CHEMICAL EQUILIBRIUM. INTRODUCTION 1. In the reaction: I 2 (g) + H 2 (g)  2 HI(g) at 150 o C, the original color of the mixture is: at 150 o C, the original.

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CHEMICAL EQUILIBRIUM

INTRODUCTION 1. In the reaction: I 2 (g) + H 2 (g)  2 HI(g) at 150 o C, the original color of the mixture is: at 150 o C, the original color of the mixture is:PURPLE a. How do we know that, no matter how much time elapses, some I 2 and H 2 remains unreacted? The purple color fades, but does not disappear.

2. Do most reactions go to COMPLETION? (Completely convert reactants to products) (Completely convert reactants to products)NO Instead of obtaining a pure product, a mixture of products AND reactants is obtained.

The PCl 5 -PCl 3 -Cl 2 System At first, the forward reaction occurs: PCl 5  PCl 3 + Cl 2 Later, the reverse reaction begins to occur & the forward reaction slows. Cl 2 + PCl 3  PCl 5 Eventually, the rates of the forward & reverse reactions are equal PCl 5  PCl 3 + Cl 2 NO MORE OVERALL CHANGE!

Define: EQUILIBRIUM State of balance in which opposing processes occur at equal rates. 3. For all experiments involving PCl 5, PCl 3, and Cl 2, the equilibrium concentrations are related by the equation: K = [PCl 3 ] x [Cl 2 ] K = [PCl 3 ] x [Cl 2 ] [PCl 5 ] [PCl 5 ]

4. Such equilibrium constants are give the symbol of K or K eq EQUILBRIUM CONSTANTS If nA + mB  xC + yD K = [C] x x [D] y [A] n x [B] m [A] n x [B] m Remember: [ ] = mol/L Remember: [ ] = mol/L

1. Write the equilibrium constant (K) for the reaction: N 2 (g) + 3 H 2 (g)  2 NH 3 (g) K = ____[NH 3 ] 2 ____ [N 2 ] x [H 2 ] 3 [N 2 ] x [H 2 ] 3 2. The value of K depends on TEMPERATURE. 3. The value of K is INDEPENDENT of: a. Initial and final concentrations b. Volume of the container

4. Write the expression for the equilibrium constant for the reaction: H 2 (g) + I 2 (g)  2 HI(g) K = ____[HI] 2 ____ [H 2 ] x [I 2 ] [H 2 ] x [I 2 ] 5. How does the amount of solid or liquid present affect the equilibrium position? Does not affect it.

1. In writing the expression for K, what happens to concentrations of LIQUIDS and SOLIDS? [ ] = 1 Ex. CaCO 3 (s)  CaO(s) + CO 2 (g) K = [CO 2 ] K = [CO 2 ] 1 or K = [CO 2 ] or K = [CO 2 ]

Write equilibrium expressions for: 1. 3 Fe(s) + 4 H 2 O(l)  Fe 3 O 4 (s) + 4 H 2 (g) 2. 2 NOCl(g)  2 NO(g) + Cl 2 (g) 3. CS 2 (g) + 3 Cl 2 (g)  S 2 Cl 2 )l) + CCl 4 (l) 1. K = [H 2 ] 4 2. K = [NO] 2 x [Cl 2 ] [NOCl] 2 [NOCl] 2 3. K = _____1________ [CS 2 ] x [Cl 2 ] 3 [CS 2 ] x [Cl 2 ] 3

b. Calculate K if reactant concentrations are M each and product concentrations are M each x x x 10 -2

VALUE OF Keq 1. If K is small (<1), then mostly reactants at equilibrium reaction to the LEFT is favored Ex. N 2 (g) + O 2 (g)  2 NO(g) K = ___[NO] 2 ___ = 1 x at 25 o C [N 2 ] x [O 2 ] [N 2 ] x [O 2 ]

2. If K is large, then mostly PRODUCTS at equilibrium; reaction to the RIGHT is favored. Ex. 2 CO(g) + O 2 (g)  2 CO 2 (g) K = ____[CO 2 ] 2 ___ = 4 x (500 o C) [CO] 2 x [O 2 ] [CO] 2 x [O 2 ]

3. If K is close to 1, then appreciable amounts of product & reactants NEITHER reaction is favored Ex. N 2 (g) + O 2 (g)  2 NO(g) K = ___[NO] 2 ___ =.10 at 2000 o C [N 2 ] x [O 2 ] [N 2 ] x [O 2 ]

LeChatelier’s Principle If a system in equilibrium is subjected to a STRESS, the equilibrium will shift to relieve the stress. a. Three factors affect the equilibrium position: TemperaturePressureConcentration

EFFECT OF CONCENTRATION a. If a gaseous (or dissolved) substance is added on the left, the reaction occurs in the FORWARD DIRECTION. The equilibrium is shifted TO THE RIGHT. b. If a gaseous (or dissolved) substance is REMOVED on the left, the reaction occurs in the BACKWARD DIRECTION. The equilibrium is shifted TO THE LEFT.

Ex. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Add N 2 ? Equilibrium shifts → [NH 3 ] _______, [H 2 ] _______ Remove NH 3 ? Equilibrium shifts → [N 2 ] ________, [H 2 ] _______ Add on the LEFT, shift → Add on the RIGHT, shift ←

EFFECT OF PRESSURE (Gases only) a. Increasing pressure favors the reaction with the lesser volume. (Look for # moles of gas) Ex. PCl 5 (g)  PCl 3 (g) + Cl 2 (g) Increasing pressure favors the reaction In the reverse direction (Equil shifts LEFT) [PCl 3 ] _____, [Cl 2 ] ______, [ PCl 5 ] _______

b. Decreasing pressure favors the reaction with the greater volume. Ex. CaCO 3 (s) → CaO(s) + CO 2 (g ) Ex. CaCO 3 (s) → CaO(s) + CO 2 (g ) Decreasing pressure favors the reaction in the forward direction. Remember: Decreasing volume of gases increases pressure & vice versa.

EFFECT OF TEMPERATURE a. Increasing temperature favors the endothermic reaction (absorbs the excess heat) PCl 5 (g)  PCl 3 (g) + Cl 2 (g) ∆ H = +22 Kcal Increasing temperature favors the forward reaction, shifting the above equilibrium to the right. [PCl 5 ] _________, [PCl 3 ] __________, [PCl 5 ] _________, [PCl 3 ] __________, [Cl 2 ] _________

b. Decreasing temperature favors the exothermic reaction, shifting the above exothermic reaction, shifting the above equilibrium to the left. equilibrium to the left. Ex. 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) ∆H = kcal ∆H = kcal Increase T, equil shifts _________ Decrease T, equil shifts _________

*CHANGING TEMPERATURE ALSO AFFECTS AND CHANGES THE VALUE OF K.* EFFECT OF A CATALYST a. A catalyst changes the RATE of the reaction but is not consumed by it b. Gets to the equilibrium point faster.

INCREASING RATE OF REACTION 1. Increasing temperature increases the rate of reaction. a. Why? More molecules have sufficient energy to react. (Have reached Activation Energy.)

2. Increase surface area (solids). 3. Increase concentration. a. More molecules collide, so more collisions with sufficient energy to react. (Activation energy) 4. Add a catalyst. a. Catalyst lowers the activation energy, so more molecules have sufficient energy to react.

HABER PROCESS N 2 (g) + 3 H 2 (g)  2 NH 3 (g) ∆H = -92 kcal How can yield be increased? a. Pressure Increase pressure Shifts equilibrium ________ Also increases RATE * PRESSURE IS 400 ATM* * PRESSURE IS 400 ATM* b. Temperature INCREASE would shift equilibrium LEFT. BUT reactions go faster at higher T. *Use MODERATE temp.: 400 o C*

c. Catalyst No effect on POSITION of equilibrium. BUT rate at which equilibrium is reached is increased. *Solid catalyst: Fe + small amounts of K 2 O & Fe 2 O 3 * Fe + small amounts of K 2 O & Fe 2 O 3 *