RI_H2_CHEM_P3_ANS Prelim

Question 1 (a) Finely powdered Ca metal has a large surface area which increases its rate of reaction with water vapour in the air to produce hydrogen gas. Ca + H2O  CaO + H2 The heat produced causes hydrogen gas to combust spontaneously with oxygen gas in the air, causing explosions. 2H2 + O2  2H2O This generates more water which spark off further explosive reactions with finely powdered Ca metal. (b)(i) Q = I x t = ne x F ne = 1×105×3×60×60 96500 =1.12×104mol Ca2+ (l) + 2e  Ca(s) Mass of Ca formed = 1.12×104 2 × 40.1 = 2.24 × 105𝑔 = 0.224 tonnes (b)(ii) From Data Booklet, 2H2O + 2e  2H2 + 2OH E=0.83V Ca2+ + 2e  Ca E=2.87 V Since E(H2O/H2) is less negative than E(Ca2+/Ca), H 2O will be preferentially reduced at the cathode instead of Ca2+. No Ca metal will be obtained. (c)(i) Ca(CH3CO2)2  CaCO3 + CH3COCH3 solid M (c)(ii) White residue contains CaCO3 and CaO. Method 1 Amount of CaCO3 formed at 450 oC = Amount of Ca(CH3CO2)2 = 2.0 158.1 = 0.01265 mol Let the amount of CO2 evolved = y mol. Amount/mol CaCO3  CaO + CO2 Initial 0.01265 - - Change -y +y +y Final 0.01265 - y y y Mass of CaCO3 in the residue = (0.01265 – y)(100.1) = (1.266 – 100.1y) g Mass of CaO in the residue = 56.1y g mass of CaCO3in the residue mass of CaO in the residue = 70 30  1.266−100.1𝑦 56.1 y = 7 3 1.266 – 100.1y = 130.9y y = 0.005483 Amount of CO2 evolved = 0.005483 mol Mass of CO2 evolved = (0.005483)(44.0) = 0.241 g

Method 2 Mass of CaCO3 formed at 450°C = 2 /158.1 x 100.1 = 1.266 g CaCO3  CaO + CO2 mass after partial decomp / g 1.266 – x – y x y Since mass of white residue = (1.266 – y) g and %(CaO in white residue) = 30% ⇒ x 0.3 1.266 y  Since amt of CaO formed = amt of CO2 formed, ⇒ xy 56.1 44 Solving simultaneous equations, y = mass of CO2 lost = 0.241 g Method 3 Let mass of white residue be x g and mass of CO2 evolved by y g. Mass of CaO = (0.3x) g Since CaCO3  CaO + CO2, amt of CaO = amt of CO2 0.3x y 56.1 44 By conservation of mass, x + y = 1.266 g Solving simultaneous equations, y = 0.241 g.) (c)(iii) Ba(CH3CO2)2 will decompose at a higher temperature. Ba 2+ ion has a larger ionic radius as compared to Ca 2+ ion resulting in a lower charge density. Ba 2+ ion therefore has weaker polarizing power, hence weakening the covalent bonds within CH 3CO2 - ions to a smaller extent as compared to Ca2+ ion, resulting in higher thermal decomposition temperature. (d)(i) Nucleophilic addition

(d)(ii) (e)(i) Compare experiments 1 and 2. When [Q] is increased 1.25 times, rate is increased 1.25 times.  rate  [Q]  Order of reaction with respect to Q = 1 Compare experiments 1 and 3. When