08 The Equilibrium Constant

Chem 317 (for Chem 305)

Name _________________________

Exercise #_____ The Equilibrium Constant When a system is at equilibrium, a constant value is established by the multiplicative product of the concentrations of the products' concentrations (each raised first to the power of its coefficient), then divided by the multiplicative product of the reactants' concentrations (each raised first to the power of its coefficient). Now that may be somewhat wordy, so here it is using a generic chemical equation: wA + xB Ù yC + zD Following the word definition above, we have this:

A, B, C, and D are the chemical substances. x, y, w and z are the coefficients. Let’s do it again, this time using a specific chemical equation: 2 SO2 + O2 Ù 2 SO3 And the answer is:

By the way, you may notice that the terms products and reactants are somewhat blurred in an equilibrium situation. The agreed upon convention is to use the reaction AS WRITTEN and call the left side "reactants" and the right side "products." Exercise #1 Write an equilibrium expression for each of the following reactions. 1) 3 O2 <===> 2 O3

4) PCl5 <===> PCl3 + Cl2

2) N2 + 3 H2 <===> 2 NH3

5) SO2 + (1/2) O2 <===> SO3

3) H2 + I2 <===> 2 HI

Adapted from “The Chem Team” http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium.html

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Chem 317 (for Chem 305)

Name _________________________

Calculating the Equilibrium Constant from Equilibrium Conditions The easiest way to explain is by an example problem or two: 1) Calculate the equilibrium constant (Keq) for the following reaction: H2 + I2 <==> 2 HI when the equilibrium concentrations at 25 °C were found to be: [H2] = 0.0505 M [I2] = 0.0498 M [HI] = 0.389 M So, how do you solve the problem? The first thing to do is write the equilibrium expression for the reaction as written in the problem. This is what to write:

Now, all you have to do is substitute numbers into the problem. Keq is what we want to find, so that's our "x." Here is what we get: x=

(0.389)2 = 60.2 [(0.0505) (0.0498)]

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

(0.0127)2 = 48.8 [(0.00560) (0.000590)]

Adapted from “The Chem Team” http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium.html

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Chem 317 (for Chem 305)

Name _________________________

Exercise #2: Calculate Keq using the same equation as above and with the following equilibrium concentrations: The answer is 48.4 [H2] = 0.00460 M [I2] = 0.000970 M [HI] = 0.0147 M

Keq=

An important point: did you remember to square the numerator. This is the number one student problem in solving this type of problem - forgetting the exponent. The number two error is wanting to change the concentrations. For example, when [HI] = 0.0147, many students will want to double it, saying “Well, there is a 2 HI in the equation.” No, No, No!! Use the concentrations as given. One more discussion point: you may have noticed the Keq 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.

Using a Known Keq If we know the value of the equilibrium constant, and the concentration of all reactants and products except one, we can use the equilibrium constant expression and a little algebra (I know that is a scary word!) to solve for the one we don’t know. Example: Find the equilibrium concentration of N2 if the equilibrium concentrations are: [NH3] = 0.157 M [H2] = 0.763 M N2 + 3 H2 Ù 2 NH3

For the reaction K eq =

[NH3 ] 2 = 6.02 x 10 -2 3 [N 2 ][H 2 ]

We need to rearrange the equation to solve for [N2]. We can do this by multiplying both sides of the equation by [N2] and dividing both sides by Keq. This results in: [N 2 ] =

[NH3 ] 2 (K eq )[H 2 ]3

=

(0.157) 2 0.02465 = = 0.922 M -2 3 (6.02 x 10 )(0.763) (0.0602)(0.4442)

Adapted from “The Chem Team” http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium.html

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Chem 317 (for Chem 305)

Name _________________________

Exercise #3 2 SO2 + O2 Ù 2 SO3

Given the equation

Keq = 4.36

Calculate the equilibrium concentration of oxygen given the following equilibrium concentrations: [SO2] = 1.50 [SO3] = 3.50

Using Keq to Predict Relative Concentrations The size of the equilibrium constant can give us information about the relative amounts of reactants and products present at equilibrium. When K << 1 ƒ The value in the numerator (representing product concentrations) is much smaller than the value in the denominator (representing reactant concentrations). ƒ The reaction lies to the left (mostly reactants) When K >> 1 ƒ The value in the denominator (representing reactant concentrations) is much smaller than the value in the numerator (representing product concentrations). ƒ The reaction lies to the right (mostly products) When K ≅ 1 ƒ The value in the denominator (representing reactant concentrations) is close to the value in the numerator (representing product concentrations). ƒ The reaction lies in the middle (mix of reactants and products) Exercise #4 Use Keq to predict the relative concentrations of reactants and products for the following equilibria. 1) HI + H2O Ù I- + H3O+ a) mostly products

b) mostly reactants

2) HF + H2O Ù F- + H3O+ a) mostly products

Keq = 2.5 x 1010 c) mix of reactants and products

Keq = 6.6 x 10-4

b) mostly reactants

c) mix of reactants and products

3) Which acid HI or HF is a “strong acid” and why?

Adapted from “The Chem Team” http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Equilibrium.html

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