Equilibrium Chapter 16

Reversible Reactions – A chemical reaction in which the products can regenerate the original reactants. Reversible Reactions – A chemical reaction in which the products can regenerate the original reactants. Forward reaction – the formation of products from the reactants Forward reaction – the formation of products from the reactants Smog – 2NO 2  N 2 O 4 as N 2 O 4 is being formed in one direction, NO 2 is being formed in the other… Smog – 2NO 2  N 2 O 4 as N 2 O 4 is being formed in one direction, NO 2 is being formed in the other… NO 2 to N 2 O 4 is the forward reaction NO 2 to N 2 O 4 is the forward reaction N 2 O 4 to NO 2 is reverse reaction N 2 O 4 to NO 2 is reverse reaction Both are reactants and both are products! Both are reactants and both are products!

Chemical Equilibrium – the rate at which the forward reaction is equal to the reverse reaction Chemical Equilibrium – the rate at which the forward reaction is equal to the reverse reaction It is the state in which the concentrations of reactants and products remain constant It is the state in which the concentrations of reactants and products remain constant rate they are formed in each reaction equals the rate they are consumed in the opposite reaction. rate they are formed in each reaction equals the rate they are consumed in the opposite reaction. ***At equilibrium the reaction does NOT stop**** ***At equilibrium the reaction does NOT stop****

Factors affecting reaction rates Concentrations – when the concentrations of the substances reacting are higher, the reaction is faster. Concentrations – when the concentrations of the substances reacting are higher, the reaction is faster. Ex. Zn + 2HCl  ZnCl 2 + H 2 (1.0m) Ex. Zn + 2HCl  ZnCl 2 + H 2 (1.0m) Ex. Zn + 2HCl  ZnCl 2 + H 2 (2.0m) Faster Ex. Zn + 2HCl  ZnCl 2 + H 2 (2.0m) Faster Temperature – higher temp. speed up endothermic reactions. Temperature – higher temp. speed up endothermic reactions.

16-2 The law of chemical equilibrium Equilibrium constant – equal to the ratio of the product concentrations (raised to the power indicated by the coefficients) to the reactant concentrations (raised to the power of the coefficients) Equilibrium constant – equal to the ratio of the product concentrations (raised to the power indicated by the coefficients) to the reactant concentrations (raised to the power of the coefficients) Law of Chemical Equilibrium – Every reversible reaction proceeds to an equilibrium state. Law of Chemical Equilibrium – Every reversible reaction proceeds to an equilibrium state.

aA + bB  cC + dD aA + bB  cC + dD Keq = [C] c [D] d [products] coefficients Keq = [C] c [D] d [products] coefficients [A] a [B] b [reactants] coefficients [A] a [B] b [reactants] coefficients **When writing K eq only use the elements in the GAS state** **When writing K eq only use the elements in the GAS state** [ ] mean concentration [ ] mean concentration

Keq is a measure of the extent to which a reaction proceeds to completion. Keq is a measure of the extent to which a reaction proceeds to completion. If Keq is >>1, a lot of the products were formed. (remember products over reactants) If Keq is >>1, a lot of the products were formed. (remember products over reactants) If Keq is <<1, the reaction did not really get going. If Keq is <<1, the reaction did not really get going. If Keq = 1, the reaction is at equilibrium when the products and reactants are equal If Keq = 1, the reaction is at equilibrium when the products and reactants are equal

Homogeneous equilibria – equilibrium conditions that deal with one state of matter. Homogeneous equilibria – equilibrium conditions that deal with one state of matter. Heterogeneous equilibria – equilibrium conditions that deal with more than one state of matter. Heterogeneous equilibria – equilibrium conditions that deal with more than one state of matter.

We only use gases because liquids and solids concentrations do not significantly change We only use gases because liquids and solids concentrations do not significantly change The Reaction Quotient..(Q) is used to determine if a reaction is at equilibrium. The Reaction Quotient..(Q) is used to determine if a reaction is at equilibrium. Solve it the same way you solve Keq but put in the numbers. Solve it the same way you solve Keq but put in the numbers.

After you solve for Q, compare the Keq value with the Q value… After you solve for Q, compare the Keq value with the Q value… If Q > Keq The reaction will proceed to the left If Q > Keq The reaction will proceed to the left (Because the denominator of the Q is too small and the numerator is too large IE there’s too many products and not enough reactants) (Because the denominator of the Q is too small and the numerator is too large IE there’s too many products and not enough reactants) If Q < Keq the reaction will proceed to the Right If Q < Keq the reaction will proceed to the Right (because the denominator is too large and the numerator is too small… There are too many reactants so more products need to be formed to reach equilibrium (because the denominator is too large and the numerator is too small… There are too many reactants so more products need to be formed to reach equilibrium If Q = Keq the system is at equilibrium. There is no shift. If Q = Keq the system is at equilibrium. There is no shift.

Consider the reaction: COCl 2   CO + Cl 2 Keq =170 Consider the reaction: COCl 2   CO + Cl 2 Keq =170 If the concentrations of CO and Cl 2 are each 0.15M and the concentration of COCl 2 is 1.1 x M, is the reaction at equilibrium? If not, in which direction will it proceed? If the concentrations of CO and Cl 2 are each 0.15M and the concentration of COCl 2 is 1.1 x M, is the reaction at equilibrium? If not, in which direction will it proceed?

1st write the Keq 1st write the Keq COCl 2 (g)  > CO (g) + Cl 2 (g) Keq = 170 COCl 2 (g)  > CO (g) + Cl 2 (g) Keq = 170 Keq= [CO][Cl 2 ] = Q = (0.15)(0.15) = 20 Keq= [CO][Cl 2 ] = Q = (0.15)(0.15) = 20 [COCl 2 ] 1.1 x [COCl 2 ] 1.1 x Q < Keq Q < Keq The reaction proceeds to the right because Q < Keq The reaction proceeds to the right because Q < Keq

16-3 Le Chatelier’s Principle When a stress is applied to a reaction at equilibrium, the reaction will shift to relieve that stress. When a stress is applied to a reaction at equilibrium, the reaction will shift to relieve that stress.

Stresses 1. Changing the amounts of reactants or products 1. Changing the amounts of reactants or products If you add more reactant, shift to right to use up reactant If you add more reactant, shift to right to use up reactant If you add more product, shift to left to use up product If you add more product, shift to left to use up product

Stresses 2. Change in pressure 2. Change in pressure If the pressure is increased, the reaction will shift to the side with less moles. If the pressure is increased, the reaction will shift to the side with less moles. Ex. 2NO 2  N 2 O 4 Ex. 2NO 2  N 2 O 4 The reaction will proceed to the right because there are half the amount of moles The reaction will proceed to the right because there are half the amount of moles

Change in temp. Reversible reactions are endothermic one way and exothermic the other. Reversible reactions are endothermic one way and exothermic the other. If you add heat, and the reaction is exothermic, the reaction will go toward completion If you add heat, and the reaction is exothermic, the reaction will go toward completion If you add heat, and the reaction is endothermic, the reaction will go toward the reactants. If you add heat, and the reaction is endothermic, the reaction will go toward the reactants.