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1301 Dynamic Equilibrium, Keq, and the Mass Action Expression 1301 Dynamic Equilibrium, Keq, and the Mass Action Expression The Equilibrium Process Dr. Fred Omega Garces Chemistry 111 Miramar College 1 Equilibrium 05.2015 Concept of Equilibrium & Mass Action Expression Extent of a Chemical Reaction Many reactions do not convert 100% of reactants to products. There is often a point in a reaction when the products will back react to form reactants. The extent of the reaction i.e., 20% or 80%, can be determined by measuring the concentration of each component in solution. In general the extent of the reaction is a function of: Temperature, Concentration and degree of organization which is monitored by some constant value called the - The equilibrium constant (Keq). 2 Equilibrium 05.2015 (Dynamic) Equilibrium Chemical Equilibrium is a dynamic state in which the rates of the forward and the reverse reaction are equal. A teeter-totter is an analogy of a system in balance. The left is balanced with the right. In a chemical reaction this means that the Left (reactant) is changing at the same rate as the right (products). Reactant Product 3 Equilibrium 05.2015 How Equilibrium is achieved Consider the reaction A → B Forward A → B rate = kf[A] Reverse B → A rate = kr[B] Overall A D B rate forward (kf[A]) = rate reverse (kr[B]) Example: N2O4 + E D 2 NO2 Effect of temperature on the N2O4/NO2 equilibrium. The tubes contain a mixture of NO2 and N2O4 . As predicted by Le Chatelier’s principle, the equilibrium favors colorless N2O4 at lower temperatures, but shifts to the darker brown NO2 at higher temperature. 4 Equilibrium 05.2015 Equilibrium: Mass Action Expression When equilibrium is establish A D B That is rate forward = rate reverse or kf[A] = kr[B] Rearranging this equation can be rearrange to kf/kb = [B] /[A] The mass action expression defines the equilibrium constant (Keq): Keq = [B] / [A] For any general chemical process at equilb. aA + bB D pP + qQ A mass action expression can be written: p q P • Q K = [ ] [ ] eq a b [ A ] • [ B ] This is also referred to as the Law of Mass Action 5 Equilibrium 05.2015 Heterogeneous Systems (1) Substances are in different phases at equilibrium i.e., solid and aqueous. Which solid is more concentrated? 100 g (1 cup) 200 g (2 cup) Concentration of a solid (and pure liquid) is always a constant; The concentration does not change ! 6 Equilibrium 05.2015 Heterogeneous Systems Consider the following reaction: CaCO3(s) D CaO (s) + CO2 (g) CaO • CO [ ] [ 2] Keq = CaCO [ 3] but, [CaO] = constant (solid) and CaCO = constant (solid) [ 3] Constant • CO [ 1 ] [ 2] Keq = Constant [ 2] K = K = CO eq p [ 2] 7 Equilibrium 05.2015 € Haber Process (D signify that the reactants are not N2(g) + 3H2 (g) D 2NH3 (g) completely (100%) converted to products. As soon as NH3 is form, some of it decomposes and form N2 and H2 the starting chemical. This takes place until the amount consumed is equal to the amount produced, i.e., equilibrium is reached. - 1.0 - -.60 - Law of Mass Action: -.20 2 !NH # " 3$ K = = K eq 3 p !N #•!H # " 2$ " 2$ 8 Equilibrium 05.2015 Meaning of Keq Which is favored: Reactant or Product? p q x [P ] • [Q ] [Product ] K eq = a = y [ A ] • [ B ]b [ Reactant ] For Keq > 1, the ratio [Product]x > [Reactant]y For Keq = 1, the ratio [Product]x = [Reactant]y For Keq < 1, the ratio [Product]x < [Reactant]y For Keq > 1, at equilibrium Product is favored. For Keq = 1, at equilibrium Product and Reactant are equal. For Keq < 1, at equilibrium Reactant is favored. 9 Equilibrium 05.2015 Calculating Keq from reaction • The mass action expression is equal to the equilibrium constant. • Equilibrium constants are generally written without units. For Keq >> 1, at equilibrium Product is favored. For Keq < 1, at equilibrium Reactant is favored. 10 Equilibrium 05.2015 Equilibrium Constants: Keq, Kc, Kp, Ka, Kb, Ksp, Kf, Kw… Kc versus Kp: For reactions that undergoes reversibility, the efficiency of the reaction (extent to which the reaction forms product) is indicated by its equilibrium constant, Keq. If the equilibrium constant (Keq) express through its mass action expression is based on molar concentration, then the equilibrium constant is written as Kc. If the equilibrium constant (Keq) express through its mass action expression is based on partial pressure, then the equilibrium constant is written as Kp. Ksp: If the rxn is a solid dissolving to ions, i.e., solubility of a salt, then Keq is written as Ksp. Kf: If the reaction is a formation of a transition complex, then Keq is written as Kf. Ka: If the reaction involves an acid losing a proton, then Keq is written as Ka. Ka2: If the rxn involves an diprotic acid losing its second proton, then Keq is written as Ka2. Ka3: If the reaction involves a polyprotic acid losing third proton, then Keq is written as Ka3 . Kb: If the reaction involves an base, then Keq is written as Kb. Kw: If the reaction involves water, then Keq is written as Kw. 11 Equilibrium 05.2015 Keq: Kc versus K p If the equilibrium constant (Keq) express through its mass action expression is based on molar concentration, then the equilibrium constant is written as Kc. If the equilibrium constant (Keq) express through its mass action expression is based on partial pressure, then the equilibrium constant is written as Kp. −Δn +Δn K c =K p(RT) K p =K c(RT) [RT is positive exponent (+Δn)] where Δn = ∑(gas moles) − ∑(gas moles) € € product reactant 12 Equilibrium 05.2015 € Derivation Between Kc and Kp Consider the Reaction: COCl2 (g) D CO (g) + Cl2 (g) [CO] [Cl ] CO * Cl K = 2 or (p ) (p 2 ) c KP = [COCl2] COCl (p 2 ) It can be shown that: € € −Δn +Δn K c =K p(RT) K p =K c(RT) [RT is positive exponent (+Δn)] where Δn = ∑(gas moles) − ∑(gas moles) € € product reactant 13 Equilibrium 05.2015 € 1. Calculating Keq: Kc - Kp relationship Carbon dioxide is placed in a 5.00-L high-pressure reaction vessel at 1400.°C. Given the following reaction, CO (g) + 1/2 O2 (g) D CO2 (g) , calculate the equilibrium constant Kc and Kp after equilibrium is achieved with 4.95 mol of CO2 , 0.0500 mol of CO, and 0.0250 mol of O2 found in the container. T = 1400°C CO + 1 O ! CO (g) 2 2 (g) 2 (g) V = 5.0 L i Δ mol e 0.050 mol 0.025mol 4.95mol Vol = 5.0 L -2 -3 -1 [e] 1.00 ⋅10 M 5.00 ⋅10 M 9.90 ⋅10 M 9.90 ⋅10-1M [CO2 ] [ ] kc = 1/2 = 1/2 CO ∗ O -2 -3 €[ ] [ 2 ] [1.00 ⋅10 M]∗[5.00 ⋅10 M] 3 kc =1.40 •10 € 14 Equilibrium 05.2015 2. Calculating Keq iCe - method At 300°C, 0.600 mol of PCl5 is placed in a 100mL reaction vessel. During the course of the reaction PCl5 converts to PCl3 and Cl2 gas. At equilibrium the amount of PCl3 formed is 0.200mol. What is the equilibrium constant for the following reaction. PCl5(g) D PCl3 (g)+ Cl2 (g) Note: 0.600mol/0.100L = 6.00 M and 0.200mol/0.100L = 2.00M PCl ↔ PCl + Cl 5 (g) 3 ( g ) 2 ( g ) i 6.00 M 0 0 V = 100 mL Δ -X +X +X = 0.100L "e$ ? 2.00 ? # % 15 Equilibrium 05.2015 Animation: PCl5 D PCl3 + Cl2 PCl D PCl + Cl C 5(g) 3 (g) 2 (g) Cl Cl P C C Cl Cl Cl C Cl P P Cl Cl Cl Cl Cl Cl Cl P C Cl Cl Cl Cl Cl P C C Cl Cl Cl Cl Cl Cl P P C Cl Cl Cl Cl Cl Cl P Cl Cl PCl ↔ PCl + Cl 5 (g) 3 ( g ) 2 ( g ) i 6.00 M 0 0 V = 100 mL Δ -X +X +X = 0.100L "e$ ? 2.00 ? # % 4.00 2.00 16 Equilibrium 05.2015 Summary: Equilibrium constant • The mass action expression is equal to the equilibrium constant. • Equilibrium constants are generally written without units. For Keq >> 1, at equilibrium Product is favored. For Keq < 1, at equilibrium Reactant is favored. 17 Equilibrium 05.2015 LeChatelier’s Principle Changing the concentration or pressure of a substance in a reaction at equilibrium is creating a stress. The removal of something creates a “deficit stress”, and adding something creates an “excess stress.” Reactions will always respond to stress by replacing the deficit (making more of what was removed) or by conversion of the excess (consuming some of what was added). This description works regardless of which side of the reaction is changed. Stress/Deficit will be in the form of: (1) Concentration, (2) Pressure and (3) Temperature variation 18 Equilibrium 05.2015 LeChatelier Principle Methods of disturbing Equilibrium and LeChatelier response. Stress on Reactant Stress on Product Relief on Reactant Relief on Product Reactant D Product Equilibrium Revisited: Equilibrium - Situation in which changes occur at equal rates so no net change is apparent. LeChatelier Principle: A stress on the a system at equilibrium causes changes to the system to reduce the stress until a new equilibrium is established. 19 Equilibrium 05.2015 Reactant Change: Stress added Consider the affect on Equilibrium as a result of increasing the concentration of the reactant. Stress on the reactant Which direction does the reaction shift towards? i) A system at ii) A stress is equilibrium. added to the reactant.
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