1].[10-6.2] = 3.38

Note: the real pH is 3.21, which is within the acceptable error for the AP Chem exam.

The solubility of NH3 is decreased.

As NH3 dissolves in water, it combines according to NH3 + H2O ⇒ NH4+ + OH-

Le Chatelier's Principle says that as concentration of an ion in solution increases, that ion's solubility decrease

Ka is the acid dissociation constant for an acid in solution. The higher the Ka the stronger the acid.

Ex: for acetic acid, CH3COOH, Ka can be calculated as:

Remember: Keq equations include only components in the gaseous or aqueous phase; components in the solid or liquid phase, like H2O, are ignored.

1. HCN (Hydrogen Cyanide)

2. HClO (Hypochlorous Acid)

3. HNO2 (Nitrous Acid)

4. HF (Hydrofluoric Acid)

5. H2SO3 (Sulfurous Acid)

6. H2S (Hydrogen Sulfide)

Remember: technically any acid that is not on the list of STRONG acids, is considered weak.

1. Strong (HCl = hydrogen chloride)

2. Strong (HNO3 = nitric acid)

3. Strong (H2SO4 = sulfuric acid)

4. Weak (HCN = hydrogen cyanide)

5. Weak (H2S = hydrogen sulfide)

1. Strong (HI = hydrogen iodide)

2. Strong (HBr = hydrogen bromide)

3. Weak (HF = hydrogen fluoride)

4. Strong (HClO4 = perchloric acid)

5. Weak (H2SO3 = sulfurous acid)

1. For a strong acid, Ka > 1.

2. For a weak acid, Ka \< 1.

The general acid equation and Ka calculation for the acid HA is:

HA ⇔ H+ + A-
Ka = [H+][A-] / [HA]

Note: In an AP Chem problem, you would be given at least two of these concentrations in order to solve.

The Ka for acid A (stronger) will be higher than acid B.

In general, the higher the Ka value: the stronger the acid.

HCl (hydrochloric acid) has 1 Ka value for the one proton it can donate.

HCl + H2O ⇒ Cl - + H3O+

H3PO4 (phosphoric acid) has 3 Ka values for the three protons it can donate.
H3PO4 + H2O ⇒ H2PO4- + H3O+
H2PO4- + H2O ⇒ HPO42- + H3O+
HPO42- + H2O ⇒ PO43- + H3O+

When a salt is dissolved in water, two separate ions are created; if either (or both) of those ions react with water to create a change in pH, that is considered hydrolysis.

Cationic hydrolysis (e.g. NH4Cl + H2O) makes a solution acidic; anionic hydrolysis (e.g. NaCH3COO + H2O) will make the solution basic.

A salt of a strong acid and strong base will never undergo hydrolysis, the resulting solution will always be neutral.

Kh is the hydrolysis constant, and quantifies how much salt can be hydrolysed in a saturated solution. This is essentially a proportion of how much water there is available to dissolve the salt in.

A higher Kh means that more salt can be hydrolyzed.

Kb is the base dissociation constant. The higher the value of Kb, the stronger the base is in solution.

from the general equation: B- + H2O ⇔ OH- + BH
Kb = [OH-][BH] / [B-]

1. NaOH

2. Ca(OH)2

3. K2O

4. CaH2

5. NaNH2

Generally, only certain hydroxide, oxide, and amide salts will be strong bases.

1. NH3

2. N(CH3)3

3. NH4OH

4. HS-

5. H2O

Remember: any base that's has not been listed/used as a strong bases is considered a weak base by default.

1. Strong (NaOH = sodium hydroxide)

2. Strong (Na2O = sodium oxide)

3. Weak (NH3 = ammonia)

4. Strong (H- = hydride ion)

5. Weak (HS- = hydrosulfide ion)

1. Weak (H2O = water)

2. Strong (NH2- = amide ion)

3. Strong (CaO = calcium oxide)

4. Weak (N(CH3)3 = trimethyl amine)

5. Strong (Ca(OH)2 = calcium hydroxide)

Base B (stronger) will have a higher Kb value.

In general: the stronger the base in solution, the higher the Kb value.

pKa = -log (Ka)

In fact: p(anything) = -log (anything).
You only need to memorize that one general equation.

1. For a strong acid, pKa \< 0.

2. For a weak acid, pKa > 0.

Acid X (stronger) will have a lower pKa than acid Y (weaker).

Recall: the stronger the acid, the higher the Ka value will be. The higher the Ka, the lower the pKa will be.

pKb = -log (Kb)

In fact: p(anything) = -log (anything).
You only need to memorize that one general equation.

Base X (stronger) will have a lower pKb value than base Y.

Recall that a stronger base will have a higher Kb value, hence a lower pKb

A chemical buffer is a mixture whose pH stays relatively constant when acid or base are added.

Buffers are typically a mixture of a weak acid and its conjugate base in solution. Added base will be neutralized by the weak acid, while added acid will be neutralized by the conjugate base. In either case, the pH will only shift marginally.

To calculate the pH of a buffered solution, you must use the Henderson-Hasselbach equations.

The buffered solution's pH response will be decreased (a lesser pH change).

Having more buffer ions will keep the pH from changing as significantly, since the buffer ions already in solution will neutralize the added acid or base and reduce its effect.

A buffer works most effectively when pH = pKa of the weak acid (or when pOH = pKb of the weak base) that makes up the buffer.

pH = pKa + log([A-] / [HA])

pOH = pKb + log([BH] / [B-])

6.3

According to Henderson-Hasselbach, pH = pKa + log ( [A-]/[HA] )

When [A-] = [HA] the final term becomes log (1) = 0.
At this point, then, pH = pKa = 6.3.

• Acetic acid (pH 3.7-5.6)

• Bicarbonate (pH 7-10)

• Sodium citrate (pH 3-5)