Presentation on theme: "1 Chemical Equilibrium Chapter 15 AP CHEMISTRY s/ap%20ch13.ppt."— Presentation transcript:
1 Chemical Equilibrium Chapter 15 AP CHEMISTRY s/ap%20ch13.ppt
2 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
3 Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2 : N 2 O 4 (g) 2NO 2 (g). At some time, the color stops changing and we have a mixture of N 2 O 4 and NO 2. Chemical equilibrium is the point at which the concentrations of all species are constant. The Concept of Equilibrium
4 NO 2 - N 2 O 4 Demonstration
5 The Concept of Equilibrium
6 As the substance warms it begins to decompose: N 2 O 4 (g) 2NO 2 (g) A mixture of N 2 O 4 (initially present) and NO 2 (initially formed) appears brown. When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g) N 2 O 4 (g). At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re- form N 2 O 4 : The double arrow implies the process is dynamic.
7 Notes on Equilibrium Expressions (EE) K does not include any pure solids or liquids The expression shows products divided by reactants Like the rate constant, k, the units of K depend on the experiment being performed For the reverse reaction K = 1/K (reactants and products switch)
8 Notes on Equilibrium Expressions (EE) For a reaction multiplied by an integer, n, K new = (K orig ) n For a given reaction, K is dependent only on temperature
10 Heterogeneous Equilibria... are equilibria that involve more than one phase. CaCO 3 (s) CaO(s) + CO 2 (g) K = [CO 2 ] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.
11 Practice Problem: Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) In trial #1, you start with 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq), and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value of the equilibrium constant for this reaction? Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Initial Change Equilibrium
12 Practice Problem Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Equilibrium
13 Practice Problem #2: Using the previous reaction: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) and the K value we determined: K = 0.33 determine if the following concentrations are at equilibrium: Initial:10.0 M Fe 3+ (aq), 8.00 M SCN - (aq), and 2.00 M FeSCN 2-
14 Reaction Quotient H 2 (g) + F 2 (g) 2HF(g) After the equilibrium constant (K) is known, we can use it to determine if a reaction is at equilibrium The reaction quotient, Q, has the same form as the equilibrium constant expression EXCEPT initial concentrations are used instead of equilibrium concentrations
15 Predicting the Direction of a Reaction Using Reaction Quotient If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K). If Q < K then the forward reaction must occur to reach equilibrium.
16 Solving Equilibrium Problems 1.Balance the equation. 2.Write the equilibrium expression. 3.List the initial concentrations. 4.Calculate Q and determine the shift to equilibrium.
17 Solving Equilibrium Problems (continued ) 5.Define the change needed to reach equilibrium. 6.Substitute equilibrium concentrations into equilibrium expression and solve. 7.Check calculated concentrations by calculating K.
18 K vs. K p For any reaction: K p = K(RT) n n = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.
19 Le Châteliers Principle... if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.
20 Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Can this be predicted? Le Châteliers Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance. Le Châteliers Principle
21 Le Châteliers Principle
22 Effects of Changes on the System 1.Concentration: The system will shift away from the added component. 2.Temperature: K will change depending upon the temperature (treat the energy change as a reactant).
23 Increase of Pressure to an Equilibrium.
24 Effects of Changes on the System (continued) 3.Pressure: a. Addition of inert gas does not affect the equilibrium position. b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.
25 Change in Reactant or Product Concentrations Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases. Le Châteliers Principle
26 Change in Reactant or Product Concentrations Le Châteliers Principle
27 The Haber Process Le Châteliers Principle
28 N 2 and H 2 are pumped into a chamber. The pre-heated gases are passed through a heating coil to the catalyst bed. The catalyst bed is kept at C under high pressure. The product gas stream (containing N 2, H 2 and NH 3 ) is passed over a cooler to a refrigeration unit. In the refrigeration unit, ammonia liquefies but not N 2 or H 2. The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized The Haber Process for producing NH 3 Le Châteliers Principle