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Equilibrium Chapter 12. Equilibrium Chemical equilibrium is a dynamic condition in which concentrations do not change and the rates of the forward and.

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Presentation on theme: "Equilibrium Chapter 12. Equilibrium Chemical equilibrium is a dynamic condition in which concentrations do not change and the rates of the forward and."— Presentation transcript:

1 Equilibrium Chapter 12

2 Equilibrium Chemical equilibrium is a dynamic condition in which concentrations do not change and the rates of the forward and reverse reactions are equal. Meaning… The reactants are making products at the same rate that products are making reactants.

3 Equilibrium Does this graph look familiar? According to this Concentration v. Time graph, where is the equilibrium? The equilibrium begins here.

4 Equilibrium Hydrogen and nitrogen combine to form ammonia at the same time ammonia is decomposing into hydrogen and nitrogen. The concentrations of hydrogen, nitrogen, and ammonia remain constant as both the forward and reverse reactions proceed at the same rate.

5 Equilibrium Only reactions that meet these three conditions can reach equilibrium: The reaction must occur in a closed container. The reaction must be a reversible reaction. The reaction must have plenty of chemicals to begin with.

6 Equilibrium How is equilibrium measured? with an Equilibrium Constant (K eq )

7 Equilibrium For a reversible reaction (one that proceeds in both the forward and reverse directions), the equilibrium constant (K eq ) is a number showing the relationship between the mathematical product of the concentrations of the products divided by the mathematical products of the concentrations of the reactants, each raised to the power of its coefficient in a balanced equation.

8 Equilibrium The equation to calculate this relationship is called an equilibrium expression. To determine the expression for any reaction: **Concentration (molarity) is represented by putting a [bracket] around a formula. [H + ] means “the concentration of hydrogen ions”

9 Equilibrium EXAMPLE: 3H 2 (g) + N 2 (g) 2NH 3 (g) Write the equilibrium expression for the reaction above: In the above K eq expression, [NH 3 ] is raised to the power of 2 because its coefficient in the balanced equation is 2. [H 2 ] is raised to the power of 3 because its coefficient is 3, and the power of [N 2 ] is 1 because its coefficient is 1.

10 Equilibrium Only use [GASES (g)] and [AQUEOUS (aq)] compounds in K eq expressions because [SOLIDS (s)] and [LIQUIDS (l)] have no concentrations, so they do not need to be included.

11 Equilibrium EXAMPLE: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) Write the equilibrium expression for the reaction above: (the [H 2 O] is not used because it’s a liquid)

12 Equilibrium What does the calculated value for K eq tell us? K eq = 1 means that AT EQUILIBRIUM the [products] = [reactants] K eq > 1 means that AT EQUILIBRIUM the [products] > [reactants] K eq < 1 means that AT EQUILIBRIUM the [products] < [reactants]

13 Equilibrium EXAMPLE: A mixture of plenty of gaseous H 2 O and solid C are placed in a large container and allowed to come to equilibrium according to the following reaction: C(s) + H 2 O(g) CO(g) + H 2 (g) The equilibrium concentrations of the species are as follows: [H 2 O] = 0.56 M, [CO] = 0.33 M, and [H 2 ] = M. What is the value of the equilibrium constant for this reaction?

14 Equilibrium EXAMPLE: C(s) + H 2 O(g) CO(g) + H 2 (g) Does this equilibrium constant show that there are more products, more reactants, or equal amounts of the products and reactants? AND, how do you know? There are more reactants; the value of K eq is smaller than 1.

15 Equilibrium EXAMPLE: C(s) + H 2 O(g) CO(g) + H 2 (g) Let us assume that you run this reaction again in a new container with fresh water and carbon. This time, you are able to measure the carbon monoxide and hydrogen gas, only, finding [CO] = 0.15 M and [H 2 ] = M. What is the equilibrium concentration of water in this scenario?

16 Equilibrium EXAMPLE: 4 HCl(aq) + MnO 2 (s) Cl 2 (g) + 2 H 2 O(l) + MnCl 2 (aq) The equilibrium concentrations for this reaction are found to be: [HCl] = 1.20 M, [Cl 2 ] = 0.78 M, and [MnCl 2 ] = M. What is the value of the equilibrium constant for this reaction?

17 Equilibrium EXAMPLE: 4 HCl(aq) + MnO 2 (s) Cl 2 (g) + 2 H 2 O(l) + MnCl 2 (aq) Does this equilibrium constant show that there are more products, more reactants, or equal amounts of the products and reactants? AND, how do you know? There are more reactants; the value of K eq is smaller than 1.

18 Equilibrium EXAMPLE: 4 HCl(aq) + MnO 2 (s) Cl 2 (g) + 2 H 2 O(l) + MnCl 2 (aq) Let us assume that you run this reaction again in a new container with fresh chemicals. At equilibrium, you measure [HCl] = 0.15 M and [Cl 2 ] = M. What is the equilibrium concentration of manganese(II) chloride in this reaction?

19 Le Chatelier’s Principle When a system (reaction) that is at equilibrium is STRESSED, the rate of either the forward or reverse reaction must change to accommodate the new change in conditions. Therefore, the system will shift its equilibrium point in order to relieve the stress, and the concentrations of reactants and products will change.

20 Le Chatelier’s Principle Think of a teeter totter…. If it is in balance, it can teeter and totter back and forth with a nice smooth, constant motion. If weight is added to one side, that side goes down towards the ground and the other side flies upwards. In order to get back in balance, one person may have to push harder or faster, or change position. Eventually, the teeter totter will balance out and go up and down smoothly again, but each of the people on the teeter totter has had to make adjustments to get back into balance.

21 Le Chatelier’s Principle Each stress alters the collisions, which alters the rate, which shifts the reaction. Stresses that may affect a chemical reaction: concentration changes energy changes (usually temperature changes) volume changes

22 Le Chatelier’s Principle Concentration Changes: Changing the concentration of a reactant or product: If you add N 2, the system will have to use it up by making more NH 3. Therefore the system will SHIFT towards the products, or to the RIGHT If you add H 2, the system will SHIFT RIGHT If you use up N 2, the system will have to make more so it will SHIFT towards the reactants, or SHIFT LEFT

23 Le Chatelier’s Principle Concentration Changes (cont.): Changing the concentration of a reactant or product: If you use up H 2, the system will SHIFT LEFT If you add NH 3, the system will SHIFT LEFT to use the excess NH 3 up If you use up NH 3, the system will SHIFT RIGHT to produce more

24 Le Chatelier’s Principle Concentration Changes (cont.): Introducing a substance not involved in the reaction: If you add O 2 to the system… even though it is not part of the reaction, O 2 will react with H 2 to form water. This will use up some H 2, so the system will SHIFT LEFT to make up for that loss.

25 Le Chatelier’s Principle Energy Changes If HEAT is a PRODUCT: If you increase the temperature of the system, it will SHIFT LEFT to use up the excess heat. If you decrease the temperature of the system, it will SHIFT RIGHT to produce more heat

26 Le Chatelier’s Principle Energy Changes (cont.) If HEAT is a REACTANT: If you increase temperature, it will SHIFT RIGHT to use up excess heat If you decrease temperature, it will SHIFT LEFT to produce more heat

27 Le Chatelier’s Principle Pressure changes Pressure changes affect ONLY GASES!! Recall that Boyles’ Law states that if you increase the pressure on a gas, it’s volume will decrease. If you decrease the pressure on a gas, it’s volume will increase.

28 Le Chatelier’s Principle Pressure changes (cont.) You must have a balanced equation before determining equilibrium shifts based on pressure.

29 Le Chatelier’s Principle Pressure changes (cont.) Increase the pressure on the above system and it will SHIFT RIGHT WHY? because there are fewer moles of gas (less volume) on the right side. (there are 2 moles of gas on the right vs. 4 moles of gas on the left)

30 Le Chatelier’s Principle Pressure changes (cont.) Decrease the pressure on the above system and it will SHIFT LEFT WHY? because that is were the greatest volume of gas exists.

31 You Try… Determine what will happen to the equilibrium system below as each stress is added. (Describe how the rates will change.) More NH 3 (aq) is added to the system. Cu(NH 3 ) 4 2+ (aq) Blue-purple Cu 2+ (aq) + Pale blue 4 NH 3 (aq) colorless More product is added so the reaction shifts LEFT.

32 You Try… Determine what will happen to the equilibrium system below as each stress is added. (Describe how the rates will change.) Solid Cu(NO 3 ) 2 is added to the system. Cu(NH 3 ) 4 2+ (aq) Blue-purple Cu 2+ (aq) + Pale blue 4 NH 3 (aq) colorless More product (Cu 2+ ) is added so the reaction shifts LEFT.

33 You Try… Determine what will happen to the equilibrium system below as each stress is added. (Describe how the rates will change.) Solid NaOH is added to the system. [Cu OH – Cu(OH) 2 ] Cu(NH 3 ) 4 2+ (aq) Blue-purple Cu 2+ (aq) + Pale blue 4 NH 3 (aq) colorless Since Cu 2+ reacts with OH – to create Cu(OH) 2 … Less product is present so the reaction shifts RIGHT.


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