### Equilibrium Problems - Advanced Placement level

Problem #1: When 0.0322 mol of NO and 1.70 g of bromine are placed in a 1.00 L reaction vessel and sealed, the mixture reacts and the following equilibrium is established:

2NO(g) + Br2(g) ⇌ 2NOBr(g)

At 25.0 °C the equilibrium of nitrosyl bromide is 0.438 atm. What is the Kp?

Solution:

1) Determine pressure exerted by initial amount of NO:

P = (nRT)/V

P = [ (0.0322 mol) (0.08206 L atm mol¯11) (298 K) ] / 1.00 L

P = 0.787415 atm (keeping a few guard digits)

2) Determine pressure exerted by initial amount of Br2:

P = (nRT)/V

P = (gRT)/(molar mass times V)

P = [ (1.70 g) (0.08206 L atm mol¯11) (298 K) ] / [ (159.808 g mol¯1) (1.00 L) ]

P = 0.260135 atm

3) Determine equilibrium pressure of NO:

From the balanced equation, the NO : NOBr ratio is 1:1. Therefore, 0.438 atm of NO reacted.

0.787415 atm - 0.438 atm = 0.349415 atm

4) Determine equilibrium pressure of Br2:

From the balanced equation, the Br2 : NOBr ratio is 1:2. Therefore, 0.438/2 = 0.219 atm of Br2 reacted.

0.260135 atm - 0.219 atm = 0.041135 atm

5) Write the Kp expression and solve it:

Kp = PNOBr2 / [ ( PNO2) (PBr2) ]

Kp = (0.438)2 / [ (0.349415)2 (0.041135) ]

Kp = 38.2 (to 3 sig fig.)

Problem #2: A 1.00 L vessel contains at equilibrium 0.300 mol of N2, 0.400 mol H2, and 0.100 mol NH3. If the temp is maintained constant, how many moles of H2 must be introduced into the vessel in order to double the equilibrium concentration of NH3?

Solution:

1) Solve for the Kc first:

Kc = [NH3]2 / ( [N2] [H2]3 )

x = (0.100)2 /[ (0.300) (0.400)3 ]

x = 0.5208333

2) Create a new equilibrium expression with [NH3] = 0.200. Use the stoichiometry of the equation to determine the [N2] when the [NH3] is doubled. Set x = [H2]. Remember the cube:

[N2] = 0.250 since 0.050 moles of N2 are required to make the added 0.100 mole of NH3

0.52083 = (0.200)2 /[ (0.250) (x)3 ]

x = 0.675 M

3) This is the [H2] present in the new equilibrium conditions (remember, we are determining how much H2 got added, which is why it is larger than 0.400). Figure out how much had to have been used (from the balanced equation). Add that to the equilibrium amount. This is the total initial [H2]:

the molar ratio between NH3 and H2 is 2:3; therefore, 0.15 mole of H2 got used up in making the added 0.100 mol of NH3 in the new equilibrium

0.675 + 0.150 = 0.825 M (this is the initial amount of H2 in the equilibrium that eventually produced 0.200 mol of NH3)

4) Subtract 0.400 to get the amount of H2 that was added:

0.825 - 0.400 = 0.425 mol of H2 added to push the [NH3] from 0.100 to 0.200 in the new equilibrium

Comment on the above solution: I posted the above answer on sci.chem many years ago. I got this response (which has an error in it):

After solving for Keq at the initial conditions, I set the ammonia concentration to 0.200 M, the nitrogen concentration to 0.250 M, and the hydrogen concentration to (0.400 + x) M. After plugging the new values into the expression for Keq and solving for x, my calculations yielded a value of 0.274746132241 moles of hydrogen needed. (aren't TI-85's wonderful). This is of course too many sig figs but when plugged into the problem it does agree with the original Keq to 10 places. My answer would then be that 0.275 moles of hydrogen are needed in order to double the concentration of the ammonia under the conditions given. (My assumption of 0.250 moles for nitrogen is based upon the balanced equation and the idea that it takes 0.050 moles of nitrogen to make 0.100 moles of ammonia.)

Someone else then posted this

And it takes 0.15 moles of H2 to make the extra 0.1 moles of ammonia. John [N.B. that's me, the ChemTeam!] got it right. The number you came up with is the change in H2 concentration observed after the extra hydrogen is added. However, 0.15 mol of the H2 added goes into making the extra 0.1 mol of ammonia. The final H2 concentration should be 0.4 + x - 0.15, or 0.25 + x.

Problem #3: Nitric oxide and bromine at initial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300. K. At equilibrium the total pressure was 110.5 torr. The reaction is as follows.

2 NO(g) + Br2(g) ⇌ 2 NOBr(g)

Determine the Kp.

Solution:

1) Set up an ICEbox:

 PNO PBr2 PNOBr Initial 98.4 41.3 0 Change - 2x -x +2x Equilibrium 98.4 - 2x 41.3 - x +2x

2) Determine the value of x:

At equilibrium, the sum of the partial pressures is equal to 110.5 torr.

(98.4 - 2x) + (41.3 - x) + 2x = 110.5

x = 29.2 torr

3) At equilibrium, the partial pressures will be:

NO: 40.0 torr
Br2: 12.1 torr
NOBr: 58.4 torr

4) Determine the Kp:

Kp = (PNOBr)2 / [ (PNO)2 (PBr2) ]

Kp = (58.4)2 / [ (40.0)2 (12.1) ]

Kp = 0.176 torr¯1

5) Additional problem: What would Kp be if the pressures had been given in atmospheres?

0.176 torr¯1 x 760 torr atm¯1 = 137 atm¯1

6) Additional problem (much harder!): What would be the partial pressures of all the species if NO and Br2, both at an initial pressure of 0.300 atm, were allowed to come to equilibrium at this temperature? Sorry, no solution will be provided.

Problem #4: Consider the following equilibrium:

2 CH3OH(g) ⇌ CH3OCH3(g) + H2O(g)

If Kp is 13.54, what is the ratio between PCH3OH and PCH3OCH3?

Solution:

1) Let the three partial pressures be:

PCH3OH = x
PCH3OCH3 = y
PH2O = y

The key assumption is that PCH3OCH3 = PH2O. This comes from assuming all the two products came from the methanol reacting. If we do not assume this, then we have no sure knowledge about the partial pressures of the two products and we cannot solve the problem.

2) Insert values into the Kp expression and solve:

Kp = [ (PCH3OCH3) (PH2O) ] / (PCH3OH)2

13.54 = [(y) (y)] / (x)2

3.68 = y/x

3) However, we wish the ratio PCH3OH : PCH3OCH3

x/y = 0.272

Problem #5: Consider the reaction:

H2 + I2 ⇌ 2HI
whose Keq = 54.8. If an equimolar mixture of the reactants gives the concentration of the product to be 0.500 M at equilibrium, determine the concentration of hydrogen.

Solution:

1) The equilibrum expression is:

Keq = [HI]2 / ([H2] [I2])

2) Substituting into this, we have:

54.8 = (0.500)2 / [(x) (x)]

Comment: we know the equilibrium concentrations for H2 and I2 are equal because of the following two reasons: (1) they started out equimolar (that is, in equal amounts) and (2) they were used up in a 1:1 ratio (that is, an equal rate of consumption for both reactants) to make HI.

3) Seeing the the right-hand side is a perfect square, we take the square root and proceed to the answer:

7.403 = 0.500 / x

7.403x = 0.500

x = 0.0675 M (to three sig fig)

Problem #6: What compound, if any, will precipitate when 80. mL of 1.0 x 10¯5 M Ba(OH)2 is added to 20. mL of 1.0 x 10¯5 M Fe2(SO4)3?

Solution:

1) This is the overall chemical equation:

3Ba(OH)2 + Fe2(SO4)3 ---> 3BaSO4 + 2Fe(OH)3

2) There are four possible scenarios:

Only barium sulfate precipitates:
Ba2+(aq) + SO42¯(aq) ---> BaSO4(s)

Only iron(III) hydroxide precipitates:

Fe3+(aq) + 3OH¯(aq) ---> Fe(OH)3(s)

Both precipitate:

Net-ionic not written.

Neither precipitates

No ionic equation would be written.

3) We need to know how many moles of each ion is initially present:

Ba2+ ---> (1.0 x 10¯5 mol/L) (0.080 L) (1) = 8.0 x 10¯7 mol

OH¯ ---> (1.0 x 10¯5 mol/L) (0.080 L) (2) = 1.6 x 10¯6 mol

Fe3+ ---> (1.0 x 10¯5 mol/L) (0.020 L) (2) = 4.0 x 10¯7 mol

SO42¯ ---> (1.0 x 10¯5 mol/L) (0.020 L) (3) = 6.0 x 10¯7 mol

4) We need to know the molarity of each ion that is initially present:

[Ba2+]o = 8.0 x 10¯6 M

[OH¯]o = 1.6 x 10¯5 M

[Fe3+]o = 4.0 x 10¯6 M

[SO42¯]o = 6.0 x 10¯6 M

Each mol amount was divided by 0.10 L.

5) Calculate a reaction quotient for each precipitation equation:

BaSO4
Q = [Ba2+]o [SO42¯]o

Q = (8.0 x 10¯6) (6.0 x 10¯6)

Q = 4.8 x 10¯11

Fe(OH)3

Q = [Fe3+]o $\text{[OH¯]}{\text{}}_{o}^{3}$

Q = (4.0 x 10¯6) (1.6 x 10¯5)3

Q = 1.6384 x 10¯20

6) Compare Q to each Ksp:

 Substance Ksp Q Result BaSO4 1.1 x 10¯10 4.8 x 10¯11 Q < Ksp Fe(OH)3 2.79 x 10¯39 1.6384 x 10¯20 Q > Ksp

BaSO4 does not precipitate. Fe(OH)3 does precipitate.

Problem #7: The solubility product constant for Cu(IO3)2 is 1.44 x 10-7. What volume of 0.0520 M S2O32- would be required to titrate a 20.00 mL sample of saturated solution of Cu(IO3)2

Solution:

1) Determine moles of iodate in 20.00 mL:

1.44 x 10¯-7 = (x) (2x)2

x = 0.003302 M

[IO3¯] = 0.006604 M

0.006604 mol/L times 0.02000 L = 1.3208 x 10-4 mol

2) Determine iodate : thiosulfate ratio:

IO3¯(aq) + 6H+(aq) + 6S2O32-(aq) ----> I¯(aq) + 3S4O62-(aq) + 3H2O(aq)

The iodate : thiosulfate ratio is 1 : 6

3) Determine volume of thiosulfate required:

0.0520 mol/L = 1.3208 x 10-4 mol / x

x = 0.00254 L = 2.54 mL (if the ratio were 1:1)

since ratio is 1:6, we do this:

2.54 x 6 = 15.24 mL (this is the answer)

4) An alternate approach:

1 mole of IO3¯ reacts with 6 moles of S2O32-

Assume 1.00 L of solution present. Therefore:

moles of S2O32- = 6.604 x 10-4 mol x 6 = 0.0396 mol

number of moles = molarity x volume

volume = 0.0396 / 0.0520 = 0.762 L = 762 mL

However, we only titrated 20.00 mL, so:

762 x 0.02 = 15.24 mL

Problem #8: Bromine chloride, BrCl, a reddish covalent gas with properties similar to those of Cl2, may eventually replace Cl2 as a water disinfectant. One (1.00) mole of chlorine and one (1.00) mole of bromine are enclosed in a 8.05 L flask and allowed to reach equilibrium at a certain temperature.

Cl2(g) + Br2(g) ⇌ 2 BrCl(g)

Kc = 11.57 x 10¯2 at the given temperature. What mass of Cl2 is present at equilibrium?

Solution:

1) Determine molarities of Cl2 and Br2:

[Cl2] = [Br2] = 1.00 / 8.05 = 0.1242236 M (I'll carry several guard digits)

2) Set up an ICEbox:

 [Cl2] [Br2] [BrCl] Initial 0.1242236 0.1242236 0 Change - x -x +2x Equilibrium 0.1242236 - x 0.1242236 - x +2x

3) Determine x:

Kc = [BrCl]2 / ([Cl2] [Br2]

11.57 x 10¯2 = (2x)2 / [(0.1242236 - x) (0.1242236 - x)]

The right-hand side is a perfect square.

0.340147 = 2x / (0.1242236 - x)

0.0422543 - 0.340147x = 2x

0.0422543 = 2.340147x

x = 0.01805626 M

4) Determine [Cl2] at equilibrum:

0.1242236 minus 0.01805626 = 0.10616734 M

5) Determine mass of Cl2 present at equilibrium:

MV = g / molar mass

(0.10616734 mol/L) (8.05 L) = x / 70.906 g/mol

x = 60.6 g (to three sig figs)

6) We can determine if this is the correct answer by attempting to recover the Kc:

Kc = (0.03611252)2 / [(0.10617) (0.10617)

The computation is left to the reader.

Problem #9: Highly toxic S2F10 gas decomposes to reversily form SF4 and SF6 gases. A 2.00 L rigid flask originally containing 0.500 M S2F10 is allowed to reach equilibrium and found to contain 0.023 M S2F10. If an additional 0.125 mol of the S2F10 is added and equilibrium re-established, what will be the concentrations of each component?

Solution:

1) Let us write the chemical equation:

S2F10 ⇌ SF4 + SF6

2) In order to calculate the second part (where the equilibrium is re-established), we need to know the Kc value. For that, we need to know the equilibrium concentrations:

S2F10
0.023 M (given in problem)

SF4

0.477 M of SF4 was produced, based on the 1:1 stoichiometry between S2F10 and SF4

SF6

0.477 M of SF6 was produced, based on the 1:1 stoichiometry between S2F10 and SF6

3) Calculate the equilibrium constant:

 [SF4] [SF6] Kc  = ––––––––– [S2F10]

 (0.477) (0.477) Kc  = –––––––––––– 0.023

Kc = 9.892

4) We know that 0.125 mol of S2F10 is added, but it gets added to a 2.00 L flask. Calculate the molarity:

0.125 mol / 2.00 L = 0.0625 M

5) Now, we set up an ICE chart. Here are the initial conditions:

 [S2F10] [SF4] [SF6] Initial 0.0855 0.477 0.477 Change Equilibrium

The initial [S2F10] came from 0.023 + 0.0625 = 0.0855. Remember, some S2F10 was added to disturb the equilibrium.

6) The completed ICEbox looks like this:

 [S2F10] [SF4] [SF6] Initial 0.0855 0.477 0.477 Change −x +x +x Equilibrium 0.0855 − x 0.477 + x 0.477 + x

The addition of S2F10 has pushed the equilibrium to the right, making more of the produces and decreasing the amount of the reactant (after it had been increased by adding some to the system when it was at equilibrium).

7) Substitute values into the equilibrium expression, then solve for x:

 (0.477 + x) (0.477 + x) 9.892  = ––––––––––––––––––– 0.0855 − x

(9.892) (0.0855 − x) = (0.477 + x)2

8) At this point, we have a choice. We can neglect the 'x' in '0.0855 − x' and, hopefully, generate an approximate solution that is reasonably close to the answer. Or, we can do a full quadratic equation. I know, let's do both! First, the approximate solution:

(9.892) (0.0855) = (0.477 + x)2

0.919655 = 0.477 + x

x = 0.443 M

Well, that's not going to work!

(9.892) (0.0855 − x) = (0.477 + x)2

0.845766 − 9.892x = 0.227529 + 0.954x + x2

x2 + 10.846x − 0.618237 = 0

Visiting a quadratic equation solver yields a negative root (rejected) and 0.0567 (accepted).

10) At the new equilibrium, the concentrations are:

[S2F10] = 0.0855 − 0.0567 = 0.0288 M
[SF4] = 0.477 + 0.0567 = 0.5337 M
[SF6] = 0.477 + 0.0567 = 0.5337 M

11) Let's see if those values work:

 (0.5337) (0.5337) Kc  = ––––––––––––––––––– 0.0288

Kc = 9.89

Yay, it works.

Problem #10: Calculate Kp for each of the two reactions (happening in the same flask):

2FeSO4(s) ⇌ Fe2O3(s) + SO3(g) + SO2(g)
SO3(g) ⇌ SO2(g) + 12O2(g)

After equilibrium is reached, total pressure is 0.836 atm and partial pressure of oxygen is 0.0275 atm

Solution:

1) Let's write the Kp expression for each reaction:

Kp1 = (PSO3) (PSO2)

 (PSO2) (PO2)1⁄2 Kp2 = ––––––––––– PSO3

2) Now comes a key point:

The PSO2 and the PSO3 values in each Kp expression will be the same numerical value. Both pressure amounts are in the same flask. You can't have two different pressures of a gas in the same flask. The pressure of the gas is the same throughout the entire flask.

3) Let's start with the first reaction and allow it to go to equilibrium. At that point, we have this:

Kp1 = (x) (x)

We do not know the pressure values yet, but we do know that they are the same because of the 1:1 stoichiometry between SO2 and SO3 in the first reaction.

Keep in mind that we are acting like the second reaction has not yet happened. Once the second reaction happens, the SO2 and SO3 values are going to change. They are only equal right now because we are acting as if the second reaction has not yet started.

4) Now, let the second reaction happen:

 PSO3 PSO2 PO2 Initial x x 0 Change −0.055 +0.055 +0.0275 Equilibrium x − 0.055 x + 0.055 0.0275

The 0.055 value comes from the 2:1 molar ratio between SO2 and O2. For every one O2 produced, two SO2 are also produced (as well as two SO3 used up.

5) We know something about the three pressures. They all add up to 0.836 atm:

(x − 0.055) + (x + 0.055) + 0.0275 = 0.836 atm

2x = 0.8085

x = 0.40425 atm

6) We can now determine the three equilibrium pressures:

PSO3 = 0.40425 − 0.055 = 0.34925 atm
PSO2 = 0.40425 + 0.055 = 0.45925 atm
PO2 = 0.0275 atm

7) The equilibrium constants:

Kp1 = (0.34925) (0.45925) = 0.160

 (0.45925) (0.0275)1⁄2 Kp2 = ––––––––––––––––– = 0.218 0.34925