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NAME: AKINPELUMI SAMUEL .O. DEPARTMENT: ANATOMY COLLEGE: MEDICAL AND HEALTH SCIENCE CHEMISTRY 101 15/MHS03/004 1. Calculate the change in PH obtained on the addition of 0.03 mole of solid NaOH to a buffer solution that consists of 0.15M sodium acetate and 0.15M acetic acid solution, if we assume that there is no change in volume (Ka = 1.8 x 10-5). 2. (a) The rate constant of a first order reaction is 2.5 ×10 -6/s and the initial concentration is 0.1moldm-3, what is the initial rate in moldm-3s-1. (b) The initial rate of a second order reaction is 5.0×10-7moldm-3s-1, and the initial concentrations of the two reacting substances are each 0.2moldm-3.What is the rate constant in dm3mol-1s-1? (c) The first order rate constant for the decomposition of a certain antibiotic in water at 250C is 1.30year-1. (i) If a 5.0×10-3moldm-3 solution of this antibiotic is stored at 250C for 30 days, what is the concentration of the antibiotic? (ii) How long will it take for the concentration of the antibiotic to reach 1.0×10-3 moldm-3? 3. A reaction mixture at 800K consists of 0.65 M H2, 0.52M, N2 and 0.24 M NH3. Calculate the concentration reaction quotient QC at this temperature for the formation of Ammonia. 4. At 500K 1 mole of NOCl was found to be 9.0% dissociated in a 1.00L vessel. 2NOCl (g) 2NO (g) +Cl2 (g) Calculate the pressure equilibrium constant Kp for the reaction at this temperature. What is the partial pressure of chlorine in the vessel? 5. Write the expressions for the concentration equilibrium constant Kc and pressure equilibrium constant Kp for the following reactions: a) NH4Cl(s) <------> NH3 (g) + HCl (g) b) CO2(g) + 2NH3 <-------> CO(NH2)2 + H2O(l) 6. The half-life of 38Cl is 37.2 mins; determine (i) the decay constant of 38Cl and (ii) the number of atoms of radioactive 38Cl in a sample that has an activity of 10 µc 7. Differentiate between ionic and covalent compounds. 8. (a) 3.0g of 2-hydroxybenzoic acid is reflux with 6.5g of ethanoic anhydride to produce aspirin and ethanoic acid. (I) Which one of the two reagents is the limiting reagent? (II) How much of excess reagent (in grams) is left at the end of the reaction? (b) 130g of ethyne was combusted in the presence of excess oxygen. Calculate the mass of CO2 produced in the process. (c) Discuss the three factors affecting ionization energy in the periodic table. 1. CH3COONa= 0.15(sodium acetate) CH3COOH= 0.15(acetic acid) Ka= 1.8 x 10-5 Mole added=0.03 pH= pKa + log [Conjugate base] [Acid] = 4.74 + log [0.15] [0.15] = 4.74 NaOH + CH3COOH-> CH3COONa + H2O Before 0.03 0.15 0.15 reaction Change -0.03 -0.03 +0.03 due to reaction After 0 0.12 0.18 reaction CH3COONa= 0.18M CH3COOH= 0.12M pH= pKa+ log [Conjugate base] [Acid] = 4.74 + log [0.18] [0.12] = 4.74 + 0.18 = 4.92 Change in pH= 4.92 - 4.74= 0.18 2. a) First Order reaction R = K [A] 1 R = 2.5 × 10-6 × 0.1 R = 2.5 × 10-7mol/dm3s b) R = K [A] 1[B] 1 Second Order Reaction K= R ÷ ([A] [B]) K = 5.0 × 10-7 ÷ (0.2 × 0.2) K = 5.0 × 10-7 ÷ (0.04) K = 1.25 × 10-5 c) K= 1 ln a t (a-x) 4.12 × 10-8 = (2,592,000)-1 × (log a ÷ (a-x)) (ln a ÷ (a-x)) = 4.12 × 10-8 × 2,592,000 (ln 0.005 ÷ (0.005-x) = 0.1068 (-2.9957 ÷ (0.005-x) = 0.1068 0.005 – X = -2.9957 ÷ 0.1068 X = -28.0496 - 0.005 X = -28.0546 ÷ -1 X= 28.0546mol/dm3 3. N2 + H2 NH3 [N2]= O.52M [H2] = 0.65M [NH3] = 0.24M QC = [NH3] [H2] [N2] QC = [0.24] [0.65] [0.52] QC = [0.24] [0.338] QC = 0.71 4. 2NOCl (g) 2NO+(g) + Cl2-(g) Recall: n = CV C=n÷v C=1÷1 C=1 But only 9% of NOCl dissociated which is equal to 9% of 1 = 0.09 2NOCl(g) <-> 2NO+(g) + Cl2-(g) 1M <-> 2(0.09) 0.09 KC = [NO+]2 [Cl2-] [NOCl]2 KC = (0.18)2(0.09) ÷ 12 KC = 0.002916 ÷ 1 KC = 0.002916 Recall: Where Kc= 0.002916, R= 8.314, T= 800, where (c+d) (a+b) are the order of reaction KP = KC RT (c+d) – (a+b) KP = 0.002916 × 8.314 (800) (2+1) – (2+0) KP = 0.002916 × 8.314×800 KP = 19.395 b) pCl = [Cl]RT(C+D)-(a+b) pCl = 0.09 × 8.314 × 800 pCl = 598.608 Pa 5. NH4Cl(s) <-> NH3 (g) + HCl (g) Kc = [NH3] [HCL] [NH4Cl] KP = PNH3 × PHCl PNH4Cl CO2 (g) + 2NH3 (g) <-> CO (NH2)2(g) + H2O (l) Kc = [CO (NH2)2] [H2O] [CO2] [NH3]2 Kp = PH2O* PCO (NH2)2 PCO2 × P (NH3)2 6. Data: = 2,232 Recall: K = 3.105 × 10-4 b) Recall = -KN N = ÷ -K N = -0.0322 7. Electrovalent Compounds Covalent compounds Involves transfer of electrons from metal atom to a Involves sharing of electrons either between non-metal atom. atoms of same or different element. Formed between metals Formed between non-metals High boiling and melting point Low melting point Soluble in polar solvents like water Soluble in non-polar solvents They conduct electricity in molten state Most covalent compounds do not conduct electricity. 8. a) (i) C7H6O3 + C4H6O3 C9H8O4 + CH3COOH nC7H6O3 = = 0.02172mol nC4H6O3 = = 0.06367mol Mole ratio= 1 From the values given, = 2.931 Since 2.9 is greater than 1, nC4H6O3 is in excess of nC7H6O3 Hence, C7H6O3 is the limiting reagent (ii) Excess nC4H6O3 = (2.931- 0.02172) mole = 2.9mole Excess C4H6O3 = 2.9 × 102.09 = 296.061g b) 2C2H2 +5O2 4CO2 + 2H2O nC2H2 = 5 Molar ratio of C2H2 to CO2 is 2: 4 Therefore nCO2 = 10 But mass = molar mass × number of moles of CO2 Mass of CO2 = 44 × 10 = 440g c) Factors affecting Ionization Energy 1. Distance of outermost electron from the nucleus: Across the period, as atomic number increases, atomic radius decreases. As the distance decreases, the attraction of the positive nucleus for the electron will increase. More energy is needed to remove the outermost electron, thus the ionization energy increases. 2. Size of the positive nuclear charge: As the nuclear charge increases, its attraction for the outermost electron increases, and so more energy is required to remove the outermost electron. Hence, the ionization energy increases. 3. Shielding and Screening effect of the inner electrons: Down a group, the shielding of outer electrons by inner electrons overcomes the influence on the increasing nuclear charge, thus the outer electron is screened or shielded from the nucleus by the repelling effect of the inner electrons. Across the group, the reverse is the case; the increasing nuclear charge has greater effect. In general, the screening effect by inner electrons is more effective, the closer they are to the nucleus.