Presentation on theme: "CHAPTER 14 CHEMICAL EQUILIBRIUM. I. Chemical Equilibrium A. Definition / Explanation When a chemical reaction occurs, the conc. of reactants decrease."— Presentation transcript:
CHAPTER 14 CHEMICAL EQUILIBRIUM
I. Chemical Equilibrium A. Definition / Explanation When a chemical reaction occurs, the conc. of reactants decrease and the conc. of products increase. Eventually chemical equilibrium is reached, in which the concentration of the products and reactants remain constant. “No net change in reaction mixture”. This does not mean [reactants] = [products] !!!
B. Achieving Equilibrium 1. Equilibrium is not reached instantaneously. How fast or slow determined by kinetics. 2.How much product and reactants are present at equilibrium is determined by thermodynamics (energetics of reaction). What is meant by “product-favored” reaction? What is meant by “reactant-favored” reaction?
C. Dynamic Equilibrium Chemical reaction reach dynamic equilibrium. State of a reaction in which its forward and reverse reactions occur at equal rates so that the conc. of the reactants and products do not change with time. ** Reactants and products formed at same rate. ** Does not mean [reactants] = [products].
D. Direction of Approach Dynamic equilibrium is independent of the direction of approach. For a given temperature, the equilibrium state will be the same whether the reaction is approached from starting materials of pure reactants or pure products. Consider rxn: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Note following diagram Explain idea
E. Catalysts Do No Effect Equilibrium Conc. They effect rate, but not equilibrium state !!!
Regardless of the starting concentrations; once equilibrium is reached … … the expression with products in numerator, reactants in denominator, where each concentration is raised to the power of its coefficient, appears to give a constant. How Can We Quantitatively Describe Equilibrium?
II. Equilibrium Constant A. Equilibrium Expression “Mass Action Expression” For rxn: aA + bB cC + dD ** No units ** Exponents always = reaction coefficients
B. Species Not in Equilibrium Expression 1. Pure solids – (NaCl(s), Na(s) 2. Pure liquids - H 2 O(l) 3. Solvents - may be involved in reaction, but its concentration does not change. (dilute solutions) **General Rule – only include substances that are (g) or (aq)
Write equilibrium expressions for the following reactions: 1. CO(g) + 3 H 2 (g) CH 4 (g) + H 2 O(g) 2.H 2 (g) + I 2 (g) 2 HI(g) 3.AgCl(s) Ag + (aq) + Cl - (aq) 4. NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq)
C. K c for Related Reactions 1. Changing Stoichiometric Coefficients N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K c = 3.5 x 10 8 What if we alter the coefficients in equation? 2 N 2 (g) + 6 H 2 (g) 4 NH 3 (g) K’ c = ? When the coefficients of an equation are multiplied by a common factor “n”, we raise the original K c value to the power “n” to obtain the new equilibrium constant.
2. Changing Direction of Reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K c = 3.5 x 10 8 What if we consider the reverse reaction? 2 NH 3 (g) N 2 (g) + 3 H 2 (g) K’ c = ? When we reverse the equation for a chemical reaction for which the equilibrium constant is K c, we invert the equilibrium constant. The reverse reaction has the equilibrium constant 1 / K c.
D. K c For Combining Reactions A + B C K c1 = 2.0 x 10 -4 C + B F K c2 = 1.0 x 10 -6 What is the K c for the reaction? A + 2 B F K c = ? When we add the equations for individual reactions to obtain an overall equation, we multiply their equilibrium constants to obtain the equilibrium constant for the overall reaction.
E. K c For Gas Phase Reactions If all reactants and products are in gas phase, the concentration terms are replaced by partial pressures. (Derived from Ideal Gas Law, PV=nRT ) When Solving, Units Must Be in atm !!!!! What is the K p expression for the rxn: 2 H 2 (g) + O 2 (g) 2 H 2 O(g)
What is the relationship between K p and K c? Δn = (sum of moles of gas products) – (sum of moles of gas reactants) As determined from coefficients in chemical reaction.
F. Types of Equilibrium Expressions K c Type Reference Table K c general rxns K a acid Acid Table K b baseBase Table K w water K sp slight Solubility Product solubility Table
III. Meaning of K c K c ValueReaction very largeproduct favored K c > 1Larger K c : more product favored very smallreactant favored K c < 1Smaller K c : more reactant favored K c = 1significant amounts of both reactants and products
1. If K c = 1, does that mean that the concentration of products equal the concentrations of reactants? 2. Which reaction has the greatest tendency to go to completion (product favored)? a. 2 NO 2 (g) N 2 O 4 (g) K c = 1.7 x 10 2 b.N 2 (g) + O 2 (g) 2 NO(g) K c = 1.7 x 10 -3
IV. Equilibrium Problems A. Solving For Equilibrium Constant 1. Guidelines a. Write equilibrium expression for reaction. b. Construct an “ICE” table with the Initial, Change, and Equilibrium concentrations of all reactants and products.
When determining concentrations for table: 1) Units of conc. = Molarity = mol / L (may use pressures if all gases and K p ) 2) You may need to use “x” to represent unknown changes in concentrations in table. 3) Remember changes in concentration of reactants and products occur in same ratio as equation coefficients. c. Always plug in equilibrium concentrations into the equilibrium expression to solve the problem.
2. Problems – Solving for K c Value 1. At equilibrium, a 2.0 L flask is found to contain 0.10 mol of CO, 0.20 mol of CO 2, and 4.8 g of carbon. C(s) + CO 2 (g) 2 CO(g) What is K c for this reaction?
2. A student placed 1.00 mol of CO and 2.00 mol of H 2 in a 2.00 L flask at 780 0 C. After reaching equilibrium, the CO concentration is found to be 0.150 M. Determine the value of the equilibrium constant for this reaction. CO(g) + 2 H 2 (g) CH 3 OH (g)
Predicting Direction of a Reaction 1. Reaction Quotient, Q “Q” has the same form as K c expression, but uses the actual concentrations in the reaction mixture. “Q” is not a constant. 2. Significance of Q By comparing the Reaction Quotient to the K c value, one can determine the direction a reaction will occur as it proceeds towards equilibrium.
If Q > K c [Product] greater than would be at equilibrium Reaction would proceed to the left. If Q < K c [Product] less than would be at equilibrium Reaction would proceed to the right. If Q = K c Reaction is at equilibrium. Dynamic equilibrium. No net change occurs.
Example Problem A 50.0 L reaction vessel contains 1.00 mol N 2, 3.00 mol H 2, and 0.500 mol NH 3. Will more ammonia, NH 3, be formed or will it dissociate when the mixture goes to equilibrium at 400 o C? The reaction is: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K c = 0.500
B. Solving for Equilibrium Concentrations 1. Guidelines a. Follow same guidelines as with K c calculations. b. Additional Comments 1) Some equilibrium concentrations may contain a number (n) plus or minus “x”, (i.e., n+x, n-x). 2) (When K c is very small or large, 10 -5 > K c > 10 5 ), “x” may be very small compared to “n” such that simplification can be made to problem. n ± x ≈n. 3) Verify assumption once “x” is determined to see that it is less than (± 5%) of “n”. 5 % RULE 4) If assumption is not true, you will need to solve problem using the quadratic equation (see later).
2. Problems 1. Problem Solved Since Perfect Square (or Quadratic Equation) A 0.20 mol sample of H 2 and a 0.20 mol sample of I 2 were placed into a 1.0 L flask. What were the equilibrium concentrations of H 2, I 2, and HI? Reaction temperature was 500K. H 2 (g) + I 2 (g) 2 HI(g) K c = 49.5
Another Example Problem: Given the equilibrium reaction at 745 K: H 2 (g) + I 2 (g) 2 HI(g) K c = 50.0. Suppose that 0.75 mol HI(g), 0.025 mol H 2 (g), and 0.025 mol I 2 (g) are placed into a 20.0 L flask and heated to 745 K. Calculate the equilibrium concentrations of all three substances.
2. Problems Solved Using Assumptions (or Quadratic Equation) A 2.0 L flask contained initial concentrations of 0.0330 M N 2 and 0.00810 M O 2. What is the equilibrium concentration of NO in the flask? N 2 (g) + O 2 (g) 2 NO(g) K c = 4.8 x 10 -31
3. Problem Solved Using Quadratic Equation (can only solve by using Quadratic Equation) Consider the following reaction at 448 0 C: I 2 (g) + Cl 2 (g) 2 ICl(g) K p = 50.5 The reaction mixture initially contains pressure of I 2 = 60.0 atm and pressure of Cl 2 = 120.0 atm. Determine the equilibrium pressure of ICl(g).
VI. Shifting Chemical Equilibriums A. Le Chatelier’s Principle 1. Definition – When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaning a new equilibrium condition that minimizes the impact of the imposed change. ** Reaction tries to maintain equilibrium state. ** Significance
2. Factors (Stress) Affecting Equilibrium a. Changing Concentrations Reaction will try to maintain equilibrium or maintain a similar K c value. A + B C + D Adding reactants: shifts equilibrium to right Adding products: shifts equilibrium to left Removing reactants? Removing products? Will solids or pure liquids affect equilibrium? Significance
Le Chatelier’s Principle Related to “Q” and “K c ”
b. Changing Pressure in Gas Equilibria Pressure increases (or volume decreases) for gaseous equilibrium cause shift in equilibrium. Think about it like a concentration factor. An increase in pressure (or decrease in volume) always drives the reaction in the direction of the fewer # of gas molecules. How will increase in pressure affect equilibrium? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H 2 (g) + I 2 (g) 2 HI(g)
Initial When pressure is increased … … two molecules of NO 2 combine … … to give one molecule of N 2 O 4, reducing the pressure increase. N 2 O 4 (g) 2 NO 2 (g)
c. Changes in Temperature When considering exothermic or endothermic reactions, consider heat as a product or reactant. (Use same principles as changes in concentration.) Add heat to exothermic: shifts equilibrium to left Add heat to endothermic: shifts right N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ΔH = -92.2 kJ N 2 (g) + 3 H 2 (g) 2 NH 3 (g) (+ heat on which side??)
B. Example Problem How will the following factors affect the final equilibrium amount of Cl 2 ? PCl 5 (g) PCl 3 (g) + Cl 2 (g) ΔH = 21 kcal/mol a. Add PCl 3 b. Remove PCl 5 c. Increase temperature d. Decrease volume e. Decrease pressure f. Adding inert gas to mixture