Intro to equilibrium abbrev alg

General Chemistry II CHEM 152 Unit 2 Week 5

Week 5 Reading Assignment Chapter 14 – Sections 14.2 (dynamic equilibrium), 14.3 (equilibrium constant), 14.5 (heterogeneous equilibria)

CHEMICAL EQUILIBRIUM How far does the reaction go? What is the final concentration of reactants and products? We have seen that when the rate forward equals the rate backwards in a chemical reaction, the system reaches a state of equilibrium. But, These are the types of questions we now want to answer.

Chemical Equilibrium Consider this reaction: If NO 2 is reddish brown and N 2 O 4 is colorless: What is happening here? What properties are changing? What is happening over time? After a long time?

The final state depends on: The chemical nature of the reactants and products The conditions of the system (temperature, pressure, volume). Get time progression Check silberberg N 2 O 4  2NO 2 Low T High T

Analyzing Reaction Progress Product-or Reactant Favored Process? If we consider the two possible chemical processes in this system: A  B k 1 = exp(-E a1 /RT) Rate forward = k 1 [A] B  A k 2 = exp(-E a2 /RT) Rate reverse = k 2 [B] Reaction progress A B E a1 E a2 .

Analyzing Reaction Progress What happens with Rate forward and Rate reverse as a function of time? When does the reaction “stop”? A  B k 1 =exp(-E a 1 /RT) Rate forward = k 1 [A] B  A k 2 =exp(-E a 2 /RT) Rate reverse = k 2 [B] TWO 1 st order reactions in OPPOSITE directions with DIFFERENT size k’s

Analyzing Reaction Progress A  B k 1 = exp(-E a1 /RT) Rate forward = k 1 [A] B  A k 2 = exp(-E a2 /RT) Rate reverse = k 2 [B] The system reaches a state of chemical equilibrium when: Rate forward =Rate reverse or k 1 [A]=k 2 [B] k 1 /k 2 =[B]/[A] = K eq (Equilibrium Constant) What does it mean?

Announcements Exams are graded will be returned in lab this week Will post powerpoints this afternoon Quiz on Friday Mechanisms and equilibrium

Chemical Equilibrium Let’s try the following system: A 100 B Probability A  B 40 % B  A 10 % Analyze the time evolution of this system How many particles of each species do we have at chemical equilibrium?

Chemical Equilibrium A 100 B A  B 40 % B  A 10 % Equilibrium? A* 0.4 = B* 0.1 A + B = 100 A = 20 B = 80 K eq = 80/20 = 4 = 0.4/0.1 The reaction keeps going but [A] and [B] remain constant. Time A B 0 100 0 A  B 40  B  A  0 1 60 40 A  B 24  B  A  4 2 40 60 A  B 16  B  A  6 3 30 70 A  B 12  B  A  7 4 25 75 Etc.

Over time the concentrations reach levels where they do NOT change This is the equilibrium condition N 2 O 4 2NO 2

Let’s Simulate: A  B What is K eq = [B]/[A] at equilibrium? Does the K eq depend on the initial values of [A] and [B]?

Conclusion? No matter what the starting concentrations at equilibrium. The ratio [B] eq /[A] eq is a constant  K eq EQUILIBRIUM CONSTANT A  B

Another Reaction N 2 O 4  2NO 2 What is the value of [NO 2 ] eq / [N 2 O 4 ] eq ? Does this value remain constant as we change the initial concentrations?

Another Reaction N 2 O 4  2NO 2 Let’s try again with [NO 2 ] 2 eq / [N 2 O 4 ] eq ? Does this value remain constant as we change the initial concentrations?

The Equilibrium Constant For a reaction of the type a A + b B c C + d D the following is a CONSTANT (at a given T) If K eq is known, then we can PREDICT concentrations of products or reactants

Your Turn . Equilibrium mix What is the value of K eq ?

K eq ~ 9 2 /(1 x 1) = 81 Your Turn

Your Turn . . Equilibrium mix What is the value of K eq ?

Your Turn . Equilibrium mix K eq ~ 2 2 /(6 x 4) = 0.17

Chemical Equilibrium Some things to keep in mind about chemical equilibrium: DYNAMIC (in constant motion); REVERSIBLE; Can be approached from either direction; After a period of time, the concentrations of reactants and products are constant; The forward and reverse reactions continue after equilibrium is attained; Catalysts do not affect equilibrium concentrations (Why?)

Let us consider the conversion of cis (A) to trans (B) butene: Effect of Catalysts A Catalyst lowers BOTH the forward and reverse reaction activation energies proportionally. So, both rates change but NOT the equilibrium distributions. Reaction progress E N E R G Y A B E a 1 E a 2 .

Adding more C(s) will NOT change the other concentrations . Does adding more solid carbon change the equilibrium? Heterogeneous Equilibrium WHY?

What happens if you add sugar to hot water? Will the sugar continue to dissolve indefinitely? Once the sugar can no longer dissolve will adding more sugar help dissolve additional sugar? The solid sugar and dissolved sugar are in equilibrium. The additional sugar has no effect on the equilibrium Solids and pure liquids do not change their “concentration” at equilibrium and there are not included in the K eq expression for a “HETEROGENEOUS EQUILIBRIUM” Heterogeneous Equilibrium

Heterogeneous Equilibrium What is the equilibrium constant for this process? How would you expect K eq to change if more H 2 , CH 4 , or C are added?

Your Turn Write the Equilibrium constant for the process: S(s) + O 2 (g) SO 2 (g) Solids and liquids NEVER appear in equilibrium expressions ( their concentration is constant during the process ).

Your Turn Write the equilibrium constant for the following process: NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH - (aq)

Manipulating K eq Consider the equilibrium: N 2 (g) + 3H 2 (g) 2NH 3 (g) K eq = 3.5 x 10 8 (25 o C) What would be the value of the equilibrium constant for these processes? 1/2 N 2 (g) + 3/2 H 2 (g) NH 3 (g) 2NH 3 (g) N 2 (g) + 3H 2 (g) If the chemical equation is multiplied by n, K c  K c n If the chemical equation is reversed, K c  1/K c

Stepwise Equilibrium (1.) N 2(g) + O 2(g)  2NO (g) (2.) 2NO (g) + O 2(g)  2NO 2(g) Manipulating K eq [NO] 2 K eq,1 = [N 2 ][O 2 ] [NO 2 ] 2 K eq,2 = [NO] 2 [O 2 ] [NO] 2 [NO 2 ] 2 K eq =  = K eq,1  K eq,2 [N 2 ][O 2 ] [NO] 2 [O 2 ] Determine the chemical equation for the overall process What is the equilibrium constant for the overall process? How does it depend on the value of Kc 1 and Kc 2 ?

Relevant Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO 3 (s) + H 2 O(liq) + CO 2 (g) Ca 2+ (aq) + 2 HCO 3  (aq)

Relevant Examples of Chemical Equilibria Phase changes such as H 2 O (s) H 2 O (liq) In order to reach equilibrium, the system should be closed

Summary Activity 1) Write the equilibrium constant for these processes: MgO(s) + SO 2 (g) + ½ O 2 (g) MgSO 4 (s) H 2 O(l) H 2 O(g) CoCl(H 2 O) 5 + (aq) + Cl  (aq) CoCl 2 (H 2 O) 2 (aq) + 3H 2 O(l)

2) If the equilibrium constant for the process H 2 (g) + 1/8 S 8 (s) H 2 S(g) K c =7.6 x 10 5 What is the equilibrium constant associated with the reaction: 8H 2 S 8H 2 (g) + S 8 (s) ?

3) Given these equilibrium reactions and constants 3/2 O 2 (g) O 3 (g) K eq,1 = 2.5x10 -29 2NO(g) + O 2 (g) 2NO 2 (g) K eq,2 = 2.25 x 10 12 Calculate the equilibrium constant for the reaction NO(g) + O 3 (g) NO 2 (g) + O 2 (g)

Apply your Knowledge 4) If K c =2.5 for the reaction A  B , is the system depicted in the figure in equilibrium? If not, how would you represent the equilibrium state?

Apply your Knowledge 5) If K eq =7 for the reaction A  2B , is the system depicted in the figure in equilibrium? If not, how would you represent the equilibrium state?

Intro to equilibrium abbrev alg

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