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Ch 1 - Basics
Ch 2 - Atoms, Molecules, and Ions
Ch 3 - Mass Relationships in Chemical Reactions
Ch 4 - Reactions in Aqueous Solutions
Ch 5 - Gases
Ch 6 - Thermochemistry
Ch 7 - Quantum Theory
Ch 8 - Periodic Relationships
Ch 9 - Chemical Bonding I
Ch_10 - Chemical Bonding II
Ch_11 - Intermolecular Forces & Liquids and Solids
Ch_12 - Physical Properties of Solutions
Ch_13 - Chemical Kinetics
Ch_14 - Chemical Equilibrium
Ch_15 - Acids and Bases
Ch_16 - Additional Equilibrium Topics
Free Response Questions
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Ch_14 - Chemical Equilibrium
Chapter 14 - Chemical Equilibrium
In this chapter we will see how the kinetics of a reaction can lead to a state of chemical equilibrium where there is no observable change in the reaction. This means that the concentrations of the reactants and products are not change and therefore remain constant. We will learn how to show this state by writing equilibrium expressions and calculating the equilibrium constant. Related to the equilibrium expression is the reaction quotient which can tell us if a reaction is moving towards forming more reactants or products at any given time. Finally, we will focus on what is called an ICE table as the primary problem solving method for solving for concentration values at equilibrium. We will immediately look at some applications of equilibrium in Chapter 16, and then go back to Chapter 15.
The Concept of Equilibrium and the Equilibrium Constant
Writing Equilibrium Constant Expressions
The Relationship Between Chemical Kinetics and Chemical Equilibrium
What does the Equilibrium Constant Tell Us?
Factors that Affect Chemical Equilibrium
Along with the embedded videos, I have included some links to some tutorials, simulations and animations. Pay attention to my worked out Practice Problems and follow along with what we are doing in class! The listed problems from the book will be due the day of the Chapter 14/16 Test. This material will be covered very quickly so keep up.
Practice Problems:14.1, 14.3, 14.6, 14.8, 14.9, 14.10, 14.11, 14.15, 14.16, 14.17, 14.18, 14.19, 14.22, 14.29, 14.30, 14.31, 14.32, 14.77, 14.33, 14.36, 14.37, 14.38, 14.40, 14.41, 14.43, 14.44, 14.48, 14.82, 14.51, 14.52, 14.53, 14.54, 14.55, 14.56, 14.58, 14.59, 14.98
14.1 - The Concept of Equilibrium and the Equilibrium Constant
We have come across the term equilibrium several times already this year. Up to this point you probably have the basic thought that equilibrium means that two processes are occurring at the same time, and while this is a very generalized definition, we need something a lot more specific in order to clear up some common misconceptions. The first thing to forever remember about chemical equilibrium is that it means that the forward and reverse rates of a chemical reaction are exactly the same. Exactly. The same. The second is that the concentration of the reactants and products do not change at equilibrium and instead remain constant. Constant. This does NOT mean however the concentration of the reactants and products is exactly the same, absolutely not. Nope. Constant, not equal. Sear these two things into you brain forever.
You can see what equilibrium looks like in the following graph of the formation of ammonia from hydrogen and nitrogen gases. It shows the disappearance of reactants (H2 & N2) and appearance of products (NH3) over time. The fact that the lines level off is what is meant by constant concentration. When this occurs, that is when equilibrium is achieved. So in the zone from A to B, the reaction is occurring and products are being made, but once you reach B and continue on through C and D, the concentrations are stable and a state of equilibrium exists.
The equilibrium constant and expression are what govern a chemical reaction at equilibrium. Unlike writing rate laws, the stoichiometric coefficients in equilibrium reactions do play a part. They become the exponents on the reactants and products in the equilibrium expression. The equilibrium expression is always listed as the products over the reactants as you can see below. This is simply the mathematical expression of the law of mass action.
Finally, the size of the equilibrium constant tells you a lot about the chemical reaction. If you have the exact same amount of reactants and products then mathematically the equilibrium expression must equal 1. If you have more products than reactants then the numerator will be larger and the equilibrium constant will be more than 1. If you have more reactants than products then the denominator will be larger and the equilibrium constant will be less than 1.
Related Problems - 14.1, 14.3
14.2 - Writing Equilibrium Constant Expressions
We will focus a lot of time now on writing equilibrium expressions as they are both fairly easy, and frequently asked for. The first thing to note is that there are reactions where everything is in the same phase (homogeneous) and those where the reactants and products are in different phases (heterogeneous). These cannot exactly be treated the same way, as they won’t have the same effect on the equilibrium. What we find is that gases and aqueous solutions greatly affect the equilibrium expression, but solid substances and pure liquids do not affect the equilibrium expression really at all. So when you look at a chemical reaction and need to write the equilibrium expression, you can ignore anything that is listed as a solid or a liquid.
Writing Equilibrium Expressions
So that leaves us with aqueous solutions and gases. Both of these can be written into an equilibrium expression based on concentration called Kc. These must have units in molarity for the concentration equilibrium expression. Unless specified otherwise, an equilibrium expression is always considered to be a concentration equilibrium expression, Kc.
For reactions that are all expressed as gases, you can also write an equilibrium expression in units of pressure. So when you have units of atm or mm Hg then you have a pressure equilibrium expression called Kp. Now an equation that is all gases can be expressed as wither a Kc or a Kp depending on whether you use molarity or atm. Surprisingly, for the exact same reaction, the Kc and Kp will not be the same number so you must be certain which thing they are asking for. The two are related however and can be converted using equation 14.5 on pg 606.
Ex 14.1 Prac Ex.png
Ex 14.2 Prac Ex.png
Ex 14.3 Prac Ex.png
Ex 14.4 Prac Ex.png
Ex 14.5 Prac Ex.png
Ex 14.6 Prac Ex.png
Converting between Kc and Kp
Finally, we have seen over several chapters how adding chemical reactions together can lead to new ways to show information. Adding chemical reactions at equilibrium allows you to also combine the equilibrium expressions. To get the new equilibrium expression all you have to do is multiply the original two equilibrium expressions. Subtracting reactions means dividing equilibrium expressions.
Ex 14.7 Prac Ex.png
Related Problems - 14.6, 14.8, 14.9, 14.10, 14.11, 14.15, 14.16, 14.17, 14.18, 14.19, 14.22, 14.29, 14.30, 14.31, 14.32, 14.77
14.3 - The Relationship Between Chemical Kinetics and Chemical Equilibrium
Remembering what we started off saying, the rate of the forward and reverse reactions at equilibrium are the same. So in theory you could set the two rate laws equal to each other. By rearranging the constants on to the same side, you can see that the ratio of the rate constants looks an awful lot like the equilibrium expressions we just finished writing. In fact, the equilibrium constant is simply the ratio of the two rate constants. Since the rate constant is dependent on temperature as we found in chapter 13, it makes sense that the equilibrium constant is also temperature dependent. So in order to write an equilibrium expression and calculate the equilibrium constant, the correctly balanced chemical equation and the temperature must be known.
Related Problems - 14.33, 14.36
14.4 - What does the Equilibrium Constant Tell Us?
We can use the equilibrium expression in several different ways. The most obvious one is to plug in concentration or pressure values and calculate the equilibrium constant. Make sure you take any exponents into account in this calculation. The second easiest thing to do is to calculate an unknown reactant or product concentration at equilibrium when the equilibrium constant and other concentrations are known. This should also be a simple calculation.
As stated earlier, the value of the equilibrium constant can tell you if reactants or products are favored in the equilibrium. But how do you know if you are at equilibrium? As we saw in the graphs in the first section, it can take a while for a reaction to reach equilibrium. If you plug in the initial concentrations of reactants and products into the equilibrium expression you can calculate something called the reaction quotient, Q. The size of the reaction quotient in comparison to the equilibrium constant will tell you if you are at equilibrium, creating products, or creating reactants. If Q = K then you are at equilibrium. If Q < K then you haven’t reached equilibrium yet and still need to create more products, moving the reaction to the right. If Q > K then you have passed equilibrium and need to go back to reactants, moving the reaction to the left.
Ex 14.8 Prac Ex.png
The most common thing you will do with an equilibrium expression is to calculate the concentration values of reactants and products given the equilibrium constant and some initial concentration values. All problems of this nature are solved the same way using something typically called an “ICE” table. ICE stands for initial, change, and equilibrium and allows you to account for all of the possible changes in an equilibrium reaction.
ICE Table example
The common scenario is to know the initial concentration of reactants and assume that there are no products in existence. This means that any change in this condition will be to reduce the concentration of reactants and add to the concentration of the products. We always make this change equal x. We use the stoichiometric coefficients from the balanced equation as coefficients on the x. Then you can just use basic algebra to plug values into the equilibrium expression and solve for x. This is the calculation you will see most often on the free response portion of the exam and the one we will practice the most often.
Ex 14.9 Prac Ex.png
Related Problems - 14.37, 14.38, 14.40, 14.41, 14.43, 14.44, 14.48, 14.82
14.5 - Factors that Affect Chemical Equilibrium
Work through this quick tutorial about a chicken (I'm not kidding) and stresses to its breathing to help you understand how "stress" affects the direction of equilibrium:
A reaction at equilibrium is something dynamic and constant movement. It is not static. As such, it can change. But something must be done to it to make it change. We term these things “stresses” and a chemical reaction at equilibrium will react to stresses according to Le Chatlier’s principle. Le Chatlier’s Principle basically states that a chemical reaction at equilibrium will react to a stress in such a way to reduce that stress and establish a new equilibrium state. It is important to realize that the applied stress prevents a reaction from achieving the same equilibrium state, it will always be a new equilibrium state. Our most common types of stress include changes in concentration, pressure, volume, and temperature.
Ex 14.11 Prac Ex.png
For changes in concentration, the first thing to remember is that changes in concentration only matter if the thing that is being changed is represented in the equilibrium expression. So changes in concentration of solids and liquids DO NOT affect the equilibrium. For everything else, you need to see if the changed item is a reactant or product. Increasing a product will shift a reaction to the reactants so the extra product gets used up. The opposite is true if more reactant is used. The addition of something shifts the reaction in such a way to use it up. You can also take something out of a reaction in which case the reaction shifts to replace it. So you can keep a reaction continually producing products if you create a way to keep removing one of the products.
Work through this
LC Virtual Lab #1
to see these types of changes.
Changing volume and pressure only affects gaseous reactants and products. In general we think of it as changes in pressure, but remember from chapter 5 that changing the volume will change the pressure. Increases in pressure will cause a reaction to shift towards the side with fewer moles of gas. If there are equal amounts of moles of gas, then there is no change with changes in pressure. Decreases in pressure will shift a reaction to the side with fewer moles of gas. Remember that a decrease in volume is an increase in pressure and vice versa. Please note that the addition of an inert gas such as helium is a favorite thing to ask about. Adding this inert gas at constant volume may increase the overall pressure but not the partial pressures of the individual gases so no change occurs.
Ex 14.12 Prac Ex.png
Work through this
LC Virtual Lab #2
to see these types of changes.
Finally, changes in temperature depend on whether a reaction is endo or exothermic. If a reaction is endothermic that means that energy is added to the reaction and we can think of heat/energy as a reactant. If a reaction is exothermic that means that energy is released from the reaction and we can think of heat/energy as a product. Then you can treat changes in temperature as simply changes in the concentration of heat/energy and it works the same way as originally discussed. If you don’t know if a reaction is endo or exothermic, then you cannot determine how changes in temperature will affect the equilibrium. But ANY change in temperature will change the value of the equilibrium constant.
Ex 14.13 Prac Ex.png
Work through this
LC Virtual Lab #3
to see these types of changes.
Finally, we need to explore how a catalyst affects equilibrium. As we found in chapter 14, the addition of a catalyst lowers the activation energy for a reaction making the rate of both the forward and reverse reactions faster. Since both the forward and reverse reactions speed up, there is actually no change in the equilibrium constant or the position of equilibrium. The only thing that does happen is that the state of equilibrium is achieved faster than would normally happen without the catalyst.
Le Chatelier's Principle
Related Problems - 14.51, 14.52, 14.53, 14.54, 14.55, 14.56, 14.58, 14.59, 14.98
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