Chapter 12 Chemical Equilibrium Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Collision Theory of Reactions A chemical reaction occurs when: Collisions between molecules have sufficient energy to break the bonds in the reactants. Molecules collide with the proper orientation. Bonds between atoms of the reactants (N 2 and O 2 ) are broken and new bonds (NO) can form. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Collision Theory of Reactions A chemical reaction does not take place if the: Collisions between molecules do not have sufficient energy to break the bonds in the reactants. Molecules are not properly aligned. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings N 2 O 2

Activation Energy The activation energy is the minimum energy needed for a reaction to take place. When a collision provides energy equal to or greater than the activation energy, product can form.

Reaction Rate and Temperature Reaction rate, Is the speed at which reactant is used up. Is the speed at which product forms. Increases when temperature rises because reacting molecules move faster providing more colliding molecules with energy of activation.

Reaction Rate and Concentration Reaction rate, Is affected by the concentration of the reactants. Is increased by adding more reactants, which increases the number of collisions.

Reaction Rate and Catalysts A catalyst: Speeds up the rate of a reaction. Lowers the energy of activation. Is not used up during the reaction.

Summary of Factors and Reaction Rate

Learning Check Indicate the effect of each factor listed on the rate of the following reaction as: 1) increases 2) decreases 3) none 2CO(g) + O 2 (g) 2CO 2 (g) A. Raising the temperature B. Adding O 2 C. Adding a catalyst D. Lowering the temperature

Reversible Reactions In a reversible reaction, there is both a forward and a reverse reaction. Suppose SO 2 and O 2 are present initially. As they collide, the forward reaction begins. 2SO 2 (g) + O 2 (g) 2SO 3 (g) As SO 3 molecules form, they also collide in the reverse reaction that forms reactants. The reversible reaction is written with a double arrow. forward 2SO 2 (g) + O 2 (g) 2SO 3 (g) reverse

Chemical Equilibrium In a system at equilibrium The rate of the forward reaction becomes equal to the rate of the reverse reaction. The forward and reverse reactions continue at equal rates in both directions. There is no further change in the amounts of reactant and product.

Equilibrium A reaction of N 2 and O 2 forms NO, which reacts in the reverse direction. At equilibrium, the amounts of N 2, O 2, and NO remain constant. N 2 (g) + O 2 (g) 2NO(g)

Learning Check Complete with 1. equal 2. not equal 3. forward 4. reverse 5. changes 6. does not change A. Reactants form products in the ________ reaction. B. At equilibrium, the reactant concentration _______. C. When products form reactants, it is the _______ reaction. D. At equilibrium, the rate of the forward reaction is ______ to the rate of the reverse reaction. E. If the forward reaction is faster than the reverse, the amount of products ________.

Equilibrium Constants For the following reaction: aA bB The equilibrium constant expression, K c, gives the concentrations of the reactants and products at equilibrium. K c = [B] b [A] a The square brackets indicate the moles/liter of each substance. The coefficients b and a are written as superscripts that raise the moles/liter to a specific power.

Writing a K c Expression In the K c expression for the following reaction at equilibrium: 2CO(g) + O 2 (g) 2CO 2 (g) STEP 1 The products are shown in the numerator and the reactants are shown in the denominator. K c = [CO 2 ] [products] STEP 2 [CO] [O 2 ] [reactants] The coefficients are written as superscripts. STEP 3 K c = [CO 2 ] 2 [CO] 2 [O 2 ]

Learning Check The K c expression for the following reaction is: N 2 (g) +3Cl 2 (g) 2NCl 3 (g) 1) [NCl 3 ]2) [N 2 ][Cl 2 ] 3 [N 2 ][Cl 2 ] [NCl 3 ] 2 3) [NCl 3 ] 2 4) [NCl 3 ] 2 [N 2 ] 3 [Cl 2 ] [N 2 ][Cl 2 ] 3

Heterogeneous Equilibrium In heterogeneous equilibrium, Solid and/or liquid states may be part of a reaction. 2NaHCO 3 (s) Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g) The concentration of solids and/or liquids is constant. The K c expression is written with only the gases. K c = [CO 2 ][H 2 O]

Calculating Equilibrium Constants When the following reaction, H 2 (g) + I 2 (g) 2HI(g) reaches equilibrium, the concentrations are: H mol/L, I mol/L, and HI 0.35 mol/L. What is the K c ? STEP 1 Write the K c expression K c = [HI] 2 [H 2 ][I 2 ] STEP 2 Substitute equilibrium concentrations. = (0.35) 2 = 8.5 x (1.2)(1.2)

Reaching Chemical Equilibrium A container filled with SO 2 and O 2 or only SO 3, Eventually contains mostly SO 3 and small amounts of O 2 and SO 3. Reaches equilibrium in both situations.

Equilibrium Can Favor Product If equilibrium is reached after most of the forward reaction has occurred, The system favors the product.

Equilibrium with a Large K c At equilibrium, A reaction with a large K c produces a large amount of product. Very little of the reactants remain. K c =[NCl 3 ] 2 = 3.2 x [N 2 ][Cl 2 ] 3 A large K c favors the products. N 2 (g) + 3Cl 2 (g) 2NCl 3 (g) When this reaction reaches equilibrium, it will essentially consist of the product NCl 3.

Equilibrium Can Favor Reactant If equilibrium is reached when very little of the forward reaction has occurred, The reaction favors the reactants.

Equilibrium with a Small K c At equilibrium, A reaction that produces only a small amount of product has a small K c. K c = [NO] 2 =2.3 x [N 2 ] [O 2 ] A small K c favors the reactants. N 2 (g) + O 2 (g) 2NO(g)

Learning Check For each K c, indicate whether the reaction at equilibrium contains mostly: (1) reactants or (2) products __A. H 2 (g) + F 2 (g) 2HF(g) K c = 1 x __B. 3O 2 (g) 2O 3 (g) K c = 1.8 x 10 -7

Using K c to Solve for Equilibrium Concentration At equilibrium, the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) has a K c of 4.2 x and contains [PCl 3 ] = [Cl 2 ] = 0.10 M. What is the equilibrium concentration of PCl 5 ?

LeChâtelier’s Principle LeChâtelier’s principle states that: Any change in equilibrium conditions upsets the equilibrium of the system. A system at equilibrium under stress will shift to relieve the stress. There will be a change in the rate of the forward or reverse reaction to return the system to equilibrium.

Summary of Changes on Equilibrium

Condition Changes and Equilibrium X X

Learning Check Indicate if each change on a reaction at equilibrium shifts 2NO 2 ( g) + heat 2NO(g) + O 2 (g) 1) toward products2) toward reactants 3) no change A. Adding NO B. Adding N 2 C. Raising the temperature D. Removing O 2 E. Increasing the volume

Saturated Solution A saturated solution: Contains the maximum amount of dissolved solute. Contains solid solute. Is an equilibrium system. rate of dissolving = rate of recrystallization solidions in solution

Solubility Product Constant The solubility product for a saturated solution: Gives the ion concentrations, which are constant at constant temperature. Is expressed as a K sp. Does not include the solid, which is constant. With Fe(OH) 2 (s) Fe 2+ (aq) + 2OH − (aq) IsK sp = [Fe 2+ ] [OH − ] 2

Learning Check Write the K sp expression for each of the following: A. FeS(s) B. Ag 2 CO 3 (s)

Calculating Solubility Product Constant Calculate the K sp of PbSO 4 with a solubility of 1.4 x M. STEP 1 PbSO 4 (s) Pb 2+ (aq) + SO 4 2− (aq) STEP 2 K sp = [Pb 2+ ][SO 4 2− ] STEP 3 = (1.4 x ) x (1.4 x ) = 2.0 x 10 -8

Solubility Product Constant