**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) + Br_{2}(g) ⇌ 2NOBr(g)

At 25.0 °C the equilibrium of nitrosyl bromide is 0.438 atm. What is the K_{p}?

**Solution:**

1) Determine pressure exerted by initial amount of NO:

P = (nRT)/VP = [ (0.0322 mol) (0.08206 L atm mol¯

^{1}K¯^{1}) (298 K) ] / 1.00 LP = 0.787415 atm (keeping a few guard digits)

2) Determine pressure exerted by initial amount of Br_{2}:

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

P = [ (1.70 g) (0.08206 L atm mol¯

^{1}K¯^{1}) (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 Br_{2}:

From the balanced equation, the Br_{2}: NOBr ratio is 1:2. Therefore, 0.438/2 = 0.219 atm of Br_{2}reacted.0.260135 atm - 0.219 atm = 0.041135 atm

5) Write the K_{p} expression and solve it:

K_{p}= P_{NOBr}^{2}/ [ ( P_{NO}^{2}) (P_{Br2}) ]K

_{p}= (0.438)^{2}/ [ (0.349415)^{2}(0.041135) ]K

_{p}= 38.2 (to 3 sig fig.)

**Problem #2:** A 1.00 L vessel contains at equilibrium 0.300 mol of N_{2}, 0.400 mol H_{2}, and 0.100 mol NH_{3}. If the temp is maintained constant, how many moles of H_{2} must be introduced into the vessel in order to double the equilibrium concentration of NH_{3}?

**Solution:**

1) Solve for the K_{c} first:

K_{c}= [NH_{3}]^{2}/ ( [N_{2}] [H_{2}]^{3})x = (0.100)

^{2}/[ (0.300) (0.400)^{3}]x = 0.5208333

2) Create a new equilibrium expression with [NH_{3}] = 0.200. Use the stoichiometry of the equation to determine the [N_{2}] when the [NH_{3}] is doubled. Set x = [H_{2}]. Remember the cube:

[N_{2}] = 0.250 since 0.050 moles of N_{2}are required to make the added 0.100 mole of NH_{3}0.52083 = (0.200)

^{2}/[ (0.250) (x)^{3}]x = 0.675 M

3) This is the [H_{2}] present in the new equilibrium conditions (remember, we are determining how much H_{2} 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 [H_{2}]:

the molar ratio between NH_{3}and H_{2}is 2:3; therefore, 0.15 mole of H_{2}got used up in making the added 0.100 mol of NH_{3}in the new equilibrium0.675 + 0.150 = 0.825 M (this is the initial amount of H

_{2}in the equilibrium that eventually produced 0.200 mol of NH_{3})

4) Subtract 0.400 to get the amount of H_{2} that was added:

0.825 - 0.400 = 0.425 mol of H_{2}added to push the [NH_{3}] 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 K_{eq}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 K_{eq}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 K_{eq}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 H_{2}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 H_{2}concentration observed after the extra hydrogen is added. However, 0.15 mol of the H_{2}added goes into making the extra 0.1 mol of ammonia. The final H_{2}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) + Br_{2}(g) ⇌ 2 NOBr(g)

Determine the K_{p}.

**Solution:**

1) Set up an ICEbox:

P _{NO}P _{Br2}P _{NOBr}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

Br_{2}: 12.1 torr

NOBr: 58.4 torr

4) Determine the K_{p}:

K_{p}= (P_{NOBr})^{2}/ [ (P_{NO})^{2}(P_{Br2}) ]K

_{p}= (58.4)^{2}/ [ (40.0)^{2}(12.1) ]K

_{p}= 0.176

5) Additional problem: What would K_{p} be if the pressures had been given in atmospheres?

0.176 x 760 torr atm¯^{1}= 137

6) Additional problem (much harder!): What would be the partial pressures of all the species if NO and Br_{2}, 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 CH_{3}OH(g) ⇌ CH_{3}OCH_{3}(g) + H_{2}O(g)

If K_{p} is 13.54, what is the ratio between P_{CH3OH} and P_{CH3OCH3}?

**Solution:**

1) Let the three partial pressures be:

P_{CH3OH}= x

P_{CH3OCH3}= y

P_{H2O}= yThe key assumption is that P

_{CH3OCH3}= P_{H2O}. 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 K_{p} expression and solve:

K_{p}= [ (P_{CH3OCH3}) (P_{H2O}) ] / (P_{CH3OH})^{2}13.54 = [(y) (y)] / (x)

^{2}3.68 = y/x

3) However, we wish the ratio P_{CH3OH} : P_{CH3OCH3}

x/y = 0.272

**Problem #5:** Consider the reaction:

Hwhose K_{2}+ I_{2}⇌ 2HI

**Solution:**

1) The equilibrum expression is:

K_{eq}= [HI]^{2}/ ([H_{2}] [I_{2}])

2) Substituting into this, we have:

54.8 = (0.500)^{2}/ [(x) (x)]Comment: we know the equilibrium concentrations for H

_{2}and I_{2}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 / x7.403x = 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 Fe_{2}(SO_{4})_{3}?

**Solution:**

1) This is the overall chemical equation:

3Ba(OH)_{2}+ Fe_{2}(SO_{4})_{3}---> 3BaSO_{4}+ 2Fe(OH)_{3}

2) There are four possible scenarios:

Only barium sulfate precipitates:Ba^{2+}(aq) + SO_{4}^{2}¯(aq) ---> BaSO_{4}(s)Only iron(III) hydroxide precipitates:

Fe^{3+}(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:

Ba^{2+}---> (1.0 x 10¯^{5}mol/L) (0.080 L) (1) = 8.0 x 10¯^{7}molOH¯ ---> (1.0 x 10¯

^{5}mol/L) (0.080 L) (2) = 1.6 x 10¯^{6}molFe

^{3+}---> (1.0 x 10¯^{5}mol/L) (0.020 L) (2) = 4.0 x 10¯^{7}molSO

_{4}^{2}¯ ---> (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:

[Ba^{2+}]_{o}= 8.0 x 10¯^{6}M[OH¯]

_{o}= 1.6 x 10¯^{5}M[Fe

^{3+}]_{o}= 4.0 x 10¯^{6}M[SO

_{4}^{2}¯]_{o}= 6.0 x 10¯^{6}MEach mol amount was divided by 0.10 L.

5) Calculate a reaction quotient for each precipitation equation:

BaSO_{4}Q = [Ba^{2+}]_{o}[SO_{4}^{2}¯]_{o}Q = (8.0 x 10¯

^{6}) (6.0 x 10¯^{6})Q = 4.8 x 10¯

^{11}Fe(OH)

_{3}Q = [Fe^{3+}]_{o}$\text{[OH\xaf]}{\text{}}_{\mathrm{o}}^{3}$Q = (4.0 x 10¯

^{6}) (1.6 x 10¯^{5})^{3}Q = 1.6384 x 10¯

^{20}

6) Compare Q to each K_{sp}:

Substance K _{sp}Q Result BaSO _{4}1.1 x 10¯ ^{10}4.8 x 10¯ ^{11}Q < K _{sp}Fe(OH) _{3}2.79 x 10¯ ^{39}1.6384 x 10¯ ^{20}Q > K _{sp}BaSO

_{4}does not precipitate. Fe(OH)_{3}does precipitate.

**Problem #7:** The solubility product constant for Cu(IO_{3})_{2} is 1.44 x 10^{-7}. What volume of 0.0520 M S_{2}O_{3}^{2-} would be required to titrate a 20.00 mL sample of saturated solution of Cu(IO_{3})_{2}

**Solution:**

1) Determine moles of iodate in 20.00 mL:

1.44 x 10¯^{-7}= (x) (2x)^{2}x = 0.003302 M

[IO

_{3}¯] = 0.006604 M0.006604 mol/L times 0.02000 L = 1.3208 x 10

^{-4}mol

2) Determine iodate : thiosulfate ratio:

IO_{3}¯(aq) + 6H^{+}(aq) + 6S_{2}O_{3}^{2-}(aq) ----> I¯(aq) + 3S_{4}O_{6}^{2-}(aq) + 3H_{2}O(aq)The iodate : thiosulfate ratio is 1 : 6

3) Determine volume of thiosulfate required:

0.0520 mol/L = 1.3208 x 10^{-4}mol / xx = 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 IO_{3}¯ reacts with 6 moles of S_{2}O_{3}^{2-}Assume 1.00 L of solution present. Therefore:

moles of S

_{2}O_{3}^{2-}= 6.604 x 10^{-4}mol x 6 = 0.0396 molnumber 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 Cl_{2}, may eventually replace Cl_{2} 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.

Cl_{2}(g) + Br_{2}(g) ⇌ 2 BrCl(g)

K_{c} = 11.57 x 10¯^{2} at the given temperature. What mass of Cl_{2} is present at equilibrium?

**Solution:**

1) Determine molarities of Cl_{2} and Br_{2}:

[Cl_{2}] = [Br_{2}] = 1.00 / 8.05 = 0.1242236 M (I'll carry several guard digits)

2) Set up an ICEbox:

[Cl _{2}][Br _{2}][BrCl] Initial 0.1242236 0.1242236 0 Change - x -x +2x Equilibrium 0.1242236 - x 0.1242236 - x +2x

3) Determine x:

K_{c}= [BrCl]^{2}/ ([Cl_{2}] [Br_{2}]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 [Cl_{2}] at equilibrum:

0.1242236 minus 0.01805626 = 0.10616734 M

5) Determine mass of Cl_{2} 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 K_{c}:

K_{c}= (0.03611252)^{2}/ [(0.10617) (0.10617)The computation is left to the reader.

**Problem #9:** Highly toxic S_{2}F_{10} gas decomposes to reversily form SF_{4} and SF_{6} gases. A 2.00 L rigid flask originally containing 0.500 M S_{2}F_{10} is allowed to reach equilibrium and found to contain 0.023 M S_{2}F_{10}. If an additional 0.125 mol of the S_{2}F_{10} is added and equilibrium re-established, what will be the concentrations of each component?

**Solution:**

1) Let us write the chemical equation:

S_{2}F_{10}⇌ SF_{4}+ SF_{6}

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

S_{2}F_{10}0.023 M (given in problem)SF

_{4}0.477 M of SF_{4}was produced, based on the 1:1 stoichiometry between S_{2}F_{10}and SF_{4}SF

_{6}0.477 M of SF_{6}was produced, based on the 1:1 stoichiometry between S_{2}F_{10}and SF_{6}

3) Calculate the equilibrium constant:

[SF _{4}] [SF_{6}]K _{c}=––––––––– [S _{2}F_{10}]

(0.477) (0.477) K _{c}=–––––––––––– 0.023 K

_{c}= 9.892

4) We know that 0.125 mol of S_{2}F_{10} 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:

[S _{2}F_{10}][SF _{4}][SF _{6}]Initial 0.0855 0.477 0.477 Change Equilibrium The initial [S

_{2}F_{10}] came from 0.023 + 0.0625 = 0.0855. Remember, some S_{2}F_{10}was added to disturb the equilibrium.

6) The completed ICEbox looks like this:

[S _{2}F_{10}][SF _{4}][SF _{6}]Initial 0.0855 0.477 0.477 Change −x +x +x Equilibrium 0.0855 − x 0.477 + x 0.477 + x The addition of S

_{2}F_{10}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) The quadratic equation way:

(9.892) (0.0855 − x) = (0.477 + x)^{2}0.845766 − 9.892x = 0.227529 + 0.954x + x

^{2}x

^{2}+ 10.846x − 0.618237 = 0Visiting a quadratic equation solver yields a negative root (rejected) and 0.0567 (accepted).

10) At the new equilibrium, the concentrations are:

[S_{2}F_{10}] = 0.0855 − 0.0567 = 0.0288 M

[SF_{4}] = 0.477 + 0.0567 = 0.5337 M

[SF_{6}] = 0.477 + 0.0567 = 0.5337 M

11) Let's see if those values work:

(0.5337) (0.5337) K _{c}=––––––––––––––––––– 0.0288 K

_{c}= 9.89Yay, it works.

**Problem #10:** Calculate K_{p} for each of the two reactions (happening in the same flask):

2FeSO_{4}(s) ⇌ Fe_{2}O_{3}(s) + SO_{3}(g) + SO_{2}(g)

SO_{3}(g) ⇌ SO_{2}(g) +^{1}⁄_{2}O_{2}(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 K_{p} expression for each reaction:

K_{p1}= (P_{SO3}) (P_{SO2})

(P _{SO2}) (P_{O2})^{1⁄2}K _{p2}=––––––––––– P _{SO3}

2) Now comes a key point:

The P_{SO2}and the P_{SO3}values in each K_{p}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:

K_{p1}= (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 SO

_{2}and SO_{3}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 SO

_{2}and SO_{3}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:

P _{SO3}P _{SO2}P _{O2}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 SO

_{2}and O_{2}. For every one O_{2}produced, two SO_{2}are also produced (as well as two SO_{3}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 atm2x = 0.8085

x = 0.40425 atm

6) We can now determine the three equilibrium pressures:

P_{SO3}= 0.40425 − 0.055 = 0.34925 atm

P_{SO2}= 0.40425 + 0.055 = 0.45925 atm

P_{O2}= 0.0275 atm

7) The equilibrium constants:

K_{p1}= (0.34925) (0.45925) = 0.160

(0.45925) (0.0275) ^{1⁄2}K _{p2}=––––––––––––––––– = 0.218 0.34925