### Calculating the Equilibrium Constant from Equilibrium Concentrations

The easiest way to explain is by looking at a few examples.

By the way, I'm going to stop using Keq and will start using Kc, where 'c' stands for 'concentration.' In the future, you will also study Kp, where 'p' stands for 'pressure.'

Example #1: Calculate the equilibrium constant (Kc) for the following reaction:

H2(g) + I2(g) ⇌ 2HI(g)
when the equilibrium concentrations at 25.0 °C were found to be:
[H2] = 0.0505 M
[I2] = 0.0498 M
[HI] = 0.389 M

Solution:

1) The first thing to do is write the equilibrium expression for the reaction as written in the problem. This is what to write:

 [HI]2 Kc = –––––––– [H2] [I2]

2) Now, all you have to do is substitute numbers into the equilibrium expression:

 (0.389)2 Kc = –––––––––––––– (0.0505) (0.0498)

3) Solving the above and rounding to the correct number of sig figs (remember those??), we get 60.2

Something of a side issue is what are the units of the Kc? And, for that matter, Kp. For reasons beyond the scope of this lesson, the answer is none. The equilibrium constant does not have any units.

Some advice: your teacher may insist on putting units on the equilibrium constant. Please do not march up to him/her and announce they are wrong because some guy on the Internet says so. That will NOT make your teacher happy!

Example #2: The same reaction as above was studied at a slightly different temperature and the following equilibrium concentrations were determined:

[H2] = 0.00560 M
[I2] = 0.000590 M
[HI] = 0.0127 M

From the data, calculate the equilibrium constant.

Solution:

1) Same technique as above, write the equilibrium expression and substitute into it. Then solve. So, we get this:

 (0.0127)2 Kc = –––––––––––––––––– (0.00560) (0.0000590)

Time for a small lecture:

Please be very careful in using your calculator to solve these problems. When I solved this problem while writing the first edition of this tutorial (on December 28, 1998), I first got some really weird looking answer that didn't feel right, so I did it again. Sure enough, I have made an entry error somewhere in the problem.

Underscoring my plea for carefulness is the difference between you and me in problem solving. The above problem is routine for me and solely on the basis of experience did I reject my first answer as being wrong (it "felt" wrong). You guys don't have that experience, so you don't have the feel. Yet!!

So, BE CAREFUL.

Here endth the lecture.

Example #3: Using the same equation as above and with the following equilibrium concentrations:

[H2] = 0.00460 M
[I2] = 0.000970 M
[HI] = 0.0147 M

Calculate the Kc.

Solution:

I'm not going to write the set-up, but I want you to write it down on your paper. Then solve it. The answer is 48.4.

An important point: remember to square the numerator. This is the number one rookie problem in solving these things - forgetting the exponent.

The number two error is wanting to change the concentrations. For example, when [HI] = 0.0147, the rookie will want to double it, saying "Well, there is a 2HI in the equation. No, No, No!! Use the concentrations as given.

One more discussion point: you may have noticed the Kc answers for #2 and #3 are slightly different when they are supposed to be the same. The answer: experimental error. One can never be perfect, so the values for Keq that get published are actually an average of many careful experiments.

These next several examples have a twist. 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 the next tutorial. You can also examine the results of this search. Note that many sources call it an ICE table. Boring!!

Example #4: 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 an ICEbox:

 [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 = –––––––––––– [0.131] [0.015]

Kc = 33.9

Example #5: 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) The Kc calculation will involve 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 = –––––––––––––––––– (0.132246) (0.117028)2

Kc = 28.5537

To three sig figs and following the rule for rounding with five, the final answer is 28.6.

Example #6: 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:

Kc = [NO2]2 / [N2O4]

Kc = (3.43)2 / 6.165

Kc = 1.91

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 #7: 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:

Kc = [HI]2 / ([H2] [I2])

Kc = (5.22)2 / [(1.19) (0.0900)]

Kc = 254 (to three sig figs)

Example #8: 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) ICEbox, baby!

 [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:

 [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 = –––––––––– (1.30) (1.30)

Kc = 0.29

Example #9: The equilibrium concentrations for the reaction between CO and molecular chlorine to form COCl2(g) at 74 °C are [CO] = 0.012 M and [Cl2] = 0.054 M and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

Solution:

1) First, the chemical reaction:

CO(g) + Cl2(g) ⇌ COCl2(g)

2) Now, the Kc expression, followed by calculating its value:

 [COCl2] Kc = ––––––––– [CO] [Cl2]

 0.14 Kc = –––––––––––– (0.012) (0.054)

Kc = 216 (to three sig figs)

2) Convert Kc to Kp. The conversion is usually shown as:

Kp = Kc(RT)Δn

where n = total moles of gas on the product side minus total moles of gas on the reactant side

3) Solving:

Δn = 1 − 2 = −1

Kp = (216.05) [(0.08206) (347)]¯1

Kp = 216.05 / 28.475

Kp = 7.59 (to three sig figs)

Example #10: Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures.

CH4(g) + H2O(g) ⇌ 3H2(g) + CO(g)

What are the equilibrium constants (both Kc and Kp) for the reaction if a mixture at equilibrium contains gases with the following concentrations:

CH4 = 0.126 M; H2O = 0.242 M; CO = 0.126 M; H2 = 1.15 M

at a temperature of 760. °C?

Solution:

1) Let's write the Kc expression and solve it:

 [H2]3 [CO] Kc = –––––––––– [CH4] [H2O]

 (1.15)3 (0.126) Kc = ––––––––––––– (0.126) (0.242)

Kc = 6.2846 (to three sig figs, 6.28)

2) Convert Kc to Kp:

Kp = Kc(RT)Δn

Δn = 4 − 2 = 2

Kp = (6.2846) [(0.08206) (1033)]2

Kp = (6.2846) (7185.61)

Kp = 45200 (to three sig figs)