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Chemical Equilibrium Chapter 14 Chemical Equilibrium “Old Chemists Never Die; they just reach EQUILIBRIUM!” All physical and chemical changes TEND toward.

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Presentation on theme: "Chemical Equilibrium Chapter 14 Chemical Equilibrium “Old Chemists Never Die; they just reach EQUILIBRIUM!” All physical and chemical changes TEND toward."— Presentation transcript:

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2 Chemical Equilibrium Chapter 14 Chemical Equilibrium “Old Chemists Never Die; they just reach EQUILIBRIUM!” All physical and chemical changes TEND toward a state of equilibrium. A (l) A (g) A (s) A (l)

3 4/13/20152 Dynamic Equilibrium The net result of a dynamic equilibrium is that no change in the system is evident. Le Chatelier’s Principle - If a change is made in a system at equilibrium, the equilibrium will shift in such a way so as to reduce the effect of the change. Apply Pressure Pressure applied to the system at equilibrium caused it to shift until a new equilibrium was established.

4 4/13/20153 Dynamic Equilibrium Evaporation Open System (No Equilibrium) Evaporation Liquid Gas (No Equilibrium) (No Equilibrium) Liquid Gas Liquid Gas (Equilibrium)

5 4/13/20154 Dynamic Equilibrium Ag + + Cl - AgCl (s) Chemical Equilibrium Ag Ag +Cl - Cl -Ag + AgCl (s) Rate of Precipitation = Rate of Dissolving HC 2 H 3 O 2 HC 2 H 3 O 2 (aq) H ++ C2H3O2 C2H3O2 C2H3O2 C2H3O2 - Rate of dissociation (ionization) = Rate of Association HC 2 H 3 O 2 H + C2H3O2 C2H3O2 C2H3O2 C2H3O2 - C2H3O2 C2H3O2 C2H3O2 C2H3O2 - H+

6 4/13/20155 CHEM 1108 Lab Experiment HC 2 H 3 O 2 H + + C 2 H 3 O 2 - Red Orange [CoCl 4 ] H 2 O (l)Co(H 2 O) Cl - Pink Blue NH 4 Cl (s) NH Cl - White Colorless Solution You can actually “see” the equilibrium shift!

7 4/13/20156 Reversible Reactions N 2 O 4 (g)2 NO 2 (g) R1R1R1R1 2 NO 2 (g) N 2 O 4 (g) R2R2R2R2 N 2 O 4 (g)2 NO 2 (g) R1R1R1R1 R2R2R2R2 [R 1 = R 2 ] Homogeneous Equilibrium

8 4/13/20157 Reversible Reactions Exp M0.0 M M M Exp M M M M Exp M M M M [N 2 O 4 ] i [NO 2 ] i [N 2 O 4 ] eq [NO 2 ] eq N2O4 (g)2 NO2 (g) Q C = [NO 2 ] 2 [N 2 O 4 ] [N 2 O 4 ] Reaction Quotient K C = [NO 2 ] 2 eq [N 2 O 4 ] eq [N 2 O 4 ] eq Equilibrium Constant

9 4/13/20158 Equilibrium Constants Equilibrium Constant - When the rates of the forward and reverse reactions are equal, the system is “at equil- ibrium” and the reaction quotient = equilibrium constant. Experiment 1 K C = [ ] 2 /[0.0202] = M Experiment 2 K C = [ ] 2 /[0.0146] = M Experiment 3 K C = [ ] 2 /[ ] = M aA + bBcC + dD KC = [C]c[D]d [A]a[B]b

10 4/13/20159 Equilibrium Constants H 2 (g) + I 2 (g)2 HI (g) K C = [HI] 2 eq [H 2 ] eq [I 2 ] eq Q C = [HI] 2 [H 2 ][I 2 ] Exp.[H 2 ] eq [I 2 ] eq [HI] eq K C Median K C = Median K C = 54.47

11 4/13/ Equilibrium Constants 4 NH 3 (g) + 3 O 2 (g)2 N 2 (g) + 6 H 2 O (g) Q C = K C = [NH 3 ] 4 [O 2 ] 3 [N 2 ] 2 [H 2 O] 6 [N 2 ] eq 2 [H 2 O] 6 [NH 3 ] eq 4 [O 2 ] eq 3 [NH 3 ] eq 4 [O 2 ] eq 3

12 4/13/ Reaction Quotient vs. Equilibrium Constant Class Problem The concentration of N 2 O 4 = concentration of NO 2 = M in a reaction vessel. The equilibrium constant for N 2 O 4 (g) = 2 NO 2 (g) is Calculate Q C and state which direction the reaction will go. Class Problem If [O 2 ] = 0.21 M and [O 3 ] = 6.0 x M, what is the value of the K C for the equilibrium, 2 O 3 (g) = 3 O 2 (g)?

13 4/13/ Chemical Equilibrium Heterogeneous Reaction - A reaction that takes place in more than one phase or state. These reactions occur at the interface between phases - on the surface of liquids and solids. At a constant temperature, the concentration of a solid or liquid component remains constant in a heterogeneous equilibrium. WHY? Since the concentration is constant, it can be considered a part of the equilibrium constant and, thus, does NOT appear in the K C expression. C (s, graphite) + CO 2 (g) = 2 CO (g) K C = [CO] 2 [C][CO 2 ] [C][CO 2 ] K eq = [CO] 2 [CO 2 ]

14 4/13/ Chemical Equilibrium Class Problem A mixture that was initially M in H 2 (g) and M in I 2 (g), and contained no HI (g), was heated at o C until equilibrium was reached. The resulting equilibrium concentration of I 2 (g) was found to be M. What is the value of the K C for this equilibrium at o C? Construct an “ICE” Table: o C o C Equation:H 2 (g) + I 2 (g)  2 HI (g) Initial (I) conc., M Change (C) in conc., M Equil. (E) conc., M

15 4/13/ Chemical Equilibrium Calculate K C : [HI] 2 ( ) 2 [H 2 ][I 2 ] ( )( ) [HI] 2 ( ) 2 [H 2 ][I 2 ] ( )( ) = K C = = 54 Class Problem 14.4a. -When mol each of H 2 O H 2 O (g) and CO (g) are introduced into an empty L vessel at 959 K and allowed to come to equilibrium, the equilibrium mixture contains mol (g). Find KC KC KC KC for (g) + CO (g)  H2 H2 H2 H2 (g) + CO 2 CO 2 (g) Construct an “ICE” Table:

16 4/13/ o C 959 o C Equation:H 2 O (g) + CO (g)  H 2 (g) + CO 2 (g) I C E Chemical Equilibrium Kc =Kc =Kc =Kc = [H 2 ][CO 2 ] [H 2 ][CO 2 ] [H 2 O][CO] = 1.88 Class Problem 14.4b. - Suppose that [H 2 O] I = 2.00 M and [CO] I = 4.00 M? What are the equilibrium concentrations of the four species? (0.578) 2 (0.578) 2 (0.422) 2 (0.422) 2 =

17 4/13/ o C 959 o C Equation:H 2 O (g) + CO (g)  H 2 (g) + CO 2 (g) I C- x - x + x+ x E x x x x Chemical Equilibrium [H 2 ][CO 2 ] [H 2 ][CO 2 ] [H 2 O][CO] Kc = = 1.88 (2.00 – x)(4.00 – x) x2x2x2x2 = x 2 = 1.88(2.00 – x)(4.00 – x) = 1.88( x – x 2 ) x 2 = 15.0 – 11.3x +1.88x 2  0 = 0.88x 2 – 11.3x +15.0

18 4/13/ = 0.88 x 2 – 11.3 x (ax 2 + bx + c) Dust off the old Quadratic Formula: -(-11.3) ± [(-11.3) 2 – 4(0.88)(15.0)] 1/2 2(0.88) = 11 and 1.5 ! Which is RIGHT?

19 4/13/ Chemical Equilibrium What is ‘x’? It is the concentration of H 2 and CO 2 at equilibrium! But…you can’t have more hydrogen gas than you have of reactants to begin with! Thus, 11 M can’t be right! 1.5 M is the only sensible answer!

20 4/13/ Chemical Equilibrium x = 1.5 M Therefore: [H2O]eq = 0.5 M [CO]eq = 2.5 M [H2]eq = [CO2]eq = 1.5 M Check: Kc = (1.5M)2/(0.5M)(2.5M) = 1.8 There are no units in this case! What if you don’t remember the quadratic formula??

21 4/13/ Use Successive Approximation!!

22 4/13/ Class Exercise 14.5: Consider the following reaction for the decomposition of hydrogen sulfide: 2 H2S  2 H2 (g) + S2 (g) KC = 1.67 x o C A L vessel initially contains 1.25 x mol of H 2 S. Find the equilibrium concentrations of H 2 and S 2. Equation: 2 H 2 S (g)  2 H 2 (g) + S 2 (g) Initial (M) 2.50 x Change (M) - 2x + 2x + x Equilibrium (M) (2.50 x – 2x) 2x x

23 4/13/ Kc =Kc =Kc =Kc = [H 2 ] 2 [S 2 ] H2SH2SH2SH2S = (2x) 2 x (2x) 2 x (2.50 x – 2x) 2 = 4x 3 4x 3 (2.50 x – 2x) 2 Assume x is NEGLIGIBLE compared to 2.50 x M. Then: 4x 3 4x 3 (2.50 x ) 2 = 1.67 x ~ 1.67 x x 3 4x x x = (1.67 x ) (6.25 x ) = 4x 3 = 1.04 x x3x3x3x3 = 2.61 x x = 1.38 x 10-3 M

24 4/13/ Is x NEGLIGIBLE compared to 1.38 x M? Plug it back in to check: 4x 3 4x 3 (2.50 x – 2x) 2 [(2.50 x ) - 2(1.38 x )] 2 4x 3 4x 3 = 4x x =1.67 x x 3 = 2.07 x >>>>> x = 2.74 x There appears to be a mistake in these calculations! Please check carefully and see if you can see where it is!

25 4/13/ Complex example of Successive Approximation

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31 4/13/ Do NOT Panic! This is NOT a typical Problem! It is a Worst Case Scenario!!! Any Exam Problem will be MUCH Shorter!!

32 4/13/ Class Exercise 14.5: In an experiment starting with [N 2 O 4 ] I = M and [NO 2 ] I = M, [N 2 O 4 ] eq = M. (a) What is [NO 2 ] eq ? (b) What is the value for K c ? Equation: N 2 O 4 (g)  2 NO 2 (g) I (M) x C (M) E (M) 4.52 x [NO 2 ] 2 [NO 2 ] 2 [N 2 O 4 ] K C = = ( ) 2 /( ) = 0.212

33 4/13/ What does the value of K c MEAN? The larger K C is, the closer to completion the rxn is! N 2 (g) + O 2 (g)  2 NO (g)K C = 1 x NH 3 (g)  N 2 (g) + 3 H 2 (g)K C = 9.5 H 2 (g) + Cl 2 (g)  2 HCl (g)K C = 1.33 x 10 34


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