Calculate Equilibrium Concentrations from Initial Concentrations.

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

By the way, I'm going to stop using K_{eq} and will start using K_{c}, where 'c' stands for 'concentration.' In the future, you will also study K_{p}, where 'p' stands for 'pressure.'

**Example #1:** Calculate the equilibrium constant (K_{c}) for the following reaction:

Hwhen the equilibrium concentrations at 25.0 °C were found to be:_{2}(g) + I_{2}(g) ⇌ 2HI(g)

[H_{2}] = 0.0505 M

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

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

(0.389) ^{2}K _{c}=–––––––––––––– (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 K_{c}? And, for that matter, K_{p}. 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:

[H_{2}] = 0.00560 M

[I_{2}] = 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}K _{c}=–––––––––––––––––– (0.00560) (0.0000590)

2) The answer is 48.8

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:

[H_{2}] = 0.00460 M

[I_{2}] = 0.000970 M

[HI] = 0.0147 M

Calculate the K_{c}.

**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 K_{c} 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 K_{eq} 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 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 an ICEbox:

[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}=–––––––––––– [0.131] [0.015] K

_{c}= 33.9

**Example #5:** 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) The K_{c} calculation will involve 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}=–––––––––––––––––– (0.132246) (0.117028) ^{2}K

_{c}= 28.5537To 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 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:

K_{c}= [NO_{2}]^{2}/ [N_{2}O_{4}]K

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

_{c}= 1.91

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 #7:** 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 Kc expression, substitute values, and solve:

K_{c}= [HI]^{2}/ ([H_{2}] [I_{2}])K

_{c}= (5.22)^{2}/ [(1.19) (0.0900)]K

_{c}= 254 (to three sig figs)

**Example #8:** 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) ICEbox, baby!

[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:

[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}=–––––––––– (1.30) (1.30) K

_{c}= 0.29

**Example #9:** The equilibrium concentrations for the reaction between CO and molecular chlorine to form COCl_{2}(g) at 74 °C are [CO] = 0.012 M and [Cl_{2}] = 0.054 M and [COCl_{2}] = 0.14 M. Calculate the equilibrium constants K_{c} and K_{p}.

**Solution:**

1) First, the chemical reaction:

CO(g) + Cl_{2}(g) ⇌ COCl_{2}(g)

2) Now, the K_{c} expression, followed by calculating its value:

[COCl _{2}]K _{c}=––––––––– [CO] [Cl _{2}]

0.14 K _{c}=–––––––––––– (0.012) (0.054) K

_{c}= 216 (to three sig figs)

2) Convert K_{c} to K_{p}. The conversion is usually shown as:

K_{p}= K_{c}(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 = −1K

_{p}= (216.05) [(0.08206) (347)]¯^{1}K

_{p}= 216.05 / 28.475K

_{p}= 7.59 (to three sig figs)

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

CH_{4}(g) + H_{2}O(g) ⇌ 3H_{2}(g) + CO(g)

What are the equilibrium constants (both K_{c} and K_{p}) for the reaction if a mixture at equilibrium contains gases with the following concentrations:

CH_{4}= 0.126 M; H_{2}O = 0.242 M; CO = 0.126 M; H_{2}= 1.15 M

at a temperature of 760. °C?

**Solution:**

1) Let's write the K_{c} expression and solve it:

[H _{2}]^{3}[CO]K _{c}=–––––––––– [CH _{4}] [H_{2}O]

(1.15) ^{3}(0.126)K _{c}=––––––––––––– (0.126) (0.242) K

_{c}= 6.2846 (to three sig figs, 6.28)

2) Convert K_{c} to K_{p}:

K_{p}= K_{c}(RT)^{Δn}Δn = 4 − 2 = 2

K

_{p}= (6.2846) [(0.08206) (1033)]^{2}K

_{p}= (6.2846) (7185.61)K

_{p}= 45200 (to three sig figs)

Calculate Equilibrium Concentrations from Initial Concentrations.