Lec 13 Feb 02 Mon

# Lec 13 Feb 02 Mon - 2.9(2 Small K(K ≤ 10−3 at...

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Unformatted text preview: 2.9 (2) Small K (K ≤ 10−3): at equilibrium, small amount of product and large amount of reactant. Reaction has small tendency to go forward. 0.50 atm of COCl2 dissociates according to: COCl2 (g) ⇔ CO (g) + Cl2 (g) K = 8.3 × 10−4 at 360 °C Find equilibrium partial pressures for all species. Step 1. Calculate Q Q = 0 because only reactants are initially present. Q < K, even though K is very small, reaction goes Step 2. Make up i.c.e. table. i. c. e. COCl2 (g) ⇔ 0.500 -x 0.500 - x CO (g) + 0 +x x Cl2 (g) 0 +x x Step 3. Substitute into K and solve for x to determine equilibrium pressures. K = 8.3 ! 10 "4 = (x)(x) 0.500 " x *** Before solving the quadratic, think about simplifying. IF K is small, x will be small Approximate: 0.500 − x ≅ 0.500 Now it’s easier, x2 K = 8.3 ! 10 = 0.500 "4 x = 2.0 ! 10 -2 atm *** Check approximation, if off by less than 5 %, then OK. If not, then go back to solve quadratic. change error = ! 100% initial 2.0 ! 10 -2 ! 100% = 4% 0.500 OK PCOCl2 = 0.500 - 0.020 atm = 0.480 atm PCO and PCl2 = 0.020 atm 2.10 (3) Large K (> 103 approximately): at equilibrium, large amount of product and small amount of reactant. Reaction goes almost to completion. At 200°C, K = 3.0 × 106 for the following reaction: 2 NO (g) + O2 (g) ⇔ 2 NO2 (g) Initially : PNO = 0.100 atm, PO2 = 0.060 atm Calculate equilibrium pressures Step 1. Calculate Q No products initially, therefore Q = 0 Q < K, Reaction will go Step 2. If you set up i.c.e. table directly, you either won’t be able to solve for x or x will be so large that you won’t be able to determine reactants remaining -- let reaction go to completion since K is so large -- then let reverse reaction (with small K) establish equilibrium (back reaction) Make up table for complete reaction. Then back rxn. Stoichiometry!! 2 NO (g) + O2 (g) → 2 NO2 (g) i. 0.100 0.060 0 c. -0.100 -0.050 +0.100 completion 0 0.010 atm 0.100 atm Back rxn +2x +x - 2x Equil. 2x 0.010+x 0.100-2x This is a limiting reactant problem. NO is limiting reactant. Now, Q =∞, Q > K, rxn← Krev = 1/K = 3.3 x 10-7, since Krev is small, x in this direction will also be small so you can make approximations. Step 3. Substitute into original K and solve for x by making approximation: (0.100 " 2x) K = 3.0 ! 10 = 2 (2x) (0.010 + x) 2 6 0.100 −2x ≅ 0.100 0.010 + x ≅ 0.010 2.11 (0.100) K = 3.0 ! 10 " 2 (2x) (0.010) 2 6 If you want full marks, you always have to check approximations : 2(2.9 ! 10 "4 ) 0.100 ! 100% = 0.6% OK OK 2.9 ! 10 "4 ! 100% = 3% 0.010 PNO2 = 0.100 - 2x = 0.100 atm PNO = 2x = 5.8 × 10−4 atm PO2 = 0.010 + 2.9× 10−4 = 0.010 atm 4. Acid and Base Equilibria (17.1 - 17.5) See handout # 2 for identiﬁcation of strong and weak acids/bases Know how to write out weak acid/base equilibria Know how to rank acids (using Ka) and bases (using Kb) Acids: the larger Ka, the more H+ is released, the stronger the weak acid All conjugates of weak acids are to be treated like weak bases A. How do determine if a salt dissolved will behave as an acid/base in solution 1. If both ions are conjugate to a strong acid or a strong base, then both ions are very weak acids/bases. Ions remain as ions in solution. The result is a neutral pH Examples: NaCl, KNO3, LiClO4 … 2. If one ion has very weak properties, and other ion is a weak acid/base, then only the weak species is active and determines pH. Examples: NaNO2, KF, NH4Br, LiCH3CO2 … 3. If both ions are weak acid/base, the pH is determined by the species with the largest equilibrium constant. Examples: NH4CH3CO2, NH4NO2, (CH3)NH3ClO … 2.12 B. Calculating equilibrium concentrations of acids and bases e.g. Calculate the pH, pOH, and the percent dissociation at equilibrium for a 1.0 M ClCH2COOH solution (pKa = 2.85). Step 1. Determining Q and K Q = 0, Ka = 10-2.85 = 1.4 x 10-3 , Q < K, rxn → (pKa = -logKa) Step 2. Small K problem, ice table then approx. I C E ClCH2COOH (aq) + H2O (l) ⇔ ClCH2COO− (aq) + H3O+ (aq) 1.0 --0 0 -x --+x +x 1.0-x --x x x2 x2 1.4 #10 = ! 1.0 " x 1.0 "3 Step 3. Sub into K, and solve for x x = [H+] = 0.037 M 0.037 !100% = 3.7% 1.0 Check approximation : pH = log[H+] =1.43 Percent dissociated = pOH = 14 - pH =12.57 amount dissociated !100% initial amount 3.7 !10 " 2 M !100% = 3.7% 1.0M Generally, -- when determining equilibrium pH, make an ice table for the weak acid or base, using its K value (provided or data sheet) -- if pH at equilibrium is known, this gives you the [H+] at the bottom of your i.c.e table, using the K value you can determine the initial concentration of the weak acid or base ...
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