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NAME: AMODU OLUWATOYIN ABIGAIL. COLLEGE: MEDICINE AND HEALTH SCIENCES. DEPARTMENT: MEDICINE AND SURGERY. COURSE CODE: CHEMISTRY 101. 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=1.8×10-5). SOLUTION: NaOH + CH3COOH → CH3COONa +H2O 0.03 0.15 0.15 (before reaction) - 0.03 0.030.03+ (change due to reaction) 0 0.12 0.18 0.15M of CH3COONa 0.15M of CH3COOH 0.03 mole of NaOH PH = pKa + log [conjugate base]/[Acid] PKa = -log Ka = - log 1.8×10-5 = 4.74. PH =4.74 + log (0.18)/(0.12) = 4.92 ΔH = 4.92-4.74 = 0.18. 2(a). the rate constant of a first order reaction is 2.5×10-6/s and the initial concentration is 0.1 moldm-3, what is the initial rate in moldm-3s-1. K =2.5×10-6/sec Concentration = 0.1moldm-3 Initial rate(R) =? R=K[] 2.5×10-6 × 0.1 = 2.5× 10-7moldm-3s-1. 2(b). The initial rate of a second order reaction is 5× 10-7moldm-3s-1, and the initial concentration of the two reacting substances are each 0.2moldm-3. What is the rate constant in dm3mol-1s-1? R2 = 5.0 × 10-7moldm-3s-1 C1 =0.2 moldm-3 C2 = 0.2 moldm-3 R=K[][] 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 × 10-5moldm-3s-1. 2c. ki= 1.30/year a=5x10^-3moldm-3 t= 30 days= 0.082years. R=k1(a-x) in a/a-x=k1t 2.303loga/a-x=k1t loga/a-x= 1.30x0.082/2.303 loga/a-x=4.63x10^-2 a/a-x= antilog of 4.63x10^-2 a/a-x= 1.112 5x10^-3/a-x =1.112 a-x = 4.5x10^-3moldm^-3 ii. a=5x10^\3 a-x=1x10^-3 t=? 2.303log a/a-x =k1t 2.303x log 5x10^-3/1x10^-3=130t 2.303log5=130t 1.61= 1.30t t= 1.61/1.30 t=1.23year-1 3. 5.0 × 10-7 = K × 0.2 × 0.2 K= 1.25× 10-5moldm-3sec-1. 3H2 + N2 → 2NH3 Kc =[NH3]/[H2]^3 [N2] = [0.24]^2/[0.65]^3 [0.52] Qc = 0.4033 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: kp=kcRT^(c+d)-(atb) Kp = 0.002916×8.314×(500)^2+1-2 Kp =12.12 4b. pCl = [Cl2]RT^(c+d)-(a+b) = 0.09×8.314×500 = 374.13pa. 5a. NH4Cl → NH3 +HCl KC = [NH3] [HCl]/[NH4Cl] Kp=p[MH3] p[HCl]/P[NH4Cl] B. CO2 +2NH3 → CO(NH2)2 +H2O KC = [CO(NH2)2] [H2O]/ [CO2] [NH3]^2. 6a. t1/2 37.2mins (37.2 × 60) = 2232 secs. Decay constant(K) ? t1/2 = 0.693/k k = 0.693/t1/2 k = 0.693/ 2232 =3.1 × 10^-4s-1. 6b. k = 10µc = 10 × 10^-6 No of atoms(n) =? Avogadro’s mole constant = 6.02 ×10^23. N=k× Avogadro’s constant N =10× 10^-6 ×6.02 ×10^23 = 6.02 ×10^18. 7. DIFFERENCE BETWEEN IONIC AND COVALENT BONDING IONIC BONDING They are formed by sharing of electrons. Bonds are formed usually between non-metals They are insoluble in polar solvent They do not dissociate into ions when dissolved in water They have low melting and boiling point COVALENT BONDING They are formed by transfer of electrons Bonds are formed between metals and non-metals They are soluble in polar solvent They dissociate into ions when dissolved in water They are of high melting and boiling point. 8. 2hydroxybenzoic acid + Ethanoic anhydride → aspirin + ethanoic acid 3.0g of hydroxybenzoic acid → 6.5g of anhydride Before reaction 3.0g → 6.5g of anhydride After reaction 3.0- 3.0 → 6.5- 3.0 0g 3.5g A. the limiting agent is hydroxybenzoic acid. B. 3.5g of the reagent was left behind after the reaction. b. 2C2H2 + 5O2 → 2H20 +4CO2. 2 mole of C2H2 = 4 mole of CO2. Number of moles of C2H2 combusted = 130/52 = 2.5 Numbers of moles of CO2 produced = Mass of C02 produced/mass of C02 =2.5 Mass of CO2 produced = 176×2.5 =440g. 8c.(i). 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 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. ii. 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. iii. 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.