Equilibrium

Question 1

A researcher mixes together 1 atm of hydrogen gas with excess iodine solid. Ample time is given so the reaction reaches equilibrium.

H2(g) + I2(g) 2 HI(g)

What is Keq if the final pressure is 1.6 atm?

Answer 1

Keq is calculated from the final pressure, because the shift in the reaction (x) is equal to the change in pressure (which is 0.6 atm in this case).

     H2(g) + I2(g)  2 HI(g) I         1.00     excess            0 C        -x        -x                  +2x E       1.00-x    who cares?     2x

Keq=(HI)^2/(H2)=(2x)^2/(1-x)=(1.2)^2/(0.4) = 3.6

Question 2

Keq?

aA + bB –> cC + dD

Answer 2

Keq=[C]^c[D]^d / [A]^a[B]^b

Do not include solids or pure liquids!

Question 3

When does Keq change?

Answer 3

ONLY with temperature

Question 4

If the forward rate constant is four times the reverse rate constant, what is the Keq after a catalyst has been added that doubles the rate of the forward reaction?

Answer 4

Adding a catalyst lowers the activation energy, so the reaction speeds up. In this example, the forward rate is doubled, because the forward rate constant is doubled. However, the activation energy is lowered for the reverse reaction as well. The reverse reaction rate is also doubled. The ratio of the forward rate to the reverse rate remains the same. This means that equilibrium is the same, so the Keq is the same. Equilibrium is achieved sooner, but the same equilibrium conditions are reached.

Question 5

Keq &laquo_space;1
Keq = 1
Keq&raquo_space; 1

Answer 5

reverse reaction favored
neither direction favored
forward reaction favored

Question 6

Keq>Qc
Keq=Qc
Keq<Qc

Answer 6

ΔG < 0; reaction proceeds forward
ΔG = 0; Dynamic equilibrium
ΔG > 0; reaction proceeds reverse

Question 7

Dissolving

Answer 7

The breakdown of intermolecular forces between molecules as a solid becomes a solute within a solvent. The molecule remains intact when dissolving into solution. An example is the dissolving of sucrose into water, where atoms in the sucrose molecule remain covalently bonded, but the forces between sucrose molecules are eliminated.

Question 8

Dissociation

Answer 8

The breakdown of ionic bonds between atoms within a lattice structure as a salt turns into a solute within the solvent. The crystal lattice of the salt breaks apart when it dissociates into solution. An example is sodium chloride dissociating into water. The ionic bonds between sodium cations and chloride anions break, and the ions are stabilized by the partial charges of water.

Question 9

Solvent

Answer 9

The species in greatest concentration into which the solute dissolves, or salt dissociates. A solvent must be a fluid (have the ability to flow).

Question 10

Solute

Answer 10

The species not in highest concentration that dissolves into the solvent, or in the case of a salt, dissociates.

Question 11

Solubility

Answer 11

A measurement of the degree of dissolving that a solute undergoes within a particular solvent. The driving force for solubility is a preference for solvation of molecules (or ions) over the lattice strength of the solid. In addition, entropy favors the dissolving process. As the solubility of a compound increases, it is deduced that either the lattice energy of the solid is decreasing, the solvation energy of the solute form is increasing, or both effects are taking place.

Question 12

Saturated

Answer 12

Describes the state of a solution at the point where no more solid (solute) can dissolve into solution. When an aqueous salt solution is saturated, the rate of dissociation of the salt equals the rate of precipitation.

Question 13

Supersaturated

Answer 13

Describes the state of a solution where the amount of solid (solute) that is dissolved into solution is beyond the maximum amount at a given temperature. The solution is actually a suspension that when disturbed can form a precipitate rapidly. This state can be achieved by first heating a solvent, then adding solute to the solution until the solution is saturated at that temperature. Slowly cooling this solution causes the amount of solute in it to exceed what should dissolve at the reduced temperature.

Question 14

Solubility product (Ksp)

Answer 14

The equilibrium constant for a dissociation reaction, determined from the molar solubility according to standard rules for calculating equilibrium constants.

Question 15

Molar solubility

Answer 15

The quantitative measurement of the maximum number of moles of solid (solute) that can dissolve into enough solvent to make one liter of solution under standard conditions. For all practical purposes, the solvent is always water in inorganic chemistry and thus the calculations are similar in nearly every example. Molar solubility can be thought of as the x-value in the calculation ofthesolubility product(KSp).

Question 16

Gram solubility

Answer 16

The quantitative measurement of the maximum number of grams of solid (solute) that can dissolve into enough solvent to make one hundred milliliters of solution under standard conditions.

Question 17

Common ion effect

Answer 17

This results in a reduction in the amount of solid (solute) that can dissolve into solution due to the presence in the solution of an ion that is also present in the solid. This concept is similar to Le Chatelier’s principle, except that with Le Chatelier’s principle, the addition of one of the products (ions) causes precipitation (reduced solubility). With the common ion effect, the ion causing the reduced solubility is present in solution at the beginning of the reaction, rather than being added once the solution has reached a solubility equilibrium.

Question 18

Solubility rules

Answer 18

GENERALLY SOLUBLE____EXCEPTIONS
1A, NH4(+) (none)
NO3(-), C2H3O2(-) (none)
Cl(-), Br(-), I(-) Ag(+), Pb(2+), Hg2(2+)
SO4(2-) Ag(+),Pb(2+),Ca(2+),Sr(2+),Ba(2+)

GENERALLY INSOLUBLE___EXCEPTIONS
OH(-), S(2-) 1A, NH4(+),Ca(2+),Sr(2+),Ba(2+)
CO3(2-), PO4(3-) 1A, NH4(+)

Question 19

What is the Ksp calculation for the following?
MX(s)
MX2(s)
MX3(s)
M2X(s)
M3X(s)

Answer 19

(x represents molar solubility)
Ksp = (x)(x)       = x^2
Ksp = (x)(2x)^2 = 4x^3
Ksp = (x)(3x)^3 = 27x^4
Ksp = (2x)^2(x) = 4x^3
Ksp = (3x)^3(x) = 27x^4

Question 20

When a question asks for the highest solubility, what is it referring to?

Answer 20

It refers to the compound that produces the greatest amount of dissociated salt, which refers to greatest molar solubility, not solubility product!

Question 21

Which way will the reaction shift when Ag+(aq) is added to a saturated aqueous silver chloride solution?

Answer 21

According to Le Chatelier’s principle, adding a product shifts the reaction to the reactant side, so a precipitate forms. This means that the salt becomes less soluble.

Question 22

Is silver chloride more soluble in pure water or a 0.10 M NaCl(aq) solution?

Answer 22

According to the common ion effect, because the sodium chloride solution has chloride ions(also found in sliver chloride) already present in solution, the solubility of the salt is reduced.

Question 23

When separating by precipitation, which compound precipitates first.

Answer 23

Be careful not to fall into the trap of thinking that the smallest Ksp is the least soluble. You must consider what type of ionic system it is. The compound with the smallest value for molar solubility (x) precipitates from solution first.

Question 24

Chelating

Answer 24

Chelating is the formation of a Lewis acid-base bond between a lone pair-donor (ligand) and a lone-pair acceptor (central atom). Chelation changes the solubility of a salt by changing the concentration of free ions in solution. When a ligand binds a central metal, there is a formation constant that measures the strength of the chelation. This allows for specific ions to be removed from solution by binding them to form a more soluble complex ions.

Question 25

ΔG relates to Keq

Answer 25

ΔG = -RT lnKeq