Le Châtelier's Principle. Chemical Equilibria & the Application of Le Châtelier s Principle to General Equilibria. Using Le Châtelier's Principle

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1 Chemical Equilibria & the Application of Le Châtelier s Principle to General Equilibria CHEM 107 T. Hughbanks Le Châtelier's Principle When a change is imposed on a system at equilibrium, the system will react in the direction that reduces the amount of change. Changes include adding or removing material, or changes in pressure or temperature. Using Le Châtelier's Principle CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) An equilibrium mixture at 1200 K contains mol CO, mol H 2, mol CH 4 and mol H 2 O, all in a 10.0 L vessel. (What is?) All of the H 2 O is somehow removed, and equilibrium is re-established. What will happen to the amount of CH 4? Set up eqn. for final amount of CH 4. Changes in Temperature The equilibrium constant is not constant with temperature. Le Châtelier s Principle would suggest: Qualitatively, if a reaction is endothermic then the equilibrium constant increases with temperature If a reaction is exothermic then the equilibrium constant decreases with temperature Changes in Pressure Consider the equilibrium (significant at 700 C): CaO(s) + CO 2 (g) CaCO 3 (s) Le Châtelier s Principle would suggest: If the pressure is suddenly increased, say by suddenly compressing the container, more CO 2 would react with CaO to produce more CaCO 3. Many equilibrium reactions occur in solutions. Weak acids & bases, etc. Example: solution of ammonia Write an equilibrium constant for this.

2 = [NH + ][OH ] 4 [NH 3 ][H 2 O] But [H 2 O] is constant, so we can incorporate it into the constant itself. = [NH + ][OH ] 4 = K [NH 3 ] b By convention, tabulated K s are written like this, without including water term. NH 3 produces OH in water and is therefore a base. for a base is called K b. For NH 3, K b = What % of NH 3 is converted to NH 4 + in a 1.0 M solution? Acids - Definition of K a HA(aq) + H 2 O H 3 O + (aq) + A (aq) = [H 3 O+ ][A ] = K [HA] a Once again, the water term is omitted in tabulating the equilibrium constants of acids. An acid, say HA, produces H 3 O + in water. for an acid is called K a. For CH 3 COOH (acetic acid), K a = What is [H 3 O + ] in a 0.1 M solution? Water self-dissociates, even in the absence of added acids or bases: H 2 O + H 2 O H 3 O + (aq) + OH (aq) = [H 3 O + ][OH ] = = K w Again, [H 2 O] is constant, so K w does not include it. In pure water, what are [H 3 O + ] & [OH ]? In all aqueous solutions, equilibrium between H 3 O + & OH is continually re-established: H 2 O + H 2 O H 3 O + (aq) + OH (aq) = [H 3 O + ][OH ] = = K w This is true even with added acids or bases. In a 1.0 M NH 3 solution, what is [H 3 O + ]? ph Scale H 3 O + concentration is conveniently measured using the logarithmic ph scale: ph = -log[h 3 O + ] In pure H 2 O, [H 3 O + ] = 10 7, ph = -log 10 7 = 7 poh is defined similarly: poh = -log[oh - ] What are the ph and poh of a 1.0 M NH 3 solution?

3 ph and poh ph = -log [H 3 O + ] poh = -log [OH - ] In aqueous solutions, [H 3 O + ][OH - ] = This is true even with added acids or bases. -log {[H 3 O + ][OH - ]} = -log {10 14 } -log [H 3 O + ] - log [OH - ] =14 ph + poh = 14 Solubility Equilibrium The process is in equilibrium when rate of ions (or molecules) leaving the solid = rate returning to the solid. That is, when dissolution rate = precipitation rate Solubility Products M p X q (s) M q+ (aq) + X p- (aq) K sp = [M q+ ] p [X p ] q Note that no concentration of the M p X q solid appears in the equilibrium constant. This when the solid salt and dissolved salt are in equilibrium, the solution is saturated. When the solution is saturated, the amount of extra salt does not affect the amount of dissolved M p X q. Solubility Products In a saturated aqueous solution of MgF 2, the concentration of Mg 2+ ions is M. Write the equilibrium constant expression, and compute K sp for MgF 2. Phase Equilibria - Examples H 2 O(l ) H 2 O(g ); CO 2 (s) CO 2 (g) Gas-Liquid Equilibrium K P = P H2 O K P = P CO2 When gases are involved in equilibria, pressures appear in the equilibrium constant. Concentrations of H 2 O(l ) or CO 2 (s) do not appear in these expressions. If at least some H 2 O(l ) or CO 2 (s) are present, the equilibrium vapor pressures of the gases will not depend on the quantity of the condensed phases.

4 Problem: H 2 O(l ) H 2 O(g ) Given: kj of heat is required to boil away 1.0 mol of water at 1.0 atm. As we know, this happens at 100 C. Q: Estimate the boiling point of water on a planet where the atmospheric pressure is twice that of earth. Summary: Varieties of Equilibrium Constants Gas Phase: K P and K C your text sticks to K C, but K P is the thermodynamic equilibrium constant. Calculations can use either. Heterogeneous (more than one phase in equilibrium): The condensed phases don t appear (they have activity = 1). This is true for solid-gas, liquid-gas, and solubility products. Species in solution: Products over Reactants, but solvents don t appear because their activities are virtually constant. For example, K a s and K b s don t include [H 2 O] in the denominator. Reconsider G In thermodynamics, when we consider a reaction, we calculate G for the forward reaction as follows: G rxn = n G f products n G f reactants = γ G f (C) + δ G f (D) α G f (A) β G f (B) Each of the reactants and products are defined to be in standard states in this eqn. Look more carefully at standard states & consider G when conditions aren t standard. Standard States 298 K, 1 atm. Each gas phase reactant and product considered to be at 1 atm. pressure under standard conditions. (eg., if A & C are gases, P A = P C = 1atm) Example: NH 3 (g) + HCl(g) NH 4 Cl(s) from data in Appendix E (all in kj): G rxn = (-16.45) - (-95.30) = kj This is G for converting 1 mol NH 3 and 1 mol HCl at 1.0 atm. pressure into solid NH 4 Cl at 298 K. Standard States Any reactant or product that exists as an ion or molecule dissolved in solution, the conc. is 1.0 mol/l (1.0 M) for the standard state. Example: NH 3 (g) + H 2 O(l) NH 3 (aq) + H 2 O(l) from data in Appendix E (all in kj): G = 26.5 ( 16.45) = kj This is G for dissolving 1 mol NH 3 gas (at 1.0 atm pressure) into water to form a 1.0 M NH 3 solution all at 298 K. G: Non-standard Conditions When concentrations (or pressures, if gases are involved) are not standard, G is related to G by the equation: G = G + RT lnq where Q is the reaction quotient defined earlier: Q = P γ δ C PD P α Q = [C]γ [D] δ β A P B [A] α [B] β for gases, pressures are used, for solutions, concentrations are used.

5 G = G + RT lnq Q = P γ δ C P D P α β A P B Q = [C]γ [D] δ [A] α [B] β Q is variable (determined by the amount of reactants and products at any time.) Check: standard conditions, Q = 1, G = G At equilibrium, Q = K. But we know that when we are at equilibrium, G = 0. 0 = G + RTlnK G = RTlnK G = - RT lnk, Example Given the following data, calculate K for the reaction: I 2 (g) + H 2 (g) 2 HI(g) From Appendix E: H f (I 2 (g)) = 62.44; H f (HI(g)) = both in kj/mol S (I 2 (g)) = ; S (H 2 (g)) = S (HI(g)) = all in J/mol K I 2 (g) + H 2 (g) 2 HI(g) G = RT lnk K = exp{- G /RT} We need G in order to get K, so... H = 2[ H f (HI(g))] H f (I 2 (g)) H f (H 2 (g)) = 2[26.48] = 9.48 kj S = 2[S (HI(g))] S (I 2 (g)) S (H 2 (g)) = 2[206.59] = J/K G = H T S = 9.48 (298)(.0218) kj G = kj I 2 (g) + H 2 (g) 2 HI(g) G = RT lnk K = exp{- G /RT} We found: G = kj/mol K = exp{ J/(8.314 J/K)(298 K)} K = (no units) I 2 (g) + H 2 (g) 2 HI(g) I 2 (g) + H 2 (g) 2 HI(g) K = But, what is this equal to? The equilibrium constant must be expressed using pressures for gases when obtained directly from thermodynamic data. K P = = (P HI ) 2 /(P I2 ) K P = = (P HI ) 2 /(P I2 ) convert to concentrations, K C? use P = (n/v)rt = [conc.] RT (P HI ) 2 /(P I2 ) = [HI] 2 /[I 2 ][H 2 ] (RT) 2 /(RT)(RT) in this case K P and K C are numerically equal This is not true generally (when n 0)