Chapter 13 Equilibrium

Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time –The concentration of each species does NOT have to be the same The forward and reverse reactions happen at the same rate Relative amounts follow stoichiometry

Chemical Equilibrium Is….. Dynamic –Reactions are always taking place –Reactant turns into product Forward Reaction –Product turns back into reactant Reverse Reaction

Progress Toward Equilibrium Reactions don’t start at equilibrium In the beginning there are (usually) only reactants –No reverse reaction As the reaction proceeds product molecules are produced and the reverse reaction begins Eventually equilibrium is reached

N 2 + 2H 2  2NH 3 (Fig 13.5)

Equilibrium Constant Expressions A mathematical equation that relates the concentration of products to the concentration of reactants for a reaction at equilibrium The equilibrium constant is Keq K for Konstant, eq for equilibrium General form of the expression is concentration of products over concentration of reactants –Pressure can also be used

The Equilibrium Constant Expression For the reaction jA + kB  lC + mD Where j, k, l, & m are coefficients and A, B, C, and D are chemicals The equilibrium constant expression is

Equilibrium Constant Expressions A species in brackets indicates concentration Only solutions and gases are included in equilibrium expressions If you have a solid or liquid leave it out –Just think of it as a 1 in the expression

The Equilibrium Constant A value that relates the position of a reaction at equilibrium Calculated by inserting values into the equilibrium constant expression Keq does NOT have units Keq > 1 –Product favored Keq <1 –Reactant favored

Things To Consider The equilibrium constant is only constant for a given temperature –If the temp changes so does the constant There are many sets of conditions that will satisfy the equilibrium constant at a given temperature –For example if K=4 = (4/1) = (16/4) = (1/.25)

Other Fun If a reaction is reversed, what happens to K? K’ = 1/K If a reaction is multiplied through by a coefficient what happens to K K’ = K n

Given the reaction 2PBr 3(g) + 3Cl 2(g)  2PCl 3(g) + 3Br 2(g) A)Write the equilibrium constant expression B)At equilibrium the concentrations are PBr 3 =2.0M Cl 2 =3.0M PCl 3 =.10M Br 2 =.15M Calculate the equilibrium constant C)Calculate the equilibrium constant if the above reaction is reversed and the concentrations are the same D)Calculate the equilibrium constant if the above reaction is divided by 2.

Homework Page 645 #’s 15,17,19ab,21ab,24,26

Equilibrium Involving Pressure Pressure of gases can be used in equilibrium expressions too. Since PV=nRT Then n/V = Concentration So P = CRT Pressure of a gas is directly related to concentration When a gas pressure is used Keq is Kp

Kp Expression For the reaction jA + kB  lC + mD Where j, k, l, & m are coefficients and A, B, C, and D are chemicals in the gas phase The equilibrium constant expression is

Write the Kp and Kc expressions for the reaction PCl 5(s)  PCl 3(g) +Cl 2(g)

Kp to Kc Kp and Kc can be determined from each other. Kp = Kc(RT) Δn Where n is (l+m) – (j+k) Sum of product coefficients minus sum of reactant coefficients

Is it equilibrium? Sometimes there are lots of chemicals in a container. Need to check if it is at equilibrium. Use reaction quotient, Q Found same way as K but has a different meaning Q = K Reaction at equilibrium Q < K Needs more product shifts right Q > K Needs more reactants shifts left

Solving Equilibrium Problems Write the equation Write equilibrium expression List what is given Determine what needs to be solved Set up a solution

#40 page 647 2NO (g) + Br 2(g)  2NOBr (g)

#43 page 647 2SO 3(g)  2SO 2(g) + O 2(g)

#45 page 647 SO 2(g) NO 2(g)  SO 3(g) + NO (g)

Homework Page 646 #’s 28,30,35,39,44,46

#50 page 647 N 2 O 4(g)  2NO 2(g) K p =.25 Initial pressure of N 2 O 4 is 4.5 atm Initial pressure of NO 2 is 0 atm

#50 page 647 N 2 O 4(g)  2NO 2(g) K c =4.0x10 -7 Initial [N 2 O 4 ] of is 0.10M Initial [NO 2 ] is 0M

My Equilibrium is Stressed A reaction at equilibrium can be stressed out by changing reaction conditions –There are three stresses for reactions Concentration Pressure Temperature

LeChatlier’s Principle When a stress is applied to a reaction at equilibrium the reaction will shift to relieve the stress placed upon it.

Concentration A reaction is stressed when the concentration of a reactant or product is increased OR decreased –Add a chemical to increase –Remove a chemical to decrease Reactions shift away from increased concentrations Reactions shift toward decreased concentrations

Example Consider the reaction 2SO 2(g) + O 2(g)  2SO 3(g) How will the reaction shift with the following –Increase the amount of SO 2 Right –Increase the amount of SO 3 Left –Decrease the amount of O 2 Left

Pressure Pressure can be increased By decreasing volume Pressure can be decreased By increasing volume Increased pressure shifts reactions away from side with more gas molecules Decreased pressure shifts reactions toward side with more gas molecules

Example Consider the reaction 2SO 2(g) + O 2(g)  2SO 3(g) How will the reaction shift with the following –Increase the pressure in the container Right –Decrease the pressure in the container Left

Temperature Direction of shift depends of whether the reaction is exothermic or endothermic –How do you know Exothermic reactions release heat –It is a product Endothermic reactions absorb heat –It is a reactant

What Type of Thermic Is It? Exothermic A + B  C + 49kJ A + B  C ΔH = -49kJ ΔH means enthalpy (energy change) Endothermic A + B + 49kJ  C A + B  C ΔH = 49kJ

Temperature Change Think of the temp as a concentration of heat –Increasing the temp. will shift the reaction away from the heat –Decreasing the temp. will shift the reaction toward the heat. Same as concentration shift Change in temp. is the only stress that changes the equilibrium constant.

Example Consider the reaction 2SO 2(g) + O 2(g)  2SO 3(g) + 197kJ How will the reaction shift with the following –Increase the temperature Left –Decrease the temperature Right

Homework Page 647 #’s 47,49,54,59,62