DHS H2 CHEM P1 ANS

2 BLANK P AGE [Turn Over

3 Section A For each question, there ar e four possible answers A, B, C, and D. Choos e the one you consider to be correct. 1 When 10.0 cm3 of a 0.10 mol dm–3 solution of alkali metal salt MXO3 was reduced with an excess of acidified potassium iodide solution, the resulting iodine required 60.0 cm 3 of 0.10 mol dm –3 sodium thiosulfate solution for its reduction. The anion could be reduced to A XO2 B XO2– C XO– D X– 2 10 cm3 of a hydrocarbon is mixed with 100 cm 3 of oxygen gas which is in excess. The mixture is exploded and after it is cooled to room temperature, the residual gase s occupy a volume of 80 cm 3. Upon passing the gases through potassium hydroxide, th is volume decreases to 50 cm3. What is the formula of the unknown hydrocarbon? A C3H6 B C3H8 C C4H8 D C4H10 Option B: CxHy + (x+y/4) O2  xCO2 + y/2 H2O Volume of CO2 = 30 cm3 By using Avogadro’s Law, Mole ratio  volume ratio  no. of mol of CO2 = 3  x = 3 Volume of O2 remaining = 50 cm3  volume of O2 reacted = 100 – 50 = 50 cm3  x+(y/4) = 50/10 5 = x+(y/4)  y = 8 Hence, formula of the hydrocarbon is C 3H8. Option D: (5-n) e + X5+  Xn+ ---(1) 2I-  I2 + 2e ---(2) I2 + 2S2O3 2-  2I- + S4O6 2- ---(3)  n = -1 No of mol of S2O3 2- = 60.0/1000 x 0.10 = 0.006  No of mol of I = ½ x 0.006 = 0.003 2  no of mol of I- = 0.003 x 2 = 0.006 No of mol of X5+ = 10/1000 x 0.10 = 0.001  6I- = X5+ (2) x 3, -6I  3I + 6e 2  5-n = 6 [Turn Over

4 3 The use of Data Booklet is relevant to this question. The successive ionis ation energies, in kJ mol –1, of an element X are given below. 870 1800 3000 3600 5800 7000 13200 What is element X? A 33As B 53I C 8O D 52Te 4 A 10.0 dm 3 sample of oxygen at a pressure of 250 kPa and 3.0 dm 3 sample of nitrogen at a pressure of 500 kPa are introduced into a 2.5 dm 3 vessel at room temperature. What is the total pressure in the vessel? A 750 kPa B 1200 kPa C 1600 kPa D 1950 kPa Option C: Using Boyle’s Law, 500(3.0) = PO2’ (2.5) P1V1 = P2V2 PN2’ = 600 kPa ’ PO2 VO2 = PO2’ Vvessel  PT = PO2’ + PN2 250(10.0) = PO2’ (2.5) = 1000 + 600 PO2’ = 1000 kPa = 1600 kPa PN2 VN2 = PN2’ Vvessel Option D: The greatest jump is between the 6th and 7th I.E. Hence element X must be in group VI. Therefore, either O or Te is the answer. But with reference to the Data Booklet, the I.E. values do not correspond to that of oxygen. Hence, element X is Te. 5 Given that the standard ent halpy change of combustion of but-1-ene and ethene are p and q kJ mol–1 respectively. What is the standard enthalpy change of the reaction: 2C2H4(g)  C4H8(g)? A 2q – p kJ mol–1 B (p + q)/2 kJ mol–1 C q – 2p kJ mol–1 D (q – p)/2 kJ mol–1 Option A: [Turn Over

5 6 The data below refers to the st andard molar enthalpy changes of combustion of some members of the alkanes. Alkane Hc /kJ mol–1 CH4 –890 C2H6 –1560 C3H8 –2220 C4H10 –2880 C5H12 –3510 C6H14 –4190 Another alkane X has a standard enthalpy c hange of combustion of –6780 kJ mol –1. The formula of X is likely to be A C9H20 B C10H22 C C11H24 D C12H26 Option B: The difference in Hc accounts for every –CH2 chain which is approximately 670 kJ mol-1. Therefore, by observing the trend, the formula of X is C10H22. 7 The data below refers to the radii and charges of six ions. Ion P2+ Q + R + T 2– U – V – Radius/nm 0.16 0.19 0.15 0.16 0.19 0.15 PT, QU and RV are ionic solids of the same lattice structure. Which one of the following gives the correct order of their la ttice energies with the lowest numerical value first? A PT RV QU B PT QU RV C RV QU PT D QU RV PT Option D: This question requires students to apply the knowledge of the lattice energy expression involving charge and ionic radius. Hence, QU should have the lowest numerical value with the largest radii (sum) and lowest charge (product). PT should have the largest numerical value with the smallest radii (sum) and largest charge (product). [Turn Over

6 8 In the presence of ultraviolet light, the “inert” xenon gas will react with fluorine gas to produce XeF4 according to the equation, Xe (g) + 2 F2 (g) XeF4 (s) What is the correct equilibrium constant Kc? A B C D 9 The decomposition of dinitrogen pentoxide N 2O5 is found to be first order with respect to the concentration of N 2O5. Which one of the following graphs confirms the results? A Rate of decomposition of N2O5 [N2O5] 0 B Rate of decomposition of N2O5 [N2O5] 0 C Rate of decomposition of N2O5 [N2O5] 0 D Rate of decomposition of N2O5 [N2O5] 0 [Turn Over Option A: Since the decomposition of N2O5 is 1st order w.r.t. [N2O5],  rate  [N2O5]  rate = k[N2O5]  straight line graph passing through the origin. Option D: In writing the Kc expression involving solid reactants or products, they do not have a concentration term. Hence, they are just given a numerical value of 1.

7 10 What is the pH of the final soluti on formed by mixing equal v olumes of two separate portions of dilute sulfuric acid of pH 2.0 and pH 4.0? A 2.3 B 2.6 C 3.0 D 3.6 Option A: pH = 2.0  [H+] = 10-2 pH = 4.0  [H+] = 10-4 [H+]total = (10-2 + 10-4)/2 = 5.05 x 10-3 Resultant pH = -lg 5.05 x 10-3 = 2.3 11 Hydrogen iodide is not pr epared by the addition of concentrated sulfuric acid to solid sodium iodide. This can be best explained by the fact that A Hydrogen iodide is not displaced by sulfuric acid. B Iodide ions are oxidised to iodine. C Sodium sulfide is formed as an impurity. D Hydrogen iodide is less volatile than concentrated sulfuric acid. Ease of oxidation of halide increases down the group. NaI + H2SO4  HI + NaHSO4 8HI + H2SO4  4I2 + H2S + 4H2O 12 Which one of the following gives t he correct definition of an acid according to the Bronsted-Lowry theory? A It dissociates in water to give H+(aq) ions. B It is a proton donor. C It is a proton acceptor. D It is an electron donor. A It dissociates in water to give H+(aq) ions  Arrhenius definition of acid B It is a proton donor  Bronsted – Lowry definition of acid C It is a proton acceptor  Bronsted – Lowry definition of base. D It is an electron donor  reducing agent [Turn Over

8 13 The graph below shows the v ariation in the boiling points for eight consecutive elements in the Period Table, all wit h atomic numbers from 10 to 20. The letters do not represent any element in the Periodic Table. Which of the following can be deduced from the graph above? A Element H forms an acidic oxide only. B Element K does not conduct electricity. C Element G can exist in two allotropic forms. D Element E and beryllium are in the same group. Boiling Point K G F E H I J L Atomic number Look for elements with proton