### Calculating Equilibrium Concentrations (and equilibrium constant)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 H2 and 0.144 mol of I2 heated together in a sealed container with a volume of 1.00 dm3. At equilibrium, 0.258 mol HI is present. Calculate Kc.

Solution:

1) Write the chemical reaction:

H2(g) + I2(g) ⇌ 2HI(g)

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

 Reaction [H2] [I2] ⇌ [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: H2 and HI are in a 1:2 molar ratio. In order to produce 0.258 M of HI, 0.129 M of H2 must be consumed. The same logic about 0.129 M being consumed is true for I2.

4) Le's add to the ICEbox:

 [H2] [I2] ⇌ [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 Kc:

 [HI]2 Kc = –––––––– [H2] [I2]

 [0.258]2 Kc = –––––––––––– = 33.9 [0.131] [0.015]

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

CO(g) + 2H2(g) ⇌ CH3OH(g)

A reaction mixture in a 5.22 L flask initially contains 26.9 g CO and 2.32 g H2. At equilibrium, the flask contains 8.65 g CH3OH

Calculate the equilibrium constant (Kc) for the reaction at this temperature.

Solution:

1) Kc 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
H2 ---> (M) (5.22 L) = 2.32 g / 2.01588 g/mol
CH3OH ---> (M) (5.22 L) = 8.65 g / 32.04226 g/mol

CO ---> 0.18398 M
H2 ---> 0.22046 M
CH3OH ---> 0.051716 M

2) Let us place these numbers in an ICEbox:

 [CO] [H2] ⇌ [CH3OH] 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 [H2]:

CO and CH3OH are in a 1:1 molar ratio. For every 1 mole of CO that reacts, 1 mole of CH3OH is produced.

Since 0.051716 M of CH3OH is produced, we conclude that the [CO] must have gone down by 0.051716 M:

0.18398 − 0.051716 = 0.132264 M

H2 and CH3OH are in a 2:1 molar ratio. For every 1 moles of CH3OH produced, 2 moles of H2 are consumed.

Since 0.051716 M of CH3OH is produced, we conclude that the [H2] 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] [H2] ⇌ [CH3OH] 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:

 [CH3OH] Kc = –––––––––– [CO] [H2]2

 0.051716 Kc = –––––––––––––––––––– = 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 Kc at this temperature.

Solution:

1) Write the chemical equation:

N2O4(g) ⇌ 2NO2(g)

2) Write an ICEbox:

 [N2O4] ⇌ [NO2] Initial 7.88 0 Change −1.715 +3.43 Equilibrium 6.165 3.43

3) Write the Kc expression, put values in, and solve:

 [NO2]2 Kc = ––––––– [N2O4]

 (3.43)2 Kc = ––––––– = 1.91 (6.165)

4) Where did the 1.715 M come from?

It comes from the stoichiometry between N2O4 and NO2

The molar ratio between N2O4 and NO2 is 1:2

1 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:

H2(g) + I2(g) ⇌ 2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2] = 3.80 M and [I2] = 2.70 M. The equilibrium concentration of I2 is 0.0900 M. What is the equilibrium constant, Kc, for the reaction at this temperature?

Solution:

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

 [H2] [I2] ⇌ [HI] Initial 3.80 2.70 0 Change Equilibrium 0.0900

2) Change:

 [H2] [I2] ⇌ [HI] Initial 3.80 2.70 0 Change −2.61 −2.61 +5.22 Equilibrium 0.0900

The [I2] change comes from 2.70 − 0.0900 = 2.61

The [H2] change comes from the fact that the H2:I2 molar ratio is 1:1.

The [HI] comes from the 1:2 molar ratio between [I2] and [HI].

3) Equilibrium:

 [H2] [I2] ⇌ [HI] Initial 3.80 2.70 0 Change −2.61 −2.61 5.22 Equilibrium 1.19 0.0900 5.22

4) Write the Kc expression, substitute values, and solve:

 [HI]2 Kc = ––––––– [H2] [I2]

 (5.22)2 Kc = ––––––––––– = 254 (1.19) (0.0900)

Example #5: A 1.00 L flask was filled with 2.00 mol SO2(g) and 2.00 mol NO2(g) and heated. After equilibrium was reached, it was found that 1.30 mol NO(g) was present. Assume that this reaction occurs:

SO3(g) + NO(g) ⇌ SO2(g) + NO2(g)

Calculate the value of the equilibrium constant, Kc, for the above reaction.

Solution:

1) RICE table, baby!

 Reaction [SO3] [NO] ⇌ [SO2] [NO2] 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 Kc is our unknown, we must fill in the rest of the ICEbox:

Based on the 1:1 stoichiometry between SO3 and NO, we can determine that 1.30 mol of SO3 was produced.

Based on the 1:1 stoichiometry between NO and SO2 and the 1:1 ratio between NO and NO2, we can determine that 1.30 mole each of NO2 and of SO2 were consumed. 0.70 mol of SO2 remains as well as 0.70 mol of NO2.

We can now fill everything in:

 Reaction [SO3] [NO] ⇌ [SO2] [NO2] 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:

 [SO2] [NO2] Kc = –––––––––– [SO3] [NO]

 (0.70) (0.70) Kc = –––––––––– = 0.29 (1.30) (1.30)

Example #6: A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and 0.1600 mol of H2O is placed in a 2.000 L vessel. The following equilibrium is established at 500 K:

CO2(g) + H2(g) ⇌ CO(g) + H2O(g)
At equilibrium, H2O(g) has a pressure of 3.51 atm.

(a) Calculate the initial partial pressures of CO2, H2, and H2O.
(b) Calculate Kp for the reaction.

Solution to (a):

PV = nRT is solved for P

PCO2 = [(0.2000 mol) (0.08206 L atm mol¯11) (500 K)] / 2.000 L = 4.103 atm
PH2 = [(0.1000 mol) (0.08206 L atm mol¯11) (500 K)] / 2.000 L = 2.0515 atm
PH2O = [(0.1600 mol) (0.08206 L atm mol¯11) (500 K)] / 2.000 L = 3.2824 atm

2) RICEbox with all the given data:

 Reaction PCO2 PH2 ⇌ PCO PH2O Initial 4.103 2.0515 0 3.2824 Change Equilibrium 3.51

3) Show the changes using an unknown:

 Reaction PCO2 PH2 ⇌ PCO PH2O 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 PCO2 PH2 ⇌ PCO PH2O 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 Kp:

 Reaction PCO2 PH2 ⇌ PCO PH2O 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

 (PCO) (PH2O) Kp  = ––––––––––– (PCO2) (PH2)

 (0.2276) (3.51) Kp  = –––––––––––––– = 0.113 (3.8754) (1.8239)

Example #7: A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700 K. These react according to:

H2(g) + Br2(g) ⇌ 2HBr(g)

At equilibrium the vessel is found to contain 0.566 g of H2.

(a) Calculate the equilibrium concentration of H2, Br2, and HBr.
(b) Calculate Kc.

Solution:

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

[H2]o = 0.340774 M
[Br2]o = 0.219983 M

[H2] = 0.140377 M

The work done was grams divided by molar mass and then that answer divided by 2.00.

2) An ICEbox with the known data:

 [H2] [Br2] ⇌ [HBr] Initial 0.340774 0.219983 0 Change Equilibrium 0.140377

3) The ICEbox with the change row filled in:

 [H2] [Br2] ⇌ [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:

 [H2] [Br2] ⇌ [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 Kc:

 [HBr]2 Kc  = ––––––––– [H2] [Br2]

 (0.400794)2 Kc  = –––––––––––––––––– = 58.4 (0.140377) (0.019586)

Example #8: A sample of S8(g) is placed in an otherwise empty rigid container at 1325 K at an initial pressure of 1.00 atm. The S8 decomposes to S2(g) by the reaction:

S8(g) ⇌ 4S2(g)
At equilibrium, the partial pressure of S8 is 0.25 atm. Calculate Kp for this reaction at 1325 K

Solution (the non-ICEbox way):

1) Determine the loss in pressure by the S8:

1.00 atm − 0.25 atm = 0.75 atm

2) Determine the equilibrium amount of S2:

For every one S8 decomposed, four S2 replace it.

The pressure gain due to S2 is 0.75 atm x 4 = 3.00 atm

3) The equilibrium expression is:

 (S2)4 Kp  = ––––– (S8)

4) Insert values and solve:

 (3.00)4 Kp  = ––––– = 324 (0.25)

Solution (the ICEbox way):

1) A filled-in ICEbox is presented:

 (S8) ⇌ (S2) Initial 1.00 0 Change −x +4x Equilibrium 1.00 − x 4x

2) However, we already know the equilibrium pressure of S8 to be 0.25 atm. That means 0.75 atm of S8 must have decomposed. 0.75 atm is the value for x.

3) Modify the above ICEbox:

 (S8) ⇌ (S2) 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 Kp = 324.

Exmple #9: A 1.00-L flask was filled with 2.00 moles of gaseous SO2 and 2.00 moles of gaseous NO2 and heated. After equilibrium was reached, it was found that 1.30 moles of NO was present. Assume the reaction:

SO2(g) + NO2(g) ⇌ SO3(g) + NO(g)

Calculate the value of the equilibrium constant, Kc, for the reaction.

Solution:

1) Molarity is used in equilibrium calculations:

[SO2]o = 2.00 mol / 1.00 L = 2.00 M
[NO2]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 [SO2]:

In order for 1.30 M of NO to be produced, 1.30 M of SO2 must be consumed.

This is due to the 1:1 molar ratio between the coefficients of SO2 and NO.

[SO2] at equilibrium = 2.00 M − 1.30 = 0.70 M

3) Determine the equilibrium amount of [NO2]:

In order for 1.30 M of NO to be produced, 1.30 M of NO2 must be consumed.

This is due to the 1:1 molar ratio between the coefficients of NO2 and NO.

[NO2] at equilibrium = 2.00 M − 1.30 = 0.70 M

4) Determine the equilibrium amount of [SO3]:

While 1.30 M of NO is produced, 1.30 M of SO3 is also produced.

This is due to the 1:1 molar ratio between the coefficients of SO3 and NO.

[SO3] at equilibrium = 1.30 M

5) The equilibrium expression is:

 [SO3] [NO] Kc  = –––––––––– [SO2] [NO2]

6) Insert values and solve:

 (1.30) (1.30) Kc  = –––––––––– = 3.45 (0.70) (0.70)

Example #10: A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0 L vessel at 300 K. The following equilibrium is established:

2NO(g) + 2H2(g) ⇌ N2(g) + 2H2O(g)

At equilibrium, [NO] = 0.062 M.

(a) Calculate the equilibrium concentrations of H2, N2, and H2O.
(b) Calculate Kc.

Solution to (a):

1) How much NO is used up?

0.10 M − 0.062 M = 0.038 M

0.10 M comes from 0.10 mol / 1.0 L = 0.10 M

2) How much H2 is used up? How much H2 remains at equilibrium?

Note that the coefficients of NO and H2 are equal.

This means that the amounts used up for each substance are equal.

[H2] used up equals 0.038 M

The equilibrium amount for H2 is this:

0.050 M − 0.038 M = 0.012 M

3) How much N2 is produced? How much N2 is present at equilibrium?

The coefficient on NO is 2, the coefficient on N2 is 1.

This means N2 produced is one-half the amount that is used up by NO.

[N2] produced is 0.038 / 2 = 0.019 M

There was zero N2 at the start, therefore 0.019 M of N2 is present at equilibrium.

4) How much H2O is produced? How much H2O is present at equilibrium?

The coefficients on NO and H2O are the same.

This means that, for every amount of NO used up, that same amount of H2O is produced.

[H2O] produced equals 0.038 M.

The [H2O] present at equilibrium is 0.10 M + 0.038 M = 0.128 M

Solution to (b):

1) The equilibrium expression is:

 [N2] [H2O]2 Kc  = –––––––––– [NO]2 [H2]2

2) Insert values and solve:

 (0.019) (0.128)2 Kc  = ––––––––––––– = 562 (0.062)2 (0.012)2

Example #11: At 80 °C, Kc = 1.87 x 10¯3 for the reaction:

PH3BCl3(s) ⇌ PH3(g) + BCl3(g)

(a) Calculate the equilibrium concentrations of PH3 and BCl3 if a solid sample of PH3BCl3 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 PH3BCl3 that must be added to the flask to achieve equilibrium?

Solution to (a):

1) Write the equilibrium expression for the above reaction:

Kc = [PH3] [BCl3]

Note that PH3BCl3(s) is not part of the equilibrium expression due to it being a solid.

2) During the reaction, . . .

. . . an unknown amount of PH3 is produced during the decomposition. Likewise, an unknown amount of BCl3 is produced.

However, from the 1:1 stoichiometry of PH3 and BCl3, 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 PH3BCl3 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:

N2(g) + 3H2(g) ⇌ 2NH3(g)

At equilibrium, 15.0 moles of H2, 24.0 moles of N2, and 12.0 moles of NH3 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:

[N2] = 24.0 mol / 3.00 L = 8.00 M
[H2] = 15.0 mol/3.00 L = 5.00 M
[NH3] = 12.0 mol / 3.00 L = 4.00 M

2) Fill in an ICEbox with the known data:

 [N2] [H2] ⇌ [NH3] Initial 0 Change Equilibrium 8.00 5.00 4.00

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

 [N2] [H2] ⇌ [NH3] 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:

 [N2] [H2] ⇌ [NH3] Initial [N2]o [H2]o 0 Change −2.00 −6.00 4.00 Equilibrium 8.00 5.00 4.00

5) The answers are these values:

[N2]o = 10.00 M
[H2]o = 11.00 M

Example #13: At 25 °C, Kp = 2.90 x 10¯3 for the reaction:

NH4OCONH2(s) ⇌ 2NH3(g) + CO2(g)

In an experiment carried out at 25 °C, a certain amount of NH4OCONH2 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 Kp expression:

Kp = (PNH3)2 (PCO2)

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 CO2 produced, two NH3 are produced.

3) Continue the solution:

4x3 = 2.90 x 10¯3

x3 = 7.25 x 10¯4

x = 0.089835 atm

4) The equilibrium concentrations are as follows:

PNH3 = 2x = 0.17967 atm
PCO2 = x = 0.089835 atm

0.17967 atm + 0.089835 atm = 0.269505 atm

Rounded 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:

Kp = (0.17967)2 (0.089835)

The value stated in the problem is recovered.

Example #14: Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia (NH3). 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 Kp value for this reaction.

Solution:

1) Construct an ICEbox with the initial data:

 PN2 PH2 ⇌ PNH3 Initial 1.01 2.05 0 Change Equilibrium

 PN2 PH2 ⇌ PNH3 Initial 1.01 2.05 0 Change −x −3x +2x Equilibrium

3) Add in the equilibrium amounts:

 PN2 PH2 ⇌ PNH3 Initial 1.01 2.05 0 Change −x −3x +2x Equilibrium 1.01 − x 2.05 − 3x 2x

4) Use Dalton's Law:

Ptot = PN2 + PH2 + PNH3

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 − 3x

2.05 − 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 Kp.

7) Solve for the Kp:

PN2 = 1.01 − x = 0.505 atm
PH2 = 2.05 − 3x = 0.535 atm
PNH3 = 2x = 1.01 atm

 (PNH3)2 Kp  = ––––––––– (PN2) (PH2)3

 (1.01)2 Kp  = ––––––––––––– = 13.2 (0.505) (0.535)3

Example #15: Consider the system:

N2(g) + 3H2(g) ⇌ 2NH3(g)

A 2.568 g sample of NH3 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 Kp for the decomposition of ammonia at 25 °C.

Solution:

1) For the initial state of NH3, determine moles, then pressure:

2.568 g / 17.0307 g/mol = 0.1507865 mol

(PNH3)o = nRT / V

(PNH3)o = [(0.1507865 mol) (0.08206 L atm / mol K) (298 K)] / 2.000 L

(PNH3)o = 1.8436575 atm

2) For the equilibrium state of NH3, determine mass, then moles, then pressure:

(2.568 g) (0.757) = 1.943976 g

1.943976 g / 17.0307 g/mol = 0.1141454 mol

(PNH3) = [(0.1141454 mol) (0.08206 L atm / mol K) (298 K)] / 2.000 L

(PNH3) = 1.395649 atm

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

 PN2 PH2 ⇌ PNH3 Initial 0 0 1.8436575 Change Equilibrium 1.395649

4) Fill in the change row in the ICEbox:

 PN2 PH2 ⇌ PNH3 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 = 1.395649

x = 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 Kp:

 (PNH3)2 Kp  = ––––––––––– (PN2) (PH2)3

 (1.395649)2 Kp  = –––––––––––––––– = 28.65 (0.22400) (0.67200)3

Example #16: An 8.00 g sample of SO3 was placed in an evacuated container, where it decomposed at 600 °C according to the following reaction:

SO3(g) ⇌ SO2(g) + 12O2(g)

At equilibrium the total pressure and the density of the gaseous mixture were 1.80 atm and 1.60 g/L, respectively. Calculate Kp 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/L

V = 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 SO3 to moles:

8.00 g/80.063 g/mol = 0.0999213 mol

3) Let 'x' = moles SO3 decomposed,

4) Therefore, at equilibrium:

moles SO3 = 0.0999213 − x
moles SO2 = x
moles O2 = x/2

moles total = 0.0999213 − x + x + x/2 = 0.0999213 + x/2

5) Use PV = nRT to determine 'x'

P = nRT / V

1.80 = [(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 SO3 ---> 0.0999213 − 0.05142 = 0.0485
moles SO2 ---> 0.05142
moles O2 ---> 0.05142 / 2 = 0.02571

moles total ---> 0.0999213 + x/2 ---> 0.0999213 + 0.02571 = 0.12563

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

PSO3 ---> (1.80 atm) (0.0485 / 0.12563) = 0.695 atm
PSO2 ---> (1.80 atm) (0.05142 / 0.12563) = 0.737 atm
PO2 ---> (1.80 atm) (0.02571 / 0.12563) = 0.368 atm

8) Calculate the Kp:

 (PSO2) (PO2)½ Kp  = ––––––––––– (PSO3)

 (0.737) (0.368)½ Kp  = ––––––––––––– = 0.643 0.695