# Equilibrium Flashcards

## From the law of mass transfer to applying Le Chatelier’s Principle to a system at equilibrium, use these cards to master the topic of equilibrium as tested in most introductory undergrad chemistry courses and on the AP Chemistry exam.

1
Q

Define:

the law of mass action

A

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.

2
Q

For the chemical reaction

aA (g) + bB (g) ⇔ cC (g)

what is the value of the reaction quotient Q?

A

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.

3
Q

For the chemical reaction

A (g) ⇔ B (g)

what does it mean about the equilibrium concentrations of A and B if:

1. keq >> 1
2. keq = 1
3. keq << 1
A

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 << 1, [A]eq >> [B]eq and, at equilibrium, A will be in excess.
4
Q

What is the equilibrium constant formula for the reaction:

NO + NO3⇔ 2NO2

A

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:

5
Q

What is the equilibrium constant keq for the following reaction:

CaCO3(s)→CaO(s)+CO2(g)

A

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.

6
Q

What is true of the relationship between Keq and Q for any chemical system which is at equilibrium?

A

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.

7
Q

What can be said about the rates of the forward and reverse reactions for any chemical system which is at equilibrium?

A

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

8
Q

If the chemical system

NO + NO3 ⇔ 2NO2

is at dynamic equilibrium, what can be said about the rates of the forward and reverse reactions?

A

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.

9
Q

Define:

Le Chatelier’s Principle

A

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

10
Q

If the chemical system

aA (g) + bB (g) ⇔ cC (g)

is in equilibrium, and additional A is added, what does Le Chatelier’s Principle predict will occur?

A

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.

11
Q

If the chemical system

aA (g) + bB (g) ⇔ cC (g)

is in equilibrium, and additional A is added, what happens to the reaction quotient Q?

A

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.

12
Q

If the chemical system

aA (g) + bB (g) ⇔ cC (g)

is in equilibrium, and additional C is added, what does Le Chatelier’s Principle predict will occur?

A

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.

13
Q

If the chemical system

aA (g) + bB (g) ⇔ cC (g)

is in equilibrium, and is endothermic, what happens if the temperature is increased?

A

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.

14
Q

If the chemical system

aA (g) + bB (g) ⇔ cC (g)

is in equilibrium, and is exothermic, what happens if the temperature is increased?

A

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.

15
Q

If the chemical system

A (g) + 2 B (g) ⇔ C (g)

is in equilibrium, what happens if the pressure is increased?

A

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.

16
Q

If the chemical system

A (g) + B (g) ⇔ 3 C (g)

is in equilibrium, what happens if the pressure is decreased?

A

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.

17
Q

What is the relationship between a substance’s Gibbs’ Free Energy of Formation and the equilibrium constant of the substance’s formation reaction?

A

ΔGº = -RT ln(Keq)

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

18
Q

Define:

Kc, the molar concentration equilibrium constant

A

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.

19
Q

Define:

Kp, the partial pressure equilibrium constant

A

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.

20
Q

How can the value of Kp be calculated, given the value of Kc?

A

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)

21
Q

At STP, if the number of moles of gas decreases in a reaction, what will be the relationship between Kp and Kc?

A

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.

22
Q

At STP, if the number of moles of gas remains the same in a reaction, what will be the relationship between Kp and Kc?

A

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.

23
Q

Define:

the solubility product constant

A

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.

24
Q

If the temperature of a solution is increased, what happens to the solute’s Ksp?

A

Ksp increases.

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

25
Q

What is the solubility product constant formula for AgCl, which dissociates as below:

AgCl(s) ⇔ Ag+(aq) + Cl-(aq)

A

Ksp = [Ag+][Cl-]

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

26
Q

What is the solubility product constant formula for BaF2, which dissociates as below:

BaF2(s) ⇔ Ba++ (aq) + 2F-(aq)

A

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-].

27
Q

Define:

the common ion effect

A

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).

28
Q

Suppose NaCl were added to a saturated aqueous solution of AgCl. What will happen?

Note: NaCl has a higher solubility than AgCl.

A
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.
29
Q

A given solution contains PbCl2 and PbSO4. How could lead (II) chloride be selectively harvested from the solution?

A

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.

30
Q

Define:

complex (or coordination complex)

A

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.

31
Q

What is the affect on solubility of a salt if:

1. there are common ions present
2. complexes form from the product ions
A
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.
32
Q

Suppose that a solution is saturated in both K2[PtCl4] and NH3. What must be true of the solubility of both species if the following reaction occurs?

PtCl4+ NH3
PtCl2(NH3)2 + 2 Cl-

A

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.

33
Q

Explain why:

1. acids are more soluble in an aqueous base than neutral water.
2. bases are more soluble in an aqueous acid than neutral water.
A
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-
34
Q
1. What is Kw?
2. What is its value at STP?
A
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

35
Q

Give the equation for:

water autoionization

A

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

36
Q

Define and give the equation for:

pH

A

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.

37
Q

At 25ºC and 1atm, what is the pH of water?

A

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

38
Q

At STP, what is the pH of an acidic solution?

A

Below 7

As a solution becomes more acidic, its pH decreases.

39
Q

If base X is weaker than base Y, what do you know about the conjugate acids of each?

A

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.

40
Q

What is the equation to calculate the pH of an acid solution?

A

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.

41
Q

What is the shortcut equation to calculate the approximate pH of an acid?

A

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.

42
Q

How does the presence of dissolved NH4Cl affect the solubility of NH3 in aqueous solution?

A

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 decreases. So NH3, which forms NH4+ in solution, will be less soluble in the NH4Cl solution than it would be in pure water.

43
Q

Define and give the general equation for:

Ka

A

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:

44
Q

What is the Ka equation for the reaction of acetic acid and water?

CH3COOH + H2O ⇒
CH3COO- + H3O+

A

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

45
Q

Please list some common weak acids.

A
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.

46
Q

Which are strong acids, and which are weak?

1. HCl
2. HNO3
3. H2SO4
4. HCN
5. H2S
A
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)
47
Q

Which are strong acids, and which are weak?

1. HI
2. HBr
3. HF
4. HClO4
5. H2SO3
A
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)
48
Q
1. What is the Ka of a strong acid?
2. What is the Ka of a weak acid?
A
1. For a strong acid, Ka > 1.
2. For a weak acid, Ka < 1.
49
Q

How do you calculate the Ka of a weak acid?

A

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.

50
Q

Acid A is stronger than acid B; what do you know about their relative Ka values?

A

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

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

51
Q

How many Ka values does each of these acids possess?

HCl, H3PO4

A

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+

52
Q

Define and give the common types for:

hydrolysis of a salt

A

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.

53
Q

Define:

Kh

A

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.

54
Q

Define and give the equation for:

Kb

A

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-]

55
Q

Please list some common strong bases.

A
1. NaOH
2. Ca(OH)2
3. K2O
4. CaH2
5. NaNH2

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

56
Q

Please list some common weak bases.

A
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.

57
Q

Which are strong bases, and which are weak?

1. NaOH
2. Na2O
3. NH3
4. H-
5. HS-
A
1. Strong (NaOH = sodium hydroxide)
2. Strong (Na2O = sodium oxide)
3. Weak (NH3 = ammonia)
4. Strong (H- = hydride ion)
5. Weak (HS- = hydrosulfide ion)
58
Q

Which are strong bases, and which are weak?

1. H2O
2. NH2-
3. CaO
4. N(CH3)3
5. Ca(OH)2
A
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)
59
Q

If base A is weaker than base B, what do you know about the relative Kb values?

A

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

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

60
Q

Give the equation for:

pKa

A

pKa = -log (Ka)

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

61
Q
1. What is the pKa of a strong acid?
2. What is the pKa of a weak acid?
A
1. For a strong acid, pKa < 0.
2. For a weak acid, pKa > 0.
62
Q

Acid X is stronger than acid Y, what is true of their pKa values?

A

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.

63
Q

Give the equation for:

pKb

A

pKb = -log (Kb)

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

64
Q

If base X is stronger than base Y, what do you know about their relative pKb values?

A

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

65
Q

Define:

A chemical buffer

A

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.

66
Q

What is the effect of higher concentration of HA and A- ions on the pH response of a buffered solution, when adding an acid or base?

A

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.

67
Q

What is the optimal pH for a buffer solution?

A

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.

68
Q

Give the following equation:

Henderson-Hasselbach

A

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

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

69
Q

The pKa of a weak acid is 6.3. What pH will indicate that there are equal concentrations of A- and HA?

A

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.

70
Q

Please name the 3 most common buffer solutions tested on the AP Chemistry exam, and their effective pH ranges.

A
• Acetic acid (pH 3.7-5.6)
• Bicarbonate (pH 7-10)
• Sodium citrate (pH 3-5)