CHM 1220 In-Class Activity: Review of Kinetics and Equilibrium Fall 2017 ANSWERS 1. For the phase transition H2O(l) ⇌ H2O(g) at 298 K, ΔH° = 44.0 kJ/mol and ΔS° = 119 J/mol·K. What is the value of ∆G at 298 K when the partial pressure of water is 19.2 torr? 2. At 2000C, the value of Kc for the reaction, N2(g) + O2(g) ⇌ 2NO(g) is 4.10×10-4. What is the value of Kc at 75C if ΔH = 180.6 kJ/mol? 3. Some reactions are so rapid that they are said to be diffusion-controlled; that is, the reactants react as quickly as they can collide. An example is the neutralization of H3O+ by OH-, which has a second-order rate constant of 1.3×1011 M-1s-1 at 25 °C. If equal volumes of 2.0 M HCl and 2.0 M NaOH are mixed instantaneously, how much time is required for 99.999% of the acid to be neutralized? H3O+(aq) + OH(aq)  2 H2O( ) Relationship for ∆G and partial pressures: Plug-in vapor pressure and ∆Gºrxn: Calculate ∆Gºrxn: 1111103.1 sMkRate law: Under initial conditions: So, Integrated Rate law: Where, Need time to reach 99.999% consumption: 0.001% remains Then: Technically…since the solutions will be mixed, initial concentration of [H3O+] = 1.0 M needs to be in atm, not torr 1 atm = 760 torr    033OH1OH1 kt  nssMsMt-- 76910 69.7 0.1 103.1999997-1111   033OH00001.0OHT1 = 2000°C = 2273 K    0303OH1OH00001.01kt    030303OH99999OH1OH00001.01kkt   OH OHrate3k   003OHOH 23OHrate kK1 = 4.10 × 10-4 QRTGG lnovapvap OHovapvap2ln PRTGG T2 = 75°C = 348 K 1OH2PQ ovapovapovapSTHG   KJ/mol 119K 298J/mol10443ovapGJ/mol 1054.83ovapG   atm 0253.0lnK 298KmolJ314.8molJ 1054.83vapGkJ/mol 570.0molJ 1070.5molJ 1011.9molJ 1054.8233vapGK2 = ? 12o1211lnTTRHKK 4321010.4lnK 22731K 3481KmolJ314.8J 106.180lnK80.786.52ln2K277.6021052.4 eKN2(g) + O2(g) + 180.6 kJ ⇌ 2NO(g) By lowering the temperature, move toward the heat term, results in shift of equilibrium to the left and a smaller KcCHM 1220 In-Class Activity: Review of Kinetics and Equilibrium Fall 2017 4. Hydrofluoric acid, HF, dissociates in water according to the following equilibrium reaction: HF(aq) ⇌ H+(aq) + F-(aq) Kc = 3.5  10-4 What is the concentration of H+ in a 0.0100 M solution of HF? 5. The light-stimulated conversion of 11-cis-retinal to 11-trans-retinal is central to the vision process in humans. This reaction also occurs (more slowly) in the absence of light. At 80 °C in heptane solution, the reaction is first order with a rate constant of 1.02 × 10-5 s-1. (a) What is the molarity of 11-cis-retinal after 6.00 h if its initial concentration is 3.50 × 10-3 M? (b) How much time does it take for 25% of the 11-cis-retinal to react? Use the quadratic formula to solve for x: × Rate law First-order integrated rate law Calculate time to reaction to proceed to 25% reacted: 75% remains First-order integrated rate law 4105.3][HF]][F[HcK   4105.301.0 xxx0105.3105.3105.3105.3642462xxxxaacbbx24202 cbxax    12105.314105.3105.36244 x MMx3334101.70or 1071.121075.3105.3Mx31070.1]H[retinal][Rate  cisk0retinal]ln[retinal][ln  cisktcis   Msscis3151050.3ln 600,211002.1retinal][ln875.5retinal][ln cisMcis31081.2retinal][0retinal][75.0retinal][  ciscis0retinal]ln[retinal][ln  cisktcis 00retinal]ln[retinal][0.75ln  cisktcis kt.750lnh 83.7 204,281002.1)75.0ln(15sstCHM 1220 In-Class Activity: Review of Kinetics and Equilibrium Fall 2017 6. Calculate the value of the equilibrium constant Kc for the following reaction: 2CH3OH(g) + H2(g) ⇌ C2H6(g) + 2H2O(g) from the following information: Show your thought-process. 7. The oxidation of iodide ion by hydrogen peroxide in an acidic solution is described by the balanced equation: Initial rate data obtained at 25 °C and a constant [H+] are listed in the following table: Experiment [H2O2] (M) [I-] (M) Initial Rate (M/s) 1 0.100 0.100 1.15 × 10-4 2 0.100 0.200 2.30 × 10-4 3 0.200 0.100 2.30 × 10-4 4 0.200 0.200 4.60 × 10-4 What is the rate law for this reaction at 25 °C, including the value of the rate constant? H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) → I3- (aq) + 2 H2O (l) 2CH4(g) ⇌ C2H6(g) + H2(g) 2CH3OH(g) + 2H2(g) ⇌ 2CH4(g) + 2H2O(g) 2CH3OH(g) + H2(g) ⇌ C2H6(g) + 2H2O(g) no change 2CH4(g) ⇌ C2H6(g) + H2(g) Kc = 9.5 × 10-13 CH4(g) + H2O(g) ⇌ CH3OH(g) + H2(g) Kc = 2.8 × 10-21 flipped and multiplied by 2 reciprocal and squared no change add together multiple together From Experiments 1 & 2: [H2O2] = constant rate doubles, [I-] doubles first order with respect to [I-] From Experiments 1 & 3: [I-] = constant rate doubles, [H2O2] doubles first order with respect to [H2O2] Using data from any experiment: Rate law:   11214221015.1 1.0 1.01015.1]][IOH[Rate sMMMsMk221K1K2211KKKc 29221131021.1108.21105.9 cK]][IOH[Rate22