from initial concentrations and one equilibrium concentration

In the following problems, you will be given initial concentrations as well as one equilibrium concentration. You have to use stoichiometry to figure out the other equilibrium concentrations from the data given.

Notice that I use a thing I call an ICEbox. I go into a bit of explanation in this tutorial. You can also examine the results of this search. Note that many sources call it an ICE table.

Along the way, I learned that some teachers call this a RICE table. In the empty box to the upper left, the word 'reaction' is used. I have no personal objection to this and I will use a RICE table from time to time, just so you have experience with seeing both.

**Example #1:** 0.260 mol H_{2} and 0.144 mol of I_{2} heated together in a sealed container with a volume of 1.00 dm^{3}. At equilibrium, 0.258 mol HI is present. Calculate K_{c}.

**Solution:**

1) Write the chemical reaction:

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

2) We know some starting and some ending concentrations. We'll set up a RICE table:

Reaction [H _{2}][I _{2}]⇌ [HI] Initial 0.260 0.144 0 Change +0.258 Equilibrium 0.258

3) To fill in the remaining equilibrium boxes, we must do a bit of stoichiometry:
H_{2} and HI are in a 1:2 molar ratio. In order to produce 0.258 M of HI, 0.129 M of H_{2} must be consumed. The same logic about 0.129 M being consumed is true for I_{2}.

4) Le's add to the ICEbox:

[H _{2}][I _{2}]⇌ [HI] Initial 0.260 0.144 0 Change −0.129 −0.129 +0.258 Equilibrium 0.131 0.015 0.258

5) We can now calculate K_{c}:

[HI] ^{2}K _{c}=–––––––– [H _{2}] [I_{2}]

[0.258] ^{2}K _{c}=–––––––––––– = 33.9 [0.131] [0.015]

**Example #2:** Consider the following reaction at a particular temperature:

CO(g) + 2H_{2}(g) ⇌ CH_{3}OH(g)

A reaction mixture in a 5.22 L flask initially contains 26.9 g CO and 2.32 g H_{2}. At equilibrium, the flask contains 8.65 g CH_{3}OH

Calculate the equilibrium constant (K_{c}) for the reaction at this temperature.

**Solution:**

1) K_{c} calculations (as far as the ChemTeam knows) always use molarities. Let us calculate the three we know:

CO ---> (M) (5.22 L) = 26.9 g / 28.0105 g/mol

H_{2}---> (M) (5.22 L) = 2.32 g / 2.01588 g/mol

CH_{3}OH ---> (M) (5.22 L) = 8.65 g / 32.04226 g/molCO ---> 0.18398 M

H_{2}---> 0.22046 M

CH_{3}OH ---> 0.051716 M

2) Let us place these numbers in an ICEbox:

[CO] [H _{2}]⇌ [CH _{3}OH]Initial 0.18398 0.22046 0 Change Equilibrium 0.051716

3) We need to determine the values for the two empty Equilibrium boxes. First [CO], then [H_{2}]:

CO and CH_{3}OH are in a 1:1 molar ratio. For every 1 mole of CO that reacts, 1 mole of CH_{3}OH is produced.Since 0.051716 M of CH

_{3}OH is produced, we conclude that the [CO] must have gone down by 0.051716 M:0.18398 − 0.051716 = 0.132264 MH

_{2}and CH_{3}OH are in a 2:1 molar ratio. For every 1 moles of CH_{3}OH produced, 2 moles of H_{2}are consumed.Since 0.051716 M of CH

_{3}OH is produced, we conclude that the [H_{2}] must have gone down by 0.051716 M multiplied by 2:0.22046 − 0.103432 = 0.117028 M

4) Let us add to our ICEbox from above:

[CO] [H _{2}]⇌ [CH _{3}OH]Initial 0.18398 0.22046 0 Change −0.051716 −0.103432 +0.051716 Equilibrium 0.132264 0.117028 0.051716

5) We are now ready to calculate the equilibrium constant:

[CH _{3}OH]K _{c}=–––––––––– [CO] [H _{2}]^{2}

0.051716 K _{c}=–––––––––––––––––––– = 28.6 (0.132246) (0.117028) ^{2}

**Example #3:** At high temperatures, dinitrogen tetroxide gas decomposes to nitrogen dioxide gas. At 500. °C, a sealed vessel containing 7.88 M dinitrogen tetroxide gas was allowed to reach equilibrium. At equilibrium, the mixture contained 3.43 M nitrogen dioxide gas. Calculate the value of K_{c} at this temperature.

**Solution:**

1) Write the chemical equation:

N_{2}O_{4}(g) ⇌ 2NO_{2}(g)

2) Write an ICEbox:

[N _{2}O_{4}]⇌ [NO _{2}]Initial 7.88 0 Change −1.715 +3.43 Equilibrium 6.165 3.43

3) Write the K_{c} expression, put values in, and solve:

[NO _{2}]^{2}K _{c}=––––––– [N _{2}O_{4}]

(3.43) ^{2}K _{c}=––––––– = 1.91 (6.165)

4) Where did the 1.715 M come from?

It comes from the stoichiometry between N_{2}O_{4}and NO_{2}The molar ratio between N

_{2}O_{4}and NO_{2}is 1:21 is to 2 as x is to 3.43

x = 1.715 M

**Example #4:** The following reaction was performed in a sealed vessel at 727 °C:

H_{2}(g) + I_{2}(g) ⇌ 2HI(g)

Initially, only H_{2} and I_{2} were present at concentrations of [H_{2}] = 3.80 M and [I_{2}] = 2.70 M. The equilibrium concentration of I_{2} is 0.0900 M. What is the equilibrium constant, K_{c}, for the reaction at this temperature?

**Solution:**

1) An ICEbox containing only the information in the problem:

[H _{2}][I _{2}]⇌ [HI] Initial 3.80 2.70 0 Change Equilibrium 0.0900

2) Change:

[H _{2}][I _{2}]⇌ [HI] Initial 3.80 2.70 0 Change −2.61 −2.61 +5.22 Equilibrium 0.0900 The [I

_{2}] change comes from 2.70 − 0.0900 = 2.61The [H

_{2}] change comes from the fact that the H_{2}:I_{2}molar ratio is 1:1.The [HI] comes from the 1:2 molar ratio between [I

_{2}] and [HI].

3) Equilibrium:

[H _{2}][I _{2}]⇌ [HI] Initial 3.80 2.70 0 Change −2.61 −2.61 5.22 Equilibrium 1.19 0.0900 5.22

4) Write the K_{c} expression, substitute values, and solve:

[HI] ^{2}K _{c}=––––––– [H _{2}] [I_{2}]

(5.22) ^{2}K _{c}=––––––––––– = 254 (1.19) (0.0900)

**Example #5:** A 1.00 L flask was filled with 2.00 mol SO_{2}(g) and 2.00 mol NO_{2}(g) and heated. After equilibrium was reached, it was found that 1.30 mol NO(g) was present. Assume that this reaction occurs:

SO_{3}(g) + NO(g) ⇌ SO_{2}(g) + NO_{2}(g)

Calculate the value of the equilibrium constant, K_{c}, for the above reaction.

**Solution:**

1) RICE table, baby!

Reaction [SO _{3}][NO] ⇌ [SO _{2}][NO _{2}]Initial 0 0 2.00 2.00 Change +x +1.30 −x −x Equilibrium x 1.30 2.00 − x 2.00 − x

2) Since the value of K_{c} is our unknown, we must fill in the rest of the ICEbox:

Based on the 1:1 stoichiometry between SO_{3}and NO, we can determine that 1.30 mol of SO_{3}was produced.Based on the 1:1 stoichiometry between NO and SO

_{2}and the 1:1 ratio between NO and NO_{2}, we can determine that 1.30 mole each of NO_{2}and of SO_{2}were consumed. 0.70 mol of SO_{2}remains as well as 0.70 mol of NO_{2}.We can now fill everything in:

Reaction [SO _{3}][NO] ⇌ [SO _{2}][NO _{2}]Initial 0 0 2.00 2.00 Change +1.30 +1.30 −1.30 −1.30 Equilibrium 1.30 1.30 0.70 0.70

3) The next thing to do is write the equilibrium constant expression, put values in, and solve:

[SO _{2}] [NO_{2}]K _{c}=–––––––––– [SO _{3}] [NO]

(0.70) (0.70) K _{c}=–––––––––– = 0.29 (1.30) (1.30)

**Example #6:** A mixture of 0.2000 mol of CO_{2}, 0.1000 mol of H_{2}, and 0.1600 mol of H_{2}O is placed in a 2.000 L vessel. The following equilibrium is established at 500 K:

COAt equilibrium, H_{2}(g) + H_{2}(g) ⇌ CO(g) + H_{2}O(g)

(a) Calculate the initial partial pressures of CO_{2}, H_{2}, and H_{2}O.

(b) Calculate K_{p} for the reaction.

**Solution to (a):**

PV = nRT is solved for PP

_{CO2}= [(0.2000 mol) (0.08206 L atm mol¯^{1}K¯^{1}) (500 K)] / 2.000 L = 4.103 atm

P_{H2}= [(0.1000 mol) (0.08206 L atm mol¯^{1}K¯^{1}) (500 K)] / 2.000 L = 2.0515 atm

P_{H2O}= [(0.1600 mol) (0.08206 L atm mol¯^{1}K¯^{1}) (500 K)] / 2.000 L = 3.2824 atm

2) RICEbox with all the given data:

Reaction P _{CO2}P _{H2}⇌ P _{CO}P _{H2O}Initial 4.103 2.0515 0 3.2824 Change Equilibrium 3.51

3) Show the changes using an unknown:

Reaction P _{CO2}P _{H2}⇌ P _{CO}P _{H2O}Initial 4.103 2.0515 0 3.2824 Change −x −x +x +x Equilibrium 3.51 I know the products increase (and the reactants decrease) because the partial pressure of the water vapor increased from initial conditions to equilibrium conditions.

4) However, we know the value of x. Solve for it, then replace the change unknowns:

3.51 − 3.2824 = 0.2276 atm <--- that's the value for x

Reaction P _{CO2}P _{H2}⇌ P _{CO}P _{H2O}Initial 4.103 2.0515 0 3.2824 Change −0.2276 −0.2276 +0.2276 +0.2276 Equilibrium 3.51

5) Fill in the equilibrium values and then solve for K_{p}:

Reaction P _{CO2}P _{H2}⇌ P _{CO}P _{H2O}Initial 4.103 2.0515 0 3.2824 Change −0.2276 −0.2276 +0.2276 0.2276 Equilibrium 3.8754 1.8239 0.2276 3.51

(P _{CO}) (P_{H2O})K _{p}=––––––––––– (P _{CO2}) (P_{H2})

(0.2276) (3.51) K _{p}=–––––––––––––– = 0.113 (3.8754) (1.8239)

**Example #7:** A mixture of 1.374 g of H_{2} and 70.31 g of Br_{2} is heated in a 2.00 L vessel at 700 K. These react according to:

H_{2}(g) + Br_{2}(g) ⇌ 2HBr(g)

At equilibrium the vessel is found to contain 0.566 g of H_{2}.

(a) Calculate the equilibrium concentration of H_{2}, Br_{2}, and HBr.

(b) Calculate K_{c}.

**Solution:**

1) Since equilibrium calculations are done in molarity, some preliminary work (not shown) is needed. The answers arrived at are these:

[H_{2}]_{o}= 0.340774 M

[Br_{2}]_{o}= 0.219983 M[H

_{2}] = 0.140377 MThe work done was grams divided by molar mass and then that answer divided by 2.00.

2) An ICEbox with the known data:

[H _{2}][Br _{2}]⇌ [HBr] Initial 0.340774 0.219983 0 Change Equilibrium 0.140377

3) The ICEbox with the change row filled in:

[H _{2}][Br _{2}]⇌ [HBr] Initial 0.340774 0.219983 0 Change −x −x +2x Equilibrium 0.140377

4) We can determine the change (that's the x) by subtraction:

0.340774 M − 0.140377 M = 0.200397 M

5) This will allow us to complete the ICEbox:

[H _{2}][Br _{2}]⇌ [HBr] Initial 0.340774 0.219983 0 Change −0.200397 −0.200397 +0.400794 Equilibrium 0.140377 0.019586 0.400794

6) Write the equilibrium constant expression and solve for K_{c}:

[HBr] ^{2}K _{c}=––––––––– [H _{2}] [Br_{2}]

(0.400794) ^{2}K _{c}=–––––––––––––––––– = 58.4 (0.140377) (0.019586)

**Example #8:** A sample of S_{8}(g) is placed in an otherwise empty rigid container at 1325 K at an initial pressure of 1.00 atm. The S_{8} decomposes to S_{2}(g) by the reaction:

SAt equilibrium, the partial pressure of S_{8}(g) ⇌ 4S_{2}(g)

**Solution (the non-ICEbox way):**

1) Determine the loss in pressure by the S_{8}:

1.00 atm − 0.25 atm = 0.75 atm

2) Determine the equilibrium amount of S_{2}:

For every one S_{8}decomposed, four S_{2}replace it.The pressure gain due to S

_{2}is 0.75 atm x 4 = 3.00 atm

3) The equilibrium expression is:

(S _{2})^{4}K _{p}=––––– (S _{8})

4) Insert values and solve:

(3.00) ^{4}K _{p}=––––– = 324 (0.25)

**Solution (the ICEbox way):**

1) A filled-in ICEbox is presented:

(S _{8})⇌ (S _{2})Initial 1.00 0 Change −x +4x Equilibrium 1.00 − x 4x

2) However, we already know the equilibrium pressure of S_{8} to be 0.25 atm. That means 0.75 atm of S_{8} must have decomposed. 0.75 atm is the value for x.

3) Modify the above ICEbox:

(S _{8})⇌ (S _{2})Initial 1.00 0 Change −0.75 +4(0.75) Equilibrium 1.00 − 0.75 = 0.25 4(0.75) = 3.00 See above for the calculation giving K

_{p}= 324.

**Exmple #9:** A 1.00-L flask was filled with 2.00 moles of gaseous SO_{2} and 2.00 moles of gaseous NO_{2} and heated. After equilibrium was reached, it was found that 1.30 moles of NO was present. Assume the reaction:

SO_{2}(g) + NO_{2}(g) ⇌ SO_{3}(g) + NO(g)

Calculate the value of the equilibrium constant, K_{c}, for the reaction.

**Solution:**

1) Molarity is used in equilibrium calculations:

[SO_{2}]_{o}= 2.00 mol / 1.00 L = 2.00 M

[NO_{2}]_{o}= 2.00 mol / 1.00 L = 2.00 M

[NO] = 1.30 mol / 1.00 L = 1.30 M

2) Determine the equilibrium amount of [SO_{2}]:

In order for 1.30 M of NO to be produced, 1.30 M of SO_{2}must be consumed.This is due to the 1:1 molar ratio between the coefficients of SO

_{2}and NO.[SO

_{2}] at equilibrium = 2.00 M − 1.30 = 0.70 M

3) Determine the equilibrium amount of [NO_{2}]:

In order for 1.30 M of NO to be produced, 1.30 M of NO_{2}must be consumed.This is due to the 1:1 molar ratio between the coefficients of NO

_{2}and NO.[NO

_{2}] at equilibrium = 2.00 M − 1.30 = 0.70 M

4) Determine the equilibrium amount of [SO_{3}]:

While 1.30 M of NO is produced, 1.30 M of SO_{3}is also produced.This is due to the 1:1 molar ratio between the coefficients of SO

_{3}and NO.[SO

_{3}] at equilibrium = 1.30 M

5) The equilibrium expression is:

[SO _{3}] [NO]K _{c}=–––––––––– [SO _{2}] [NO_{2}]

6) Insert values and solve:

(1.30) (1.30) K _{c}=–––––––––– = 3.45 (0.70) (0.70)

**Example #10:** A mixture of 0.10 mol of NO, 0.050 mol of H_{2}, and 0.10 mol of H_{2}O is placed in a 1.0 L vessel at 300 K. The following equilibrium is established:

2NO(g) + 2H_{2}(g) ⇌ N_{2}(g) + 2H_{2}O(g)

At equilibrium, [NO] = 0.062 M.

(a) Calculate the equilibrium concentrations of H_{2}, N_{2}, and H_{2}O.

(b) Calculate K_{c}.

**Solution to (a):**

1) How much NO is used up?

0.10 M − 0.062 M = 0.038 M0.10 M comes from 0.10 mol / 1.0 L = 0.10 M

2) How much H_{2} is used up? How much H_{2} remains at equilibrium?

Note that the coefficients of NO and H_{2}are equal.This means that the amounts used up for each substance are equal.

[H

_{2}] used up equals 0.038 MThe equilibrium amount for H

_{2}is this:0.050 M − 0.038 M = 0.012 M

3) How much N_{2} is produced? How much N_{2} is present at equilibrium?

The coefficient on NO is 2, the coefficient on N_{2}is 1.This means N

_{2}produced is one-half the amount that is used up by NO.[N

_{2}] produced is 0.038 / 2 = 0.019 MThere was zero N

_{2}at the start, therefore 0.019 M of N_{2}is present at equilibrium.

4) How much H_{2}O is produced? How much H_{2}O is present at equilibrium?

The coefficients on NO and H_{2}O are the same.This means that, for every amount of NO used up, that same amount of H

_{2}O is produced.[H

_{2}O] produced equals 0.038 M.The [H

_{2}O] present at equilibrium is 0.10 M + 0.038 M = 0.128 M

**Solution to (b):**

1) The equilibrium expression is:

[N _{2}] [H_{2}O]^{2}K _{c}=–––––––––– [NO] ^{2}[H_{2}]^{2}

2) Insert values and solve:

(0.019) (0.128) ^{2}K _{c}=––––––––––––– = 562 (0.062) ^{2}(0.012)^{2}

**Example #11:** At 80 °C, K_{c} = 1.87 x 10¯^{3} for the reaction:

PH_{3}BCl_{3}(s) ⇌ PH_{3}(g) + BCl_{3}(g)

(a) Calculate the equilibrium concentrations of PH_{3} and BCl_{3} if a solid sample of PH_{3}BCl_{3} is placed in a closed vessel at 80 °C and decomposes until equilibrium is reached.

(b) If the flask has a volume of 0.250 L, what is the minimum mass of PH_{3}BCl_{3} that must be added to the flask to achieve equilibrium?

**Solution to (a):**

1) Write the equilibrium expression for the above reaction:

K_{c}= [PH_{3}] [BCl_{3}]Note that PH

_{3}BCl_{3}(s) is not part of the equilibrium expression due to it being a solid.

2) During the reaction, . . .

. . . an unknown amount of PH_{3}is produced during the decomposition. Likewise, an unknown amount of BCl_{3}is produced.However, from the 1:1 stoichiometry of PH

_{3}and BCl_{3}, we know these unknown amounts to be equal.Let us call this amount 'x.'

3) Therefore:

1.87 x 10¯^{3}= (x) (x)x = 0.0432 M

**Solution to (b):**

0.0432 M is 0.0432 moles per 1.00 L.(0.0432 mol/L) (0.250 L) = 0.0108 mol

molar mass of PH

_{3}BCl_{3}is 151.1677 g/mol(151.1677 g/mol) (0.0108 mol) = 1.63 g

**Example #12:** An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid 3.00 L container at a certain temperature by the reaction:

N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g)

At equilibrium, 15.0 moles of H_{2}, 24.0 moles of N_{2}, and 12.0 moles of NH_{3} were found to be present. What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

**Solution:**

1) Convert moles to molarity:

[N_{2}] = 24.0 mol / 3.00 L = 8.00 M

[H_{2}] = 15.0 mol/3.00 L = 5.00 M

[NH_{3}] = 12.0 mol / 3.00 L = 4.00 M

2) Fill in an ICEbox with the known data:

[N _{2}][H _{2}]⇌ [NH _{3}]Initial 0 Change Equilibrium 8.00 5.00 4.00

3) Fill in an ICEbox with the changes from the initial consitions:

[N _{2}][H _{2}]⇌ [NH _{3}]Initial 0 Change −x −3x +2x Equilibrium 8.00 5.00 4.00 This is known to be true:

2x = 4.00 M

Therefore:

x = 2.00 M

4) Fill in an ICEbox with the revised changes:

[N _{2}][H _{2}]⇌ [NH _{3}]Initial [N _{2}]_{o}[H _{2}]_{o}0 Change −2.00 −6.00 4.00 Equilibrium 8.00 5.00 4.00

5) The answers are these values:

[N_{2}]_{o}= 10.00 M

[H_{2}]_{o}= 11.00 M

**Example #13:** At 25 °C, K_{p} = 2.90 x 10¯^{3} for the reaction:

NH_{4}OCONH_{2}(s) ⇌ 2NH_{3}(g) + CO_{2}(g)

In an experiment carried out at 25 °C, a certain amount of NH_{4}OCONH_{2} is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.

**Solution:**

1) Write the K_{p} expression:

K_{p}= (P_{NH3})^{2}(P_{CO2})

2) An unknown amount of the two reactants is produced:

2.90 x 10¯^{3}= (2x)^{2}(x)We know the ammonia amount is double the carbon dioxide amount because of the 2:1 stoichiometry of the reaction. For every one CO

_{2}produced, two NH_{3}are produced.

3) Continue the solution:

4x^{3}= 2.90 x 10¯^{3}x

^{3}= 7.25 x 10¯^{4}x = 0.089835 atm

4) The equilibrium concentrations are as follows:

P_{NH3}= 2x = 0.17967 atm

P_{CO2}= x = 0.089835 atm

5) Add them and round off for the final answer:

0.17967 atm + 0.089835 atm = 0.269505 atmRounded off to three sig figs gives 0.270 atm for the final answer.

5) The partial pressure answers can be checked by putting them back into the equilibrium expression:

K_{p}= (0.17967)^{2}(0.089835)The value stated in the problem is recovered.

**Example #14:** Nitrogen gas (N_{2}) reacts with hydrogen gas (H_{2}) to form ammonia (NH_{3}). At 200 °C in a closed container, 1.01 atm of nitrogen gas is mixed with 2.05 atm of hydrogen gas. At equilibrium the total pressure is 2.05 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the K_{p} value for this reaction.

**Solution:**

1) Construct an ICEbox with the initial data:

P _{N2}P _{H2}⇌ P _{NH3}Initial 1.01 2.05 0 Change Equilibrium

2) Add in the changes:

P _{N2}P _{H2}⇌ P _{NH3}Initial 1.01 2.05 0 Change −x −3x +2x Equilibrium

3) Add in the equilibrium amounts:

P _{N2}P _{H2}⇌ P _{NH3}Initial 1.01 2.05 0 Change −x −3x +2x Equilibrium 1.01 − x 2.05 − 3x 2x

4) Use Dalton's Law:

P_{tot}= P_{N2}+ P_{H2}+ P_{NH3}2.05 = 1.01 − x + 2.05 − 3x + 2x

0 = 1.01 − 2x

x = 0.505 atm

5) The partial pressure of the hydrogen:

2.05 − 3x2.05 − 3(0.505)

0.535 atm

6) You will find a problem similar to the one above here. It does not go on to solve for the K_{p.
}

7) Solve for the K_{p}:

P_{N2}= 1.01 − x = 0.505 atm

P_{H2}= 2.05 − 3x = 0.535 atm

P_{NH3}= 2x = 1.01 atm

(P _{NH3})^{2}K _{p}=––––––––– (P _{N2}) (P_{H2})^{3}

(1.01) ^{2}K _{p}=––––––––––––– = 13.2 (0.505) (0.535) ^{3}

**Example #15:** Consider the system:

N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g)

A 2.568 g sample of NH_{3} was placed in a 2.000 liter flask at 25 °C. When the equilibrium was reached at that temperature, it was determined that mass of the ammonia was reduced to 75.7% of its original value.

Calculate K_{p} for the decomposition of ammonia at 25 °C.

**Solution:**

1) For the initial state of NH_{3}, determine moles, then pressure:

2.568 g / 17.0307 g/mol = 0.1507865 mol(P

_{NH3})_{o}= nRT / V(P

_{NH3})_{o}= [(0.1507865 mol) (0.08206 L atm / mol K) (298 K)] / 2.000 L(P

_{NH3})_{o}= 1.8436575 atm

2) For the equilibrium state of NH_{3}, determine mass, then moles, then pressure:

(2.568 g) (0.757) = 1.943976 g1.943976 g / 17.0307 g/mol = 0.1141454 mol

(P

_{NH3}) = [(0.1141454 mol) (0.08206 L atm / mol K) (298 K)] / 2.000 L(P

_{NH3}) = 1.395649 atm

3) Write an ICEbox and fill in the initial & equilibrium values for ammonia:

P _{N2}P _{H2}⇌ P _{NH3}Initial 0 0 1.8436575 Change Equilibrium 1.395649

4) Fill in the change row in the ICEbox:

P _{N2}P _{H2}⇌ P _{NH3}Initial 0 0 1.8436575 Change +x +3x 1.8436575 − 2x Equilibrium 0.22400 0.67200 1.395649 See just below for the value of 'x' as well as the partial pressures of nitrogen and hydrogen.

5) We can determine the value for 'x' and then the partial pressures for nitrogen and hydrogen:

1.8436575 − 2x = 1.395649x = 0.22400 atm (this is the partial pressure of the nitrogen at equilibrium)

(3) (0.22400 atm) = 0.67200 atm (this is the partial pressure of the hydrogen at equilibrium)

6) Calculate the K_{p}:

(P _{NH3})^{2}K _{p}=––––––––––– (P _{N2}) (P_{H2})^{3}

(1.395649) ^{2}K _{p}=–––––––––––––––– = 28.65 (0.22400) (0.67200) ^{3}

**Example #16:** An 8.00 g sample of SO_{3} was placed in an evacuated container, where it decomposed at 600 °C according to the following reaction:

SO_{3}(g) ⇌ SO_{2}(g) +^{1}⁄_{2}O_{2}(g)

At equilibrium the total pressure and the density of the gaseous mixture were 1.80 atm and 1.60 g/L, respectively. Calculate K_{p} for this reaction.

**Solution:**

1) The volume of the container is not made explicit. However, the density is provided and that is the avenue to the volume of the container.

8.00 g / V = 1.60 g/LV = 5.00 L

We know that the gases at equilibrium total 8.00 grams in mass. We know this because of the Law of Conservation of Mass.

2) Convert initial grams of SO_{3} to moles:

8.00 g/80.063 g/mol = 0.0999213 mol

3) Let 'x' = moles SO_{3} decomposed,

4) Therefore, at equilibrium:

moles SO_{3}= 0.0999213 − x

moles SO_{2}= x

moles O_{2}= x/2moles total = 0.0999213 − x + x + x/2 = 0.0999213 + x/2

5) Use PV = nRT to determine 'x'

P = nRT / V1.80 = [(0.0999213 + x/2) (0.08206) (873)] / 5.00

9.00 = (0.0999213 + x/2) (0.08206) (873)

0.12563098 = 0.0999213 + x/2

x/2 = 0.12563098 − 0.0999213 = 0.02570968

x = 0.05142 atm

6) Therefore, at equilibrium

moles SO_{3}---> 0.0999213 − 0.05142 = 0.0485

moles SO_{2}---> 0.05142

moles O_{2}---> 0.05142 / 2 = 0.02571moles total ---> 0.0999213 + x/2 ---> 0.0999213 + 0.02571 = 0.12563

7) Total pressure times mole fraction gives each partial pressure:

P_{SO3}---> (1.80 atm) (0.0485 / 0.12563) = 0.695 atm

P_{SO2}---> (1.80 atm) (0.05142 / 0.12563) = 0.737 atm

P_{O2}---> (1.80 atm) (0.02571 / 0.12563) = 0.368 atm

8) Calculate the K_{p}:

(P _{SO2}) (P_{O2})^{½}K _{p}=––––––––––– (P _{SO3})

(0.737) (0.368) ^{½}K _{p}=––––––––––––– = 0.643 0.695