Chemical Equilibrium When some types of chemical reactions occur in the gas or solution phases, these reaction attain chemical equilibrium, i.e., the reaction does not go to completion, but the reaction vessel will contain both reactant species and product species mixed together. Chemical Equilibrium This occurs when the concentrations of the reactants stop decreasing, and the concentrations of the products stop increasing. NO N O 4 (I will use to indicate an equilibrium process in my lecture notes) Chemical Equilibrium NO is rown gas while N O 4 is colorless NO N O 4 At any given time in a container of NO, some fraction of the gas will be in the form of NO, and some fraction will be in the form of N O 4. Chemical Equilibrium Chemical equilibrium is a dynamic process an individual molecule will repeatedly move from the NO form to the N O 4 form, the overall concentrations of NO and N O 4 do not change at a given temperature NO N O 4 The equilibrium constant, K eq, for a chemical reaction indicates whether the reactants or the products will be favored in an equilibrium process The equilibrium constant in terms of concentrations is defined as: aa bb cc dd [C] [D] [A] [B] For the NO N O 4 reaction, the equilibrium constant is given as: [NO4 ] [NO ] 1
Suppose we have 1.00 atm NO and 6.06 x 10-3 atm N O 4. Determine the value of K C. NO (g) N O 4 (g) From the Ideal Gas Law: n P [X] = = V RT 1.00 atm [NO ] = = 0.0409 M (.0806 L atm/molk)(98k) -3 6.06 x 10 atm [NO 4] = =.48 x 10 M (.0806 L atm/molk)(98k) Suppose we have 1.00 atm NO and 6.06 x 10-3 atm N O 4. Determine the value of K C. NO (g) N O 4 (g) [NO 4].48 x 10 KC = = = 0.148 [NO ] (.0409) We usually don t worry about the units on equilibrium constants because they vary depending on the stoichiometry of the chemical reaction. If the reaction involves a pure solid or pure liquid, these species do not appear in the equilibrium constant expression: CH 4 (g) H O(l) CO(g) 3 H (g) 3 [CO(g)] [H (g)] K C = [CH4 (g)] Note that H O (l) does not appear in the denominator. If we have two or more equilibrium chemical reactions, we can combine their equilibrium constant expressions to get an overall equation for the net chemical reaction Example H (aq) H O(aq) H- (aq) H 3 O (aq) H- (aq) H O(aq) CO - 3 (aq) H 3 O (aq) s H (aq) H O(l) H- (aq) H 3 O (aq) H- (aq) H O(l) - (aq) H 3 O (aq) Net: H (aq) H O(l) - (aq) H 3 O (aq) s For the first reaction: H (aq) H O(l) H- (aq) H 3 O (aq) - [HCO3 [H3O -7 KC = = 4. x 10 1 [H CO3
s For the second reaction: H- (aq) H O(l) - (aq) H 3 O (aq) - [CO [H 3 3O -11 KC = = 4.8 x 10 [HCO3 s For the net reaction: H (aq) H O(l) - (aq) H 3 O (aq) - [CO [H3O K 3 C = [H CO3-17 = KC KC =.0 x 10 1 and Pressure How does the expression for the equilibrium constant change if pressure is used as the variable instead of concentration? Using the Ideal Gas Law: nrt PA = = [A]RT V P [A] = A RT and Pressure For the generic reaction aa bb cc dd we can write the equilibrium constant in terms of pressure P P C K D P = P P A B and Pressure Substituting the relationship between pressure and concentration gives: P P C D ([C]RT) ([D]RT) KP = = P P ([A]RT) ([B]RT) A B [C] [D] (c d)-(a b) Δn = (RT) = K C (RT) [A] [B] where Δn is the change in the number of moles of gas phase molecules and Pressure Determine K P for the reaction N (g) 3 H (g) NH 3 (g) 3.5 x 10 8 at 5 o C n(prod) = n(react) = 4 Δn = 4 = - Δn KP = KC (RT) 8 - = 3.5 x 10 {(.081)(98)} 5 = 5.8 x 10 3
Using s Determine [OCN - ] when a solution of 0.10 M HOCN (aq) is prepared: Step 1 write alance chemical equation HOCN(aq) H O(l) OCN - (aq) H 3 O (aq) Step write an expression for the equilibrium constant - [OCN [H3O = 3.5 x 10 [HOCN Using s Determine [OCN - ] when a solution of 0.10 M HOCN (aq) is prepared: Step 3 determine the unknowns We can do this one of two ways: (1) let the reactants react, in which case we write: HOCN(aq) H O(l) OCN - (aq) H 3 O (aq) 0.10 x x x () let the reaction go to completion, and then let the some of the product go back to reactants: HOCN(aq) H O(l) OCN - (aq) H 3 O (aq) x 0.10 x 0.10 - x Using s Determine [OCN - ] when a solution of 0.10 M HOCN(aq) is prepared: Step 4 solve equilibrium constant expression for unknowns - [OCN [H 3O x = = 3.5 x 10 [HOCN 0.10 x Using the quadratic equation to solve for x, we get x = 0.0057 M = [OCN - ] Using s H SO 3 is formed in an equilibrium reaction between SO and H O SO (g) H O(g) H SO 3 (aq) SO has an average concentration of.006 ppm, and H O has a vapor pressure of ~0 Torr Determine the amount of H SO 3 in the troposphere Step 1 write an expression for the equilibrium constant PH SO3 Δn KP = = KC (RT) PSO P HO 3 8.47 x 10 = = 3.46 x 10 (.081) (98) Using s Determine the amount of H SO 3 in the troposphere SO (g) H O(g) H SO 3 (aq) Step determine pressures of reactants (pressures must be given in units of atm because the R used has atm units H O: 0 Torr =.06 atm SO : (.006 ppm)(1 atm) = 6 x 10-9 atm Step 3 solve expression for P HSO3 P H SO = (3.46 x 10 ) P P 3 SO H O -9-8 = (3.46 x 10 )(6 x 10 )(.06) = 5.4 x 10 atm Interpreting s If K C >> 1, then the reaction is strongly product-favored, i.e., the mixture will contain more products than reactants If K C << 1, then the reaction is strongly reactant-favored, i.e., the mixture will contain more reactants than products If K C 1, the mixture will contain approximately equal amount of reactant and products 4
Interpreting s Acetic acid, CH 3 COOH, has a K C of 1.75 x 10-5 Determine the relative concentrations of CH 3 COOH, CH 3 COO -, and H in an aqueous solution balanced equation: CH 3 COOH CH 3 COO - H - [CH3COO ][H ] -5 = 1.75 x 10 [CH3COOH] Interpreting s : build concentration table [CH 3 COOH] [CH 3 COO - ] [H ] initially 1.00 0.00 0.00 equilibrium 1.00 x x x solve equilibrium constant expression for unknown concentrations x x -5 = 1.75 x 10 1.00 - x Interpreting s : assume x is small relative to the initial concentration of the acetic acid (reactantfavored condition because K C << 1) 1.00 - x 1.00-5 x = 1.75 x 10 x =.004 M check assumptions: 1.00 -.004 1.00 the acetic acid in solution is.4% acetate ion (CH 3 COO - ) with the remainder as acetic acid (CH 3 COOH) Le Chatelier's Principle If a stress, such as a change in concentration, pressure, temperature, etc., is applied to a system at equilibrium, the equilibrium will shift in such a way as to lessen the effect of the stress. Production of Ammonia catalysis N (g) 3 H (g) NH 3 (g) heat high pressure and temperature Haber-Bosch Process 5
Increase in Concentration or Partial Pressure N (g) 3 H (g) NH 3 (g) an increase in N and/or H concentration or total pressure, will cause the equilibrium to shift towards the production of NH 3 reactant side of reaction has 4 moles of gas while product side has only moles of gas equilibrium will shift to product to reduce effects of increased pressure Decrease in Concentration or Partial Pressure N (g) 3 H (g) NH 3 (g) likewise, a decrease in NH 3 concentration or pressure will cause more NH 3 to be produced Changes in Temperature N (g) 3 H (g) NH 3 (g) heat an increase in temperature will cause the reaction to shift back towards reactants because the reaction is exothermic Increase in Volume N (g) 3 H (g) NH 3 (g) heat an increase in volume, causes the equilibrium to shift to the left where there are more gaseous molecules Decrease in Volume N (g) 3 H (g) NH 3 (g) heat a decrease in volume, causes the equilibrium to shift to the right where there are fewer gaseous molecules Stress on Equilibrium Consider the reaction: cis--butene trans--butene [trans] = 1.5 (at 600K) [cis] 6
Shifting of Equilibrium For the reaction: NO (g) N O 4 (g) low pressure high pressure 7