125ch15b(1)

# 125ch15b(1) - 15 15.23 Find K for a reaction that is the...

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Unformatted text preview: 15 15.23) Find K for a reaction that is the sum of two reactions. A (aq) + B (aq) W K1 = 1.9 x 10-4 C (aq) C (aq) + D (aq) W E (aq) + A (aq) ----------------------------------------------------B (aq) + D (aq) W E (aq) K2 = 8.5 x 102 K3 If two rxn’s (eqn’s) can be added to give a 3rd rxn (eqn) then K3 is the product of the K1 and K2. Rx1 + Rx2 = Rx3 K1 * K2 = K3 [C] [E] [A] [E] ---------- * ---------- = ---------[A] [B] [C] [D] [B] [D] K3 = K1 * K2 = (1.9 x 10-4) (8.5 x 102) = 0.1615 = 0.16 ([C] and [A] cancel out) 16 15.25) 2 Hg2O (s) º 4 Hg (R) + O2 (g) What would Kc look like if you included everything (including pure solid and liq)? [Hg]4 [O2] Kc = -------------[Hg2O]2 However, we normally exclude pure solids and liquids. Their molar conc. are essentially constant so their conc. are incorporated into the equilibrium constant. Kc [Hg2O]2 K = --------------- = [O2] [Hg]4 ’ c Kc = [O2] a) The same thing is done for Kp (ignore solids and liquids). Kp = PO 2 b) As noted above, the molar concentrations for solids and liquids are constant. Their concentration is a ratio of moles to volume occupied (which is related to their density). This doesn’t change whether you have a small or large amounts. NOTE: PURE SOLIDS and LIQUIDS do NOT appear in the expression for K or Q. (Aqueous and gaseous substances should be in K or Q.) 17 15.27) 2 HI (g) º H2 (g) + I2 (g) At 425°C, [HI] = 3.53 x 10&3 M [H2] = 4.79 x10&4 M [I2] = 4.79 x 10&4 M [H2] [I2] (4.79 x10&4) (4.79 x10&4) Kc = ------------- = -------------------------------- = 0.018 41 = 0.0184 [HI]2 (3.53 x 10&3)2 18 15.30) 15.32) Given initial conc. and asked for K Set up an Equilibrium Table (ICE table) This table should be done in Molarity (M) or Pressure (atm) If you use moles in the table you must convert the moles to conc. (M) or pressure (atm) before substituting into the expression for K. * See Next Page * 19 15.32) (cont.) 20 15.34) 21 15.36) 22 15.38) 23 15.39) SO2Cl2 (g) º SO2 (g) + Cl2 (g) [SO2] [Cl2] Kc = ---------------- = 0.078 [SO2Cl2] at 100°C Given Kc and molarities of SO2Cl2 and SO2. Want partial pressure of Cl2. [SO2] = 0.052M [SO2Cl2] = 0.108 M Easiest thing to do is to find conc. of Cl2, [Cl2], and then its partial pressure. [SO2Cl2] Kc (0.108) (0.078) [Cl2] = ---------------- = --------------------- = 0.162 M [SO2] [Cl2] (0.052) Use Ideal Gas Law, PV = nRT PCl = (n/V) RT = (M) RT = [Cl2] RT = (0.162 M)(0.08206 atm/MCK)(373K) 2 = 4.958 atm = 5.0 atm 15.41) 24 15.41) (cont.) b) H2 (g) + I2 (g) º 2 HI (g) Kc = 55.3 at 700 K An equilibrium mixture in a 2.00-L flask has 0.056 g H2 and 4.36 g I2. What is the mass of HI in the flask? 0.056 g H2 1 mol H2 [H2] = --------------- × ------------------ = 0.013889 M = 0.014 M 2.00 L 2.01588 g H2 4.36 g I2 1 mol I2 [I2] = ------------- × -------------------- = 0.008589 M = 0.00859 M 2.00 L 253.80894 g I2 [HI]2 Kc = -----------[H2] [I2] [HI] = {Kc [H2] [I2]}1/2 [HI] = {(55.3) (0.013889) (0.008589)}1/2 [HI] = 0.081223 = 0.081 M 0.081223 mol HI 127.9124 g HI ? g HI = 2.00 L × ----------------------- × -------------------1 L soln 1 mol HI = 20.779 = 21 g HI 25 15.44) 26 15.46) Solid NH4HS is placed in a closed container and decomposes until equilibrium is reached. What are the concentrations of NH3 and H2S at equilibrium? NH4HS (s) W NH3 (g) + H2S (g) initial C 0M 0M change -x +x +x ------------------------------------------------------------------------------equil constant +x +x Kc = [NH3] C [H2S] = 1.2 x 10-4 x2 = 1.2 x 10-4 NH4HS (s) doesn’t appear in K (pure solids and liquids do not appear in K) x = (1.2 x 10-4)1/2 = 1.0954 x 10-2 M [NH3] = [H2S] = 1.1 x 10-2 M 15.47) a) Solid CaSO4 is placed in a closed container and decomposes until equilibrium is reached. What are the concentrations of Ca2+ and SO42& at equilibrium? CaSO4 (s) W Ca2+ (aq) + SO42& (aq) initial C 0M 0M change -x +x +x ------------------------------------------------------------------------------equil constant +x +x Kc = [Ca2+] C [SO42&] = 2.4 x 10-5 x2 = 2.4 x 10-5 CaSO4 (s) doesn’t appear in K (pure solids and liquids do not appear in K) x = (2.4 x 10-5)1/2 = 4.8989 x 10-3 M [Ca2+] = [SO42&] = 4.9 x 10-3 M 27 15.47) (cont.) b) What is the minimum mass of CaSO4(s) needed to reach equil. in 3.0 L of soln.? A saturated soln of CaSO4(aq) is 4.9 x 10-3 M (this is the dissolved amount of CaSO4, which is [Ca2+] = [SO42&]) 4.9 x 10-3 mol 136.14 g CaSO4 ? g CaSO4 = 3.0 L soln × ------------------- × ---------------------- = 2.0 g CaSO4 1 L soln 1 mol CaSO4 15.49) I2 (g) + Br2 (g) º 2 IBr (g) Kc = 280 at 150°C The reaction is started with 0.500 mol IBr in a 1.00-L flask. Since no I2 or Br2 are present initially the reaction proceeds in the reverse direction (right to left) toward the reactants. Need to set up an ICE table. Since the reaction goes in the reverse direction the negative x is on the right-hand side of the table and the plus x’s go on the left-hand side. [IBr] = 0.500 mol/1.00 L = 0.500 M I2 + Br2 W 2 IBr initial 0.000 0.000 0.500 change +x +x & 2x -------------------------------------------------------------equil x x 0.500 & 2x [IBr]2 (0.500 & 2x)2 (0.500 & 2x)2 Kc = ------------- = ------------------- = ------------------ = 280 [I2] [Br2] (x) (x) x2 Even though the reaction is going right to left (reverse direction), you still write Kc for the balanced equation as written. That is the equation the given Kc refers to. * continued on next page * 28 15.49) (cont.) (0.500 & 2x)2 ------------------ = 280 perfect square (take square root of both sides) x2 (0.500 & 2x) ----------------- = (280)1/2 = 16.733 x 0.500 & 2x = (16.733) (x) 0.500 = (16.733) x + 2x 0.500 = (18.733) x x = 0.02669 M [I2] = [Br2] = 0.027 M [IBr] = 0.500 & 2 (0.02669) = 0.44661 M = 0.447 M ...
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