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. ↔ 2 NO2 N2O4 (I will use ↔ to indicate an equilibrium process in my lecture notes)
1 Chemical Equilibrium
•NO2 is a brown gas while N2O4 is colorless ↔ 2 NO2 N2O4 • At any given time in a container of
NO2, some fraction of the gas will be in the form of NO2, and some fraction will be in the form of N2O4.
Chemical Equilibrium • Chemical equilibrium is a dynamic process—an individual molecule will
repeatedly move from the NO2 form to the N2O4 form, the overall concentrations of NO2 and N2O4 do not change at a given temperature ↔ 2 NO2 N2O4
• The “equilibrium constant”, Keq, 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 = [C] [D] KC [A]a [B]b
Equilibrium Constant ↔ •For the 2 NO2 N2O4 reaction, the equilibrium constant is given as:
[N O ] K = 2 4 C 2 [NO2]
3 Equilibrium Constant • If the reaction involves a pure solid or pure liquid, these species do not appear in the equilibrium constant expression: Example: ↔ CH4(g) + H2O(l) CO(g) + 3 H2(g) [CO(g)] [H (g)]3 = 2 Note that H2O( KC l [CH (g)] ) does not 4 appear in the denominator.
Equilibrium Constant • 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 ↔ - + H2CO3(aq) + H2O(aq) HCO3 (aq) + H3O (aq) - ↔ 2- + HCO3 (aq) + H2O(aq) CO3 (aq) + H3O (aq)
4 Equilibrium Constants Example (con’t.) ↔ H2CO3(aq) + H2O(l) - + HCO3 (aq) + H3O (aq) - ↔ HCO3 (aq) + H2O(l) 2- + CO3 (aq) + H3O (aq) ↔ Net: H2CO3(aq) + 2 H2O(l) 2- + CO3 (aq) + 2 H3O (aq)
Equilibrium Constants Example (con’t.) For the first reaction: ↔ H2CO3(aq) + H2O(l) - + HCO3 (aq) + H3O (aq) + [HCO- (aq)][H O (aq)] K = 3 3 = 4.2 x 10-7 C1 [H2CO3(aq)]
5 Equilibrium Constants Example (con’t.) For the second reaction: - ↔ HCO3 (aq) + H2O(l) 2- + CO3 (aq) + H3O (aq) 2- + [CO (aq)][H3O (aq)] K = 3 = 4.8 x 10-11 C2 − [HCO3 (aq)]
Equilibrium Constants Example (con’t.) For the net reaction: ↔ H2CO3(aq) + 2 H2O(l) 2- + CO3 (aq) + 2 H3O (aq) + [CO2-(aq)][H O (aq)]2 = 3 3 KC [H2CO3(aq)] = K K = 2.0 x 10-17 C1 C2
6 Equilibrium Constant 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 P = = [A]RT A V P ∴ [A] = A RT
Equilibrium Constant and Pressure • For the generic reaction aA + bB ↔ cC + dD we can write the equilibrium constant in terms of pressure Pc Pd = C D KP Pa Pb A B
7 Equilibrium Constant and Pressure • Substituting the relationship between pressure and concentration gives: Pc Pd c d = C D = ([C]RT) ([D]RT) KP Pa Pb ([A]RT)a ([B]RT)b A B c d = [C] [D] (c+d)-(a+b) = Δn (RT) KC (RT) [A]a [B]b where n is the change in the number of moles of gas phase molecules
Equilibrium Constant and Pressure Example:
Determine KP for the reaction ↔ N2(g) + 3 H2(g) 2 NH3(g) 8 o KC = 3.5 x 10 at 25 C n(prod) = 2 n(react) = 4 n = 2 – 4 = -2 = Δn KP KC (RT) = 3.5 x 108 {(.0821)(298)}-2 = 5.8 x 105
8 Using Equilibrium Constants
Example: 2- Determine [SO4 ] when a solution of 1.00 M H2SO4(aq) is prepared: • Step 1—write a balance chemical equation - ↔ 2- + HSO4 (aq) + H2O(l) SO4 (aq) + H3O (aq) • Step 2—write an expression for the equilibrium constant
2- + = [SO4 (aq)] [H3O (aq)] = -2 KC - 1.2 x 10 [HSO4(aq)]
Using Equilibrium Constants
Example: 2- Determine [SO4 ] when a solution of 1.00 M H2SO4(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: - ↔ 2- + HSO4 (aq) + H2O(l) SO4 (aq) + H3O (aq) 1.00 – x x 1.00 + x (2) let the reaction go to completion, and then let the some of the product go back to reactants: - ↔ 2- + HSO4 (aq) + H2O(l) SO4 (aq) + H3O (aq) x 1.00 – x 2.00 - x
9 Using Equilibrium Constants
Example: 2- Determine [SO4 ] when a solution of 1.00 M H2SO4(aq) is prepared: • Step 4—solve equilibrium constant expression for unknowns
2- + [SO (aq)] [H3O (aq)] x(1.00 + x) 4 = = 1.2 x 10-2 - 1.00 - x [HSO4 (aq)] x2 + 1.012x - 0.012 = 0
Using the quadratic equation to solve for 2- x, we get x = 0.012 M = [SO4 ]
Using Equilibrium Constants
Example:
H2SO3 is formed in an equilibrium reaction between SO2 and H2O ↔ SO2(g) + H2O(g) H2SO3(aq)
SO2 has an average concentration of .006 ppm, and H2O has a vapor pressure of ~20 Torr
Determine the amount of H2SO3 in the troposphere Step 1—write an expression for the equilibrium constant PH SO Δ K = 2 3 = K (RT) n P P P C SO2 H2O 8.47 x 103 = = 3.46 x 102 (.0821) (298)
10 Using Equilibrium Constants
Example:
Determine the amount of H2SO3 in the troposphere ↔ SO2(g) + H2O(g) H2SO3(aq) Step 2—determine pressures of reactants (pressures must be given in units of atm because the R used has atm units
H2O: 20 Torr = .026 atm -9 SO2: (.006 ppm)(1 atm) = 6 x 10 atm
Step 3—solve expression for PH2SO3
P = (3.46 x 102) P P H2SO3 SO2 H2O = (3.46 x 102)(6 x 10-9)(.026) = 5.4 x 10-8 atm
Interpreting Equilibrium Constants
•If KC >> 1, then the reaction is strongly product-favored, i.e., the mixture will contain more products than reactants
•If KC << 1, then the reaction is strongly reactant-favored, i.e., the mixture will contain more reactants than products ≈ •If KC 1, the mixture will contain approximately equal amount of reactant and products
11 Interpreting Equilibrium Constants
Example: -5 Acetic acid, CH3COOH, has a KC of 1.8 x 10 Determine the relative concentrations of - + CH3COOH, CH3COO , and H in an aqueous solution ↔ - + balanced equation: CH3COOH CH3COO + H - + = [CH3COO ][H ] = -5 KC 1.8 x 10 [CH3COOH]
Interpreting Equilibrium Constants
Example (con’t.): build concentration table - + [CH3COOH] [CH3COO ] [H ] initially 1.00 0.00 0.00 equilibrium 1.00 – x x x
solve equilibrium constant expression for unknown concentrations x ⋅ x = 1.8 x 10-5 1.00 - x
12 Interpreting Equilibrium Constants
Example (con’t.): assume x is small relative to the initial concentration of the acetic acid (reactant- favored condition because KC << 1) 1.00 - x ≈ 1.00 check assumptions: 2 = -5 x 1.8 x 10 1.00 - .0042 ≈ 1.00 x = .0042 M the acetic acid in solution is .42% - acetate ion (CH3COO ) with the remainder as acetic acid (CH3COOH)
Interpreting Equilibrium Constants
Example (con’t.): assume x is small relative to the initial concentration of the acetic acid (reactant- favored condition because KC << 1) 1.00 - x ≈ 1.00 check assumptions: 2 = -5 x 1.8 x 10 1.00 - .0042 ≈ 1.00 x = .0042 M the acetic acid in solution is .42% - acetate ion (CH3COO ) with the remainder as acetic acid (CH3COOH)
13 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 N2(g) + 3 H2(g) ↔ 2 NH3(g) + heat high pressure and temperature
14 Haber-Bosch Process
Increase in Concentration or Partial Pressure For the process: ↔ N2(g) + 3 H2(g) 2 NH3(g) an increase in N2 and/or H2 concentration or pressure, will cause the equilibrium to shift towards the
production of NH3
15 Decrease in Concentration or Partial Pressure For the process: ↔ N2(g) + 3 H2(g) 2 NH3(g) likewise, a decrease in NH3 concentration or pressure will cause more NH3 to be produced
Changes in Temperature
For the process: ↔ N2(g) + 3 H2(g) 2 NH3(g) + heat an increase in temperature will cause the reaction to shift back towards reactants because the reaction is exothermic
16 Increase in Volume
For the process: ↔ N2(g) + 3 H2(g) 2 NH3(g) + heat an increase in volume, causes the equilibrium to shift to the left where there are more gaseous molecules
Decrease in Volume
For the process: ↔ N2(g) + 3 H2(g) 2 NH3(g) + heat a decrease in volume, causes the equilibrium to shift to the right where there are fewer gaseous molecules
17 Stress on Equilibrium Consider the reaction: cis-2-butene ↔ trans-2-butene [trans] K = = 1.5 (at 600 K) C [cis]
Shifting of Equilibrium ↔ For the reaction: 2 NO2(g) N2O4(g)
low pressure high pressure
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