Characteristics of Chemical Equilibrium. Equilibrium is Dynamic. The Equilibrium Constant. Equilibrium and Catalysts. Chapter 14: Chemical Equilibrium

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1 Characteristics of Chemical Equilibrium Chapter 14: Chemical Equilibrium 008 Brooks/Cole Brooks/Cole Equilibrium is Dynamic Equilibrium is Independent of Direction of Approach Reactants convert to products a A + b B c C + d D N (g) + 3 (g) N 3 (g) Species do not stop forming OR being destroyed Rate of formation rate of removal Concentrations are constant. 008 Brooks/Cole Brooks/Cole 4 Equilibrium and Catalysts The Equilibrium Constant For the -butene isomerization: 3 C CC C 3 3 C CC C 3 At equilibrium: rate forward rate in reverse An elementary reaction, so: k forward [cis] k reverse [trans] 008 Brooks/Cole Brooks/Cole 6 1

2 The Equilibrium Constant At equilibrium the concentrations become constant. The Equilibrium Constant We had: k forward [cis] k reverse [trans] k forward k reverse or [trans] [cis] k forward k reverse [trans] K c 1.65 (at 500 K) [cis] c for concentration based 008 Brooks/Cole Brooks/Cole 8 The Equilibrium Constant For a general reaction: a A + b B k forward k reverse K c [C] c [D] d [A] a [B] b c C + d D Products raised to stoichiometric powers divided by reactants raised to their stoichiometric powers The Equilibrium Constant N (g) + O (g) NO(g) K c [NO] [N ] [O ] 1 S 8 (s) + O (g) SO (g) K [SO] 8 c [O ] 008 Brooks/Cole Brooks/Cole 10 Equilibria Involving Pure Liquids and Solids [Solid] is constant throughout a reaction. Equilibria in Dilute Solutions pure solid concentration density mol. wt g / L g / mol [S 8 ] d S 8 / mol. wt S 8 d and mol. wt. are constants, so [S 8 ] is constant. This constant factor is absorbed into K c. Pure liquids are omitted for the same reason. N 3 (aq) + O(l) K c [N 4 + ][O - ] [N 3 ] N 4+ (aq) + O - (aq) 1.8 x 10-5 (at 5 C) (Units are customarily omitted from K c ) 008 Brooks/Cole Brooks/Cole 1

3 The Equilibrium Constant What is the equilibrium constant for: Si 4 (g) + O (g) SiO (s) + O(g)? K c for Related Reactions N (g) + 3 (g) N 3 (g) K c [N 3] [N ][ ] 3 Omit SiO (a solid) Include O (a gas) [ O] [Si 4 ][O ] Write down K c for: NaO(aq) + SO 4 (aq) Na SO 4 (aq) + O(l) Change the stoichiometry change K c ½ N 3 (g) + (g) N 3 (g) K c reaction divided by [N 3 ] [N ] ½ [ ] 3 Omit O (aq. solution) [Na SO 4 ] [ SO 4 ] [NaO] This new K c is the square root of the original K c. Multiply an equation by a factor Raise K c to the power of that factor. 008 Brooks/Cole Brooks/Cole 14 K c for Related Reactions N (g) + 3 (g) N 3 (g) K c [N 3] [N ][ ] 3 Reverse a reaction Invert K c : N 3 (g) N (g) + 3 (g) K c [N ][ ] 3 [N 3 ] 008 Brooks/Cole 15 K c for a Reaction that Combines Reactions If a reaction can be written as a series of steps: N (g) + O (g) NO(g) NO(g) + O (g) NO (g) N (g) + O (g) NO (g) The overall K c is the product of the steps: K c K c (step1) x K c (step) [NO] [NO ] [N ][O ] [NO] [O ] [NO] [N ][O ] K c [NO ] [NO] [O ] K c [NO ] [N ][O ] 008 Brooks/Cole 16 [NO ] [N ][O ] Equilibrium Constants in Terms of Pressure In a constant-v reaction, partial pressures change as concentrations change. Equilibrium Constants in Terms of Pressure At constant T, P is proportional to molar conc.: Ideal gas: PV nrt p for pressure based K p P C c P D d P A a P B b and for gas A: P A V n A RT P A n A RT [A]RT V P is easily measured, so another K is used In general, K p K c. Your text shows: K p K c (RT) Δngas Δn gas (c + d) (a + b) (moles of gaseous products) (moles of gaseous reactants) 008 Brooks/Cole Brooks/Cole 18 3

4 Equilibrium Constants in Terms of Pressure Calculate K p from K c for the reaction: N (g) + 3 (g) N 3 (g) K c 5.8 x 10 5 at 5 C. K p K c (RT) Δngas T K Δn gas (3 + 1) K p 5.8 x L atm mol -1 K -1 (98 K) x 10 mol L - atm - R must have L/mol units. ([ ] has mol/l units). mol and L units are assumed to cancel (units omitted from K c ). K p 9.7 x 10 atm - If all the equilibrium concentrations are known it s easy to calculate K c for a reaction. For: a A + b B k forward k reverse K c c C + d D [C] c [D] d [A] a [B] b At equilibrium [A].0 M & [B] 4.0 M. What is K c? K c [B]/[A] 4.0/(.0) Brooks/Cole Brooks/Cole 0 In other cases stoichiometry is used to find some concentrations. (g) + I (g) I(g) [ ] initial Concentrations change use stoichiometry: [ ] change -x -x x Stoichiometry: 1 I If I made x, lost x 1I I I lost x Let I made x 008 Brooks/Cole Brooks/Cole Add [ ] initial and [ ] change to get [ ] equilib (g) + I (g) I(g) [ ] equilib x x x Since [I ] M at equilibrium x M - x x M Calculate [ ] equilib and then K c [ ] equilib (0.0040) M [I ] equilib (0.0040) M [I] equilib (0.0040) M [I] K c ( ) [ ][I ] ( )( ) K c Brooks/Cole Brooks/Cole 4 4

5 Meaning of the Equilibrium Constant When: Predicting the Direction of a Reaction For the reaction: aa + bb cc + dd Q [C]c [D] d [A] a [B] b 008 Brooks/Cole Brooks/Cole 6 Predicting the Direction of a Reaction Predicting the Direction of a Reaction Reactants Q [Products] [Reactants] Products K c [Products] equilib [Reactants] equilib K c is a constant (only changes if T changes). So: 008 Brooks/Cole Brooks/Cole 8 Predicting the Direction of a Reaction SO (g) + O (g) SO 3 (g) K c 45 at 1000 K Predict the direction of the reaction if SO (0.085 M), O (0.100 M) and SO 3 (0.50 M) are mixed in a reactor at 1000 K. If K c is known, [ ] equilib can be calculated. Q [ SO 3 ] [SO ] [O ] (0.50) (0.085) (0.100) Brooks/Cole Brooks/Cole 30 5

6 (g) + I (g) I(g) [ ] initial Change on reaching equilibrium: [ ] change +x +x 1I I -x 1 I Amount lost [ ] equilib +x +x x Since K c 76 [I] K c ( x) [ ][I ] x 76 x ( x) Take the square root of both sides: x ±( x) [I] K c ( x) [ ][I ] (x)(x) Ignore the negative root (x cannot be negative). 008 Brooks/Cole Brooks/Cole 3 So: [ ] equilib x M [I ] equilib x M [I] equilib x M It s a good idea to check your work: Matches K c given in the problem [I] K c ( M) 76. [ ][I ] ( M)( M) Consider: A(g) B(g) + C(g) K c.500 A is added to an empty container. [A] initial M. What are [A], [B] and [C] at equilibrium? 008 Brooks/Cole Brooks/Cole 34 A B + C [ ] Initial [ ] change -x x x [ ] equilib (0.100 x) x x ax + bx + c 0 has two roots (solutions): x ere: -b ± b - 4ac a [B][C] K c [A] (x)(x) ( x) x.500 ( x) x x A quadratic equation x x M or M 008 Brooks/Cole Brooks/Cole 36 6

7 Ignore the negative root (negative concentration!) Then x M, and: [ A ] x M [ B ] x M [ C ] x M Most of the A is converted to products. Le Chatelier s Principle If a system is at equilibrium and the conditions are changed so that it is no longer at equilibrium, the system will react to reach a new equilibrium in a way that partially counteracts the change A system at equilibrium resists change. If pushed, it pushes back Check: (0.0963)(0.0963)/ Brooks/Cole Brooks/Cole 38 Changing Concentrations Changing Concentrations a A + b B c C + d D Add A (or B) and the system adjusts to remove it. more product is created. the reaction shifts forward. 3 C CC C 3 3 C CC C 3 Remove A (or B) the system adjusts to add more. more reactant is created. the reaction shifts backward. 008 Brooks/Cole Brooks/Cole 40 Changing V or P in Gaseous Equilibria Consider: N O 4 (g) NO (g) Changing V or P in Gaseous Equilibria Consistent with Le Chatelier N O 4 (g) NO (g) V doubled lower concentration. Minimize change by: Making more molecules (increase concentration). Shifting toward products - convert 1 molecule into. K c Q [ NO ] [ N O 4 ] At equilibrium (½[ NO ]) ¼[ NO Q ] ½[ NO ] ½[ N ½K c O 4 ] ½[ N O 4 ] [ N O 4 ] After V change, Q < K c equilibrium shifts to products P doubled. Minimize the change by: Removing molecules (decrease P). Shift toward reactants - convert molecules into Brooks/Cole Brooks/Cole 4 7

8 Changing V or P in Gaseous Equilibria P and V changes have no effect if Δn gas 0: Changing V or P in Gaseous Equilibria (g) + I (g) I(g) If V is doubled, the system cannot adjust gas molecules each side. K c Q [I] [ ][I ] At equilibrium (½[I]) ¼[I] Q [I] ½[ K c ]½[I ] ¼[ ][I ] [ ][I ] After V change, Q K c still at equilibrium! P A n A V RT [A]RT 008 Brooks/Cole Brooks/Cole 44 Changing Temperature Changing Temperature N (g) + O (g) NO(g) Δ kj N (g) + O (g) NO(g) Δ kj T increased? Minimize by removing heat. Forward reaction is endothermic Equilibrium shifts forward, removing heat. T ( C) K c x x x 10-3 K c increases as T increases. Explained by LeChatelier T reduced? Add heat. Reverse reaction is exothermic Equilibrium shifts in reverse, releasing heat 008 Brooks/Cole Brooks/Cole 46 Changing Temperature The aber-bosch Process Production of N 3 for fertilizer from atmospheric N N (g) + 3 (g) N 3 (g) Process developed by Fritz aber (chemist) and Carl Bosch (engineer). Perfected in Germany could not import guano from S. America and needed to make explosives 008 Brooks/Cole Brooks/Cole 48 8

9 The aber-bosch Process The aber-bosch Process N (g) + 3 (g) N 3 (g) Δ -9. kj It is: Exothermic product favored at low-t, but very slow at room-t, and must be heated. Carried out at high-p (00 atm), to favor product. Uses a catalyst to speed the reaction. Run at 450 C; N 3 is continually liquefied and removed. 008 Brooks/Cole Brooks/Cole 50 9

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