ChemTeam: Calculating Equilibrium Concentrations (and equilibrium constant) from initial concentrations and one equilibrium concentration

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 K_c.

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 K_c:

[HI]²
K_c = ––––––––
[H₂] [I₂]

[0.258]²
K_c = –––––––––––– = 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 (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
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]
K_c = ––––––––––
[CO] [H₂]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:

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 K_c expression, put values in, and solve:

[NO₂]²
K_c = –––––––
[N₂O₄]

(3.43)²
K_c = ––––––– = 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, K_c, 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 K_c expression, substitute values, and solve:

[HI]²
K_c = –––––––
[H₂] [I₂]

(5.22)²
K_c = ––––––––––– = 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, K_c, 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 K_c 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]
K_c = ––––––––––
[SO₃] [NO]

(0.70) (0.70)
K_c = –––––––––– = 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 K_p for the reaction.

Solution to (a):

PV = nRT is solved for P
P_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 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 K_c.

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 K_c:

[HBr]²
K_c = –––––––––
[H₂] [Br₂]

(0.400794)²
K_c = –––––––––––––––––– = 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 K_p 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
K_p = –––––
(S₈)

4) Insert values and solve:

(3.00)⁴
K_p = ––––– = 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 K_p = 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, K_c, 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]
K_c = ––––––––––
[SO₂] [NO₂]

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 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 K_c.

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
K_c = ––––––––––
[NO]² [H₂]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:

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:

K_c = [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, K_p = 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 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 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:

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 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:

K_p = (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 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 − 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 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:

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 K_p 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
(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 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
(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 = 1.395649
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 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 SO₃ was placed in an evacuated container, where it decomposed at 600 °C according to the following reaction:

SO₃(g) ⇌ SO₂(g) + ¹⁄₂O₂(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/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:

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

Return to Equilibrium Menu

Reaction	[H₂]	[I₂]	⇌	[HI]
Initial	0.260	0.144		0
Change				+0.258
Equilibrium				0.258

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

	[CO]	[H₂]	⇌	[CH₃OH]
Initial	0.18398	0.22046		0
Change
Equilibrium				0.051716

	0.051716
K_c =	––––––––––––––––––––	= 28.6
	(0.132246) (0.117028)²

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

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

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

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

	(0.70) (0.70)
K_c =	––––––––––	= 0.29
	(1.30) (1.30)

Reaction	P_CO₂	P_H₂	⇌	P_CO	P_H₂O
Initial	4.103	2.0515		0	3.2824
Change
Equilibrium					3.51

	(P_CO) (P_H₂O)
K_p =	–––––––––––
	(P_CO₂) (P_H₂)

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

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

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

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

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

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

	[N₂]	[H₂]	⇌	[NH₃]
Initial				0
Change	−x	−3x		+2x
Equilibrium	8.00	5.00		4.00

	P_N₂	P_H₂	⇌	P_NH₃
Initial	1.01	2.05		0
Change	−x	−3x		+2x
Equilibrium

	(P_NH₃)²
K_p =	–––––––––
	(P_N₂) (P_H₂)³

Calculating Equilibrium Concentrations (and equilibrium constant)from initial concentrations and one equilibrium concentration

Calculating Equilibrium Concentrations (and equilibrium constant)
from initial concentrations and one equilibrium concentration

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

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

	(P_SO₂) (P_O₂)^½
K_p =	–––––––––––
	(P_SO₃)

	(0.737) (0.368)^½
K_p =	–––––––––––––	= 0.643
	0.695