**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 torr¯^{1}

5) Additional problem: What would K_{p} 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 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:** When NaF is added slowly to a solution that is 0.025 M Ba^{2+} and 0.025 M Ca^{2+} what will the concentration of calcium be when the barium just begins to precipitate? K_{sp} (BaF_{2}) = 1.0 x 10¯^{7}; K_{sp} (CaF_{2}) = 1.7 x 10¯^{10}.

**Solution:**

1) What is [F¯] when BaF_{2} just begins to precipitate?

1.0 x 10¯^{7}= (0.025) (x)^{2}x = 0.0020 M

2) What is [Ca^{2+}] when [F¯] = 0.0020 M?

1.7 x 10¯^{10}= (x) (0.0020)^{2}x = 4.25 x 10¯

^{5}M

**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 (from Yahoo Answers):

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.