Practice Exam. SCH 4UI: Grade 12 University Chemistry Teacher: Mr. Winter. Name: IMPORTANT INSTRUCTIONS PLEASE READ CAREFULLY

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1 Page 1 of 7 Practice Exam SH 4UI: Grade 12 University hemistry Teacher: Mr. Winter Name: IMPORTANT INSTRUTIONS PLEASE REA AREFULLY 1. Aids Permitted: Non-Programmable alculator 2. For Multiple hoice Questions: hoose the most correct answer and fill in your selection on the SANTRON card provided. 3. For True or False Questions: Indicate whether the statements are true or false and then fill in your selection on the SANTRON card provided. 4. For Full Solution Questions: Write answers on the foolscap paper provided. LEARLY mark the question being answered. Alongside incorrect answers, marks will be deducted for errors in chemical formulas, and lack of units. raw a box around your final answer for calculations. 5. Make sure your exam has 4 pages. 6. The mark distribution: Part A: Multiple hoice 15 marks Part : True or False 10 marks Part : Full Solution 56 marks Total Marks 81 marks 7. onstant Values: c for water = 4.18 J/g O Kw = 1.00 x Hf O carbon dioxide = kj/mol Kb ammonia = 1.8 x 10-5 Hf O water vapour = kj/mol K b H3OO - (acetate) = 5.6 x Hf O water liquid = kj/mol Ka acetic acid = 1.8 x 10-5 Hf O butane = kj/mol Ksp u2s(aq) = 6.9 x 10-8

2 Page 2 of 7 Part A: Multiple hoice (15 marks) hoose the most correct answer and record your selection on the SANTRON provided. 1. onsider the following three equations: target: O(g) + H2(g) à (s) + H2O(g) ΔHtarget =? 2. 2H2O(g) à 2H2(g) + O2(g) ΔH2 3. 2(s) + O2(g) à 2O2(g) ΔH3 The enthalpy change for the target reaction, is best found through: a. ½ (ΔH3) + ½ (ΔH2). c. 2(ΔH3) 2(ΔH2). b. 6(ΔH3) + 3(ΔH2). d. ½ (ΔH2) ½ (ΔH3). 2. The decomposition of hydrogen peroxide is a first order reaction. If the initial rate of consumption is found to be 2.0 mol/l s at a given concentration, then the rate of consumption if the given concentration were doubled would be a. 2.0 mol/l s. c. 4.0 mol/l s. b. 16 mol/l s. d. 8.0 mol/l s. 3. For the following equilibrium, what is the effect of adding O2(g) to the system, at constant volume and temperature? 2NO(g) + O2(g) ßà 2NO2(g) a. No change in [NO(g)] or [NO2(g)] c. A decrease in [NO2(g)] b. An increase in [NO (g)] d. An increase in [NO2(g)] A 4. The molar enthalpy of combustion for methane gas is 243 kj/mol. In order to produce 759 kj of heat, how many grams of methane gas are needed? a g c g b g d g 5. The poh of a 0.45 mol/l Hl solution is a c b d If the Ksp of zinc chloride in water to be 1.7 x 10-10, then the molar solubility of the salt is a. 5.5 x 10-4 mol/l. c. 1.3 x 10-5 mol/l. b. 4.4 x 10-4 mol/l. d. 3.5 x 10-4 mol/l. 7. For the following aqueous chemical reaction at equilibrium, 2A + 2 ßà + 2, the equilibrium concentrations were found to be [A] = 1.0 mol/l, [] = 2.0 mol/l, [] = 3.0 mol/l and [] = 4.0 mol/l. The value of the equilibrium constant is a. 12 c b. 6 d In a saturated solution of Al(OH)3 the measured concentration of OH -1 is found to be 1.3 x 10-5 mol/l. The concentration of aluminum ions is a. 1.3 x 10-5 mol/l. c. 3.9 x 10-5 mol/l. b. 4.3 x 10-6 mol/l. d. 6.5 x 10-6 mol/l. 9. The central atom in l3 will undergo what type of hybridization? a. sp. c. sp 3. b. sp 2. d. sp 3 d. 10. The electron configuration for an Alkali Metal is best represented by a. 1s 2 2s 2 2p 5. c. 1s 2 2s 2. b. 1s 2 2s 2 2p 6 3s 1. d. 1s 2 2s 2 2p Arsenic is able to form an ion with a positive three charge because a. it is a metalloid. c. it has empty d orbitals. b. it has five valence electrons. d. it has a half full p orbital allowing for easy loss.

3 12. The electron configuration for an atom of nickel is best represented by a. [Ar]3d 8. c. [Ne]3s 2 3p 5. b. [Ar]4s 2 3d 8. d. [Ne]3s 2 3p 6 4s 2 3d 7. Page 3 of 7 A 13. When benzene and a molecule of r2 are reacted, a(n) will occur. a. substitution reaction c. addition reaction b. hydration reaction d. hydrohalogenation reaction 14. What type of intermolecular forces are found between molecules of ethane? a. ipole dipole c. london dispersion forces (LF) b. Hydrogen bonding d. All of the above 15. When two primary alcohols are combined in a condensation reaction, is formed. a. a ketone c. an ether b. an aldehyde d. an alkene Part : True and False (10 marks) Indicate whether the following statements are true or false and mark your choice on the SANTRON card provided. TRUE 16. TRUE 17. FALSE 18. FALSE 19. TRUE 20. TRUE 21. FALSE 22. TRUE 23. FALSE 24. FALSE 25. Lead metal has a standard enthalpy of formation equal to zero. The activation energy is the minimum needed for particles in a system to react. When a student heated 120 ml of distilled water in an electric kettle from 15 to 100.0, the quantity of heat flowing into the water was 70 kj. (correct answer would be 42.6KJ) At equilibrium, the concentration of the reactants is always equal to the concentration of products. After a titration is performed, the sodium acetate salt formed has a slightly basic ph. The rate law for any reaction is determined by the slowest rate-determining step. Molecules are isoelectric if they share the same chemical formula. The secondary quantum number (l) refers to the type of orbital in an energy level. When a secondary alcohol undergoes controlled oxidation, an aldehyde is produced. The functional group of ethanoic acid is a carbonyl group.

4 Part : Full Solution (56 marks) Answer the following questions on the foolscap provided. LEARLY mark the question being answered. Alongside incorrect answers, marks will be deducted for errors in chemical formulas, and improper units. Page 4 of The following data is collected during the formation of ammonia. a. alculate the rate law for the system. (3 marks) Trials 1 and 2: H2 is held constant while N2 doubles When N2 doubles rate increases 4x [2] 2 = 4, therefore reaction is second order with respect to [N2] Trials 3 and 4: N2 is held constant while H2 doubles When H2 doubles rate increases 8x [2] 3 = 8, therefore reaction is 3 rd order with respect to [H2] Therefore: rate = k[n2] 2 [H2] 3 b. alculate the rate constant for the system. (3 marks) Using data from trial 1 rate = k[n2] 2 [H2] mol/ L s = k (0.0200) 2 (0.0800) 3 k = 1.95 x 10 5 L4 mol4 s-1 c. What is the overall order of this system? (1 mark) overall order = sum of all individual orders in rate law (ie. Add all exponents) = 5 Reaction is 5 th order overall d. If the system is closed, use Lehatelier s Principle to list and explain two disturbances that could be introduced in order to increase the production of ammonia. (2 marks) Add more N2 will push equilibrium towards products to use up N2 Add more H2 will push equilibrium towards products to use up H2 Remove NH3 will push equilibrium towards products to replace NH3 Increase pressure will move towards products since lower number of gas particles ecrease volume will move towards products since lower number of gas particles Trial Table 1: Rate of Reaction ata for the prodcuction of ammonia N2(g) (mol/l) N2(g) + 3H2(g) à 2NH3(g) H2(g) (mol/l) Initial Rate of Reaction (mol/(l s))

5 Page 5 of a. Using standard enthalpy of formation values, calculate the enthalpy change when a student completely combusts one mole of butane gas (4H10) with oxygen to produce carbon dioxide and water vapour. (3 marks) 2 4H10 (g) + 13 O2 (g) à 8 O2(g) + 10 H2O(v) H rxn = Σ H f products - Σ H f reactants H rxn = [8 H f O2(g) + 10 H f H2O(v)] [2 H f 4H10 (g) + 13 H f O2 (g) ] H rxn =[8(-393.5kJ) + 10(-241.8kJ)] [2(-125.6kJ) + 13(zero)] H rxn = [-5566kJ] [-251.2] H rxn = kJ (Our reaction as written above is for 2 moles, so divide this value by 2) H = kJ/mol b. Write the balanced, thermochemical equation to include the heat as an energy term within the equation. (2 marks) formula reduced to show combustion of 1 mole of butane. 4H10 (g) + 13/2 O2 (g) à 4 O2(g) + 5 H2O(v) kJ c. If 2.5 mol of butane were combusted in a calorimeter, and the resulting change in water temperature was determined to be 35, calculate the mass of water that must have been contained in the calorimeter. (3 marks) H = kJ/mol x 2.5 moles = kJ Q = - H = kJ Q = mc t kJ = m(4.185g/j )(35 ) m = 45.4g 28. a. alculate the ion product quotient, Qsp for the reaction that occurs when two students combine 14.5 ml of mol/l copper(i) nitrate solution with 19.5 ml of mol/l magnesium sulfide solution. (4 marks) uno3 (aq) + MgS(aq) à u2s(s) + MgNO3(aq) u2s(s) à 2 u 1+ (aq) + S 2- (aq) Ksp = 6.9 x 10-8 oncentrations before mixing: 0.078M uno3 means M u 1+, and 0.078M NO M MgS means M Mg 2+, and M S 2- oncentrations after mixing (diluting) 1V1=2V2 (total volume after mixing 14.5mL mL = 34mL) [u 1+ ] [S 2- ] 1V1=2V2 (0.078M)(14.5mL) = 2(34mL) 2 =.033mol/L Qsp = [u 1+ ] 2 [S 2- ] Given above: Ksp = 6.9 x 10-8 Qsp = (.033) 2 (.0085) Qsp = 9.3 x V1=2V2 (0.0148M)(19.5mL) = 2(34mL) 2 =.0085mol/L b. o they observe a precipitate? Explain how you know. (2 marks) Qsp > Ksp, so therefore a precipitate will form!

6 29. Name the following organic molecules according to IUPA rules. (7 marks) Page 6 of 7 a) H3H2H=H2 b) 1-butene Ethoxy butane c) O d) // H3H2 \ H propanal 1,2-cyclobutane-diol O e) // f) H3H2 O - H2H3 Ethyl propanoate g) O // cyclopentanone H3H2H2 \ OH utanoic acid ompound 30. omplete the following chart (12 marks) Lewis iagram VSEPR Notation and Name of the Shape of the Molecule VSEPR: AX 6E 0 Is the Molecule Polar or Non- Polar? Non - polar Name all types of Intermolecular Attraction present Solubility in Water: relatively high or low? LF low low Melting Point: relatively high or low? PF 5 Shape Name: octahedral XeF 4O VSEPR: AX 5E 1 Shape Name: Square Pyramidal Polar LF ipole- ipole higher medium

7 Page 7 of A 0.25 mol/l hydrofluoric acid solution is made in the lab. alculate the ka of the acid if at equilibrium it has a ph of (5 marks) HF(aq) + H2O ßà H3O + (aq) + F - (aq) I x +x +x E x x x [H3O + ] = 10 -ph [HF] = x [H3O + ] = [HF] = [H3O + ] =.035 mol/l = x [HF] = mol/l Ka = [H + ][F - ] [HF] Ka = [.035][.035] Ka = 5.7 x 10-3 [.215] **Note** - This Ka value will not match the value in your textbook for HF. Recall that K values are specific for a certain temperature. We can assume that this equilibrium was reached at a different temperature than the one that is given in your Ka reference chart ml of 1.5 M KOH was used to titrate exactly ml of a solution of hydrochloric acid with unknown concentration. alculate the concentration of the acid. (3 marks) Hl(aq) + KOH(aq) H2O(aq) + Kl(aq) Mole ratio is 1:1, so ava = bvb a = (1.5M)(16.45mL) (10.00mL) a = 2.5 M [Hl] = 2.5 mol/l 33. alculate the ph of a 0.35M solution of NH4l. (5 marks) NH4l à NH4 + (aq) + l - (aq) l - is the conjugate base of Hl (a strong acid), so it will be neutral. NH4 + is the conjugate acid of NH3 (a weak base), so it will be acidic. NH4 + (aq) + H2O(l) ßà H3O + (aq) + NH3(aq) I 0.35M 0 0 -x +x +x E 0.35 x (ignore x) x x NH3 Kb = 1.8 x 10-5 Ka = [H3O + ][NH3] [NH4 + ] Ka x Kb = 1.0 x x = (x)(x) Ka = 1.0 x x 10-5 x = 1.4 x 10-5 = [H3O + ] Ka = 5.6 x ph = -log[h3o + ] ph = -log(1.4 x 10-5 ) ph = 4.85