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Download 2 Chemical equilibrium occurs when a reaction and its reverse
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Chemical equilibrium occurs when a reaction and its reverse proceed at the same rate. 2 At equilibrium, the forward and reverse reactions proceed at the same rate. 3 At equilibrium the amount of each reactant and product is constant. 4 1 Depicting Equilibrium N2O4 (g) 2 NO2 (g) 5 The Equilibrium Constant Forward reaction: N2O4 (g) → 2 NO2 (g) Rate law: Rate = kf [N2O4] 6 The Equilibrium Constant Reverse reaction: 2 NO2 (g) → N2O4 (g) Rate law: Rate = kr [NO2]2 7 2 The Equilibrium Constant Therefore, at equilibrium Ratef = Rater kf [N2O4] = kr [NO2]2 kf [NO2]2 = kr [N2O4] 8 The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes: Keq = kf kr = [NO2]2 [N2O4] 9 The Equilibrium Constant In General: aA + bB cC + dD The equilibrium expression for this reaction would be: Kc = [C]c[D]d [A]a[B]b 10 3 Questions: What are the equilibrium expressions for these equilibria? 1. SnO2 (s) + 2CO (g) ↔ Sn (s) + 2CO2 (g) 2. CaCO3 (s) ↔ CaO (s) + CO2 (g) 3. Zn (s) + Cu2+ (aq) ↔ Cu (s) + Zn2+ (aq) 4. 2O3 (g) ↔ 3O2 (g) 5. 2NO (g) + Cl2 (g) ↔ 2NOCl (g) 11 The Equilibrium Constant NOTE: pressure is proportional to concentration for gases in a closed system: Kp = (PC)c (PD)d (PA)a (PB)b 12 Relationship between Kc and Kp From the ideal gas law we know that PV = nRT Rearranging it, we get P= n RT V L*atm / Where R is 0.0821 L*atm mol*K 13 4 Relationship between Kc and Kp Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes: Kp = Kc (RT)∆n Where ∆n = (moles of gaseous product) − (moles of gaseous reactant) 14 Questions 6. If, N2 + 3H2 ↔ 2NH3 Has a Kc = 9.60 at 300˚C, what is Kp? 7. If 2SO3 ↔ 2SO2 + O2 Has a Kc = 4.08*10-3 at 1000 K, what is Kp? 15 Equilibrium Can Be Reached from Either Direction The ratio of [NO2]2 to [N2O4] remains constant (within error) at this temperature no matter what the initial concentrations of NO2 and N2O4 are. 16 5 Data from the last two trials from the table on the previous slide. 17 18 What does the magnitude of the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. 19 6 Magnitude of the Value of K Mean? If K << 1, the reaction is reactant; reactant predominates at equilibrium. 20 The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. N2O4 (g) 2 NO2 (g) [NO ]2 2 NO2 (g) Kc = [N O2 ] = 0.212 at 100°C 2 4 [N2O4] N2O4 (g) Kc = [NO 2 2] = 1 0.212 = 4.72 at 100°C 21 Questions 8. N2 + O2 ↔ 2NO has a Kc of 1*10-30 at 25˚C, is this a good reaction if you intend to collect NO? 9. What is the value of Kc for the reverse reaction and what is the equilibrium expression? 22 7 If a reaction doubled, you square the constant. In general you raise the constant to the power of the coefficient: [NO ]2 N2O4 (g) 2 NO2 (g) Kc = [N O2 ] = 0.212 at 100°C 2 4 2 N2O4 (g) [NO ] 4 NO2 (g)Kc = [N O2 ]2 = (0.212)2 at 100°C 2 4 4 23 The equilibrium constant for a net reaction is the product of the equilibrium constants for the individual steps. Step 1: NO2 + NO2 → NO3 + NO Step 2: NO3 + CO → NO2 + CO2 Kc1= [NO2][CO2] [NO3][NO] Kc2= [NO3][CO] [NO2]2 [NO][CO2] Kc= [NO2][CO] 24 Questions 10. HF(aq) ↔ H+(aq) + F- (aq) Kc = 6.8*10-4 H2C2O4 (aq) ↔ 2H+ (aq) + C2O42-(aq) Kc = 3.8*10-6 Find Kc for 2HF(aq) + C2O42-(aq) ↔ 2F-(aq) + H2C2O4 (aq (aq)) 25 8 Summary 1. At eq. ∆[conc] stops but reactions don’t 2. At eq. the ratio of [prod] and [react] are equal to a coefficient (Kc) 3. Kc and Kp have no units 4. Kp = Kc(RT)∆n 5. Krev = (Kfor)-1 6. Kc of a reaction that has been multiplied is raised to that power. 7. Kc of combined reactions is the product of those reactions. 26 9