PORTMORE COMMUNITY COLLEGE

PORTMORE COMMUNITY COLLEGE ASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY RESIT EXAMINATIONS SEMESTER JUNE, 011 COURSE NAME: CODE: CHEMISTRY FOR ENGINEEERS CHY-1008 GROUP: ADET DATE: June 7, 011 TIME: DURATION: 1:00 pm HOURS INSTRUCTIONS: 1. This paper consists of SIX questions.. Candidates must attempt ANY FOUR questions on this paper. 3. All working MUST be CLEARLY shown. 4. Keep all parts of the same question together. 5. The use of non-programmable calculators is permitted. 1

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Instructions: Answer any FOUR (4) questions. Question 1 A. Define the following terms - Bond energy - Enthalpy (H) (6mks) B. Use the following information on bond energies to find H 0 for the reaction H (g) + Br ( g) HBr (g) H H = 436 KJ / mol Br Br = 193 KJ / mol H Br = 366KJ /mol (5mks) C. State Hess s law (3mks) D. Diborane, B H 6 is a highly reactive boron Hydride. Calculate H for the synthesis of diborane from it s elements according to the equation: B (s) + 3H (g) B H 6 (g) Using the following data: Reaction H A. B (s) + 3/ O (g) B O 3 (s) -173 KJ B. B H 6 (g) + 3O (g) B O 3 (s) + 3H O (g) -035KJ C. H (g) + ½ O (g) H O (L) -86 KJ D. H O (l) H O (g) -44KJ (11mks) (Total 5mks)

Question A. What is a redox reaction? B. Draw and label the galvanic cell using the following cell diagram: Zn (s) Zn t (1M) CU t (1M) Cu (s) (3mks) (5mks) (ii) Write the half reactions. (mks) (iii) Calculate the EMF of the cell in B using the following data: E 0 Cu + /Cu = 0.34 V E 0 Zn t /Zn = -0.76 V C. What is the function of a salt bridge? D. What is electrolysis? (mks) (mks) E. Describe the principles of the lead storage Battery. (5mks) 3

Question 3 a. What is meant by the term dynamic equilibrium? (1 mark) b. Consider the following system in dynamic equilibrium N O( g g) NO ( g) 3NO( ) 155.7kJ Explain the effect if any of the following changes on the above system in equilibrium i. The temperature is increased ii. The pressure is reduced iii. The concentration of N O is reduced (6 marks) c. The partial pressure of NOand N O4 at equilibrium are 0.4 atm and 0.6 atm respectively. 1 NO ( g) N O ( ) 58kJmol 4 g i. Write and expression for K p ii. Calculate the value for K p stating units (1 mark) ( marks) d. Define the term ph (1 mark) e. Calculate the ph of the following solutions with the given concentrations. 4 i. [ H O ] = 1. 5x 10 M (1 mark) 3 ii. [ H SO 4 ] = 0.047M ( mark) iii. [NaOH]= 0.5M ( marks) f. Draw an electron-dot diagram for the structure of water and explain how water can act as both an acid and a base. (4 marks) 4

Question 4 a. What is meant by the rate of a chemical reaction? ( marks) b. Explain how the following affect the rate of a chemical reaction and illustrate where necessary using diagrams. i. Temperature (3 marks) ii. Concentration ( marks) iii. Catalyst (3 marks) c. Give two examples from your everyday life of the effects of temperature on the rate of reaction ( marks) d. One reaction which occurs in air polluted with nitrogen oxide is shown below. NO( g) O ( g) NO ( g) Five experiments were carried out to find the relationship between the initial concentrations of NO and O and the initial rate of formation of NO 3 Initial concentrations / moldm Initial rate of formation of 3 1 Experiment No. [NO] [O NO / moldm ] s 1 0.001 0.001 7 x 10 0.001 0.00 14 x 10 3 0.001 0.003 1 x 10 4 0.00 0.003 84 x 10 5 0.003 0.003 189 x 10 i. What is the order of the reaction with respect to a. NO b. O ( + marks) Show working or give appropriate reasoning ii. What is the order of the reaction iii. Write the rate equation for the reaction iv. Calculate the rate constant, k, for the reaction stating units. (1 mark) (1 mark) ( marks) 5

Question 5 a) Carbonic acid (H CO 3 ) can be produced by dissolving CO in water. When 3g KOH is added to 10ml of a 4.5M carbonic acid solution, the only salt produced was K CO 3. i) Determine what would be the limiting reagent? Give your reason (3 marks) ii) Determine the maximum yield of K CO 3 ( marks) iii) If only 3g of K CO 3 is collected at the end of the reaction, determine the percentage yield of the reaction. ( marks) b) Combustion analysis of 45.6g toluene gives 35.67g of H O and 15.5g CO. What is the empirical formula of toluene? (5 marks) c) Arrange the following in order of increasing ionic / atomic radius: Mg, Br -, Na, Mg +. Give your reason. (4 marks) d) Explain the general trend in first ionization energy across the period and down the group. (4 marks) Question 6 1. (a) List the THREE subatomic particles of an atom. [3] (b) Calculate the average isotopic mass of krypton given: [4] Mass/ a.m.u. % 77.90 0.350 79.916.7 81.913 11.56 8.914 11.55 83.91 56.90 85.911 17.37 6

(c) Complete the table below [4] Substance Proton Neutron Electron Mass Number Cu 9 64 K + 19 0 14 14 8 O - 10 16 (d) i. Define the following terms: Empirical Formula o Molecular Formula [] ii. Vitamin C (ascorbic acid) contains 40.9% carbon,4.58% hydrogen and 54.50% oxygen by mass. Calculate the empirical formula. If the molecular formula mass of ascorbic acid is 176g, what is the molecular formula? [5] iii. Balance the following equation: Al + Fe O 3 ----- Al O 3 + Fe [] ***** END OF PAPER ****** 7

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QUESTION 1 SOLUTION (A) Bond energy The enthalpy change required to break a bond in a mole of gaseous molecules. Enthalpy A thermodynamic quantity used to describe heat changes to king place at constant pressure. (B) H H + H H = 436 KJ /mol Br Br + Br H = 193 KJ / mol H 1 = 436 KJ / mol + 193 KJ / mol = 69 KJ /mol Bond forming step: H = H + Br HBr H = x 366 KJ / mol Overall: H = H 1 + H H = [69 73] KJ / mol = -103 KJ / mol C. Hess s law states that H is not dependent on the reaction pathway, eg in going from a particular set of reactions to a particular set of products the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. D. 1. B (s) + 3/ O (g) B O 3 (s) H = -173KJ. B O 3 (s) + 3 H O (g) B H 6 (g) + 3O (g) H = +035KJ Add equation 1 and 3. B(s) + 3H O (g) B H 6 (g) + 3/ O (g) H = 76 KJ Multiply equation c by 3 we get: 4. 3H (g) + 3/ O (g) 3 H O (l) 3 x -66 KJ Add equations 3 and 4 we get: 5. B(s) + 3 H (g) + 3 H O (g) B H 6 (g) + 3 H O (l) H = -96KJ Multiply equation D by 3 we get: 6. 3H O (l) 3H O (g) H = 44 x 3 KJ Add equation 5 and 6 we get: B (s) + 3H (g) B H 6 (g) H = + 36 KJ 9

H for the synthesis of 1 mole of diborane is +36 KJ. SOLUTION QUESTION 3 Redox reaction involves a transfer of electrons from the reducing agent to the oxidizing agent; oxidation involves the loss of electrons and results in an increase in oxidation number whereas reduction is going of electrons and this result in a decrease in oxidation number. Half reactions: Anode: Zn (s) Zn + (aq) + e- (oxidation) Cathode: e - + Cu + (aq) Cu (s) (reduction) Overall equation; Zn(s) + Cu + (aq) Zn + (aq) + Cu (s) Diagram of galvanic cell: Volt Meter e- e- Salt Bridge Zinc anode Copper cathode Zn z+ Cu z+ Zn + is oxidized to Zn z+ at anode Cu + is reduced to Cu at cathode 10

EMF = E cathode E anode = 0.34 V (-0.76V) = +1.10 V Salt bridge: Prevents the two solutions from mixing with each other while at the same time allowing anions and cations to move across. [Allows ions to flow between the half cells] Electrolysis is the process in which electrical energy is used to cause a non spontaneous chemical reaction to occur. Lead Battery: Lead Storage Battery commonly used in automobiles consists of 6 identical cells joined together in series. Each cell has a lead anode and a cathode made of lead dioxide (PbO ) packed on a metal plate. Both anode and cathode are immersed in an aqueous solution of (H SO 4 ) sulfuric acid which acts as the electrolyte. The cell reactions are: Anode: Pb (s) + SO - 4 PbSO 4 (s) + e- Cathode: PbO (g) + 4H + + SO - 4 + e- PbSO 4 + H O (l) Each cell can produce V; a total of 1 V from the six cells and it is used to power the ignition circuit of automatic and other electrical systems. The lead storage battery can deliver large amounts of current for a short time. The lead battery is rechargeable. a) i) KOH + H CO 3 K CO 3 + H O Solution: Moles KOH = 3g/(39 + 16 + 1) gmol -1 = 0.0536mol Moles H CO 3 = (4.5mol/1000ml) * 10ml = 0.045mol 0.045mol H CO 3 needs (* 0.045mol) KOH to react with but only 0.0536mol is present. Hence KOH is the limiting reagent. ii) Yield = 0.5 *0.0536mol * [(39*) + 1 + (3*16)] = 3.698g iii) %Yield = 3g/3.698g * 100 = 81.1% b) Moles C produced = 15.5g/(44.01g/mol) = 3.465mol Moles H produced = 35.67g/(18.016g/mol) * = 3.96mol Total mass collected = (3.465mol C * 1.01g/mol) + (3.96mol H * 1.008) = 45.61g Hence toluene consists of only C and H. Ratio: 3.465mol C : 3.96mol H 11

(1mol C : 1.143mol H) * 7 7mol C : 8mol H Emperical Formula : C 7 H 8 c) Mg + < Mg < Na < Br - Reason: Br is in period 3 and hence would have the largest atomic radius. When an atom gains an electron there is a drastic increase in atomic radius; and hence Br - must be the biggest of the set. Also, across the period atomic radius decrease, hence Na atomic radius is larger than Mg. Mg + is smaller than Mg since atomic radius decreases upon loss of electrons. d) As you move across the period, a proton is added and an electron is added to the same valence shell. The addition of a proton causes Zeff to increase. Also, there is no significant change in the shielding of the valence electrons as a result of adding electrons to the same valence shell, and hence the effective nuclear charge (Zeff) is not greatly affected. Hence, the net effect is that Zeff increases, and therefore the energy required to remove an electron from an atom must increase. Down the group shielding drastically increases since valence electrons are added to a new energy level. Consequently, there is a net decrease in Zeff and first ionization energy decreases. Question 3 a. Dynamic equilibrium- the rate of forward is equal to the rate of the reverse reaction 1 mark b. (i). increase temperature will cause a shift in the position of the equilibrium from left to right so as to decrease the temperature and maintain equilibrium marks as to (ii). Reducing the pressure by increasing the volume will shift position from left to right so increase the number of moles of nitrogen oxide marks (iii) Reducing the concentration of N O will cause a shift from right to left so as to replenish the N O removed and so maintained equilibrium marks PN O4 c. (i). K p 1 mark P ) ( NO (0.4) (ii). K P marks (0.6) =1.11 1 atm d. ph log[ H ] 1 mark 1

e. (i) ph 3. 8 1 mark (ii) ph 0. 07 marks (iii) ph 14 poh = 14-0.6 marks =13.4 Question 4 a. Rate of reaction- the change in concentration of reactants or products with time. ( marks) b. Increase temperature, increase the number of molecules with the minimum amount of energy, (activation energy) causing more frequent effective collision hence increasing the rate of reaction ( marks) Students should draw Boltzman s curve (1 mark) Increase in concentration, increase the number of molecules per unit volume and hence the frequency of effective collision thus increasing the rate ( marks) A catalyst speeds up the rate of a reaction by providing an alternate pathway, one with a lower activation energy ( marks) Students should draw the graph showing the effect of a catalyst (1 mark) c. i. spoiling of milk 13

d. ii. Weathering of rocks Any other two examples NO Rate Rate 4 3 84 *10 1*10 4 x O x Rate Rate y 1 Conc Conc 5 6 1 14 *10 7 *10 y 6 6 4 3 x 0.00 0.001 Conc Conc 1 y x 0.00 0.001 y ( marks) ( marks) ( marks) ii. Order of reaction = +1 = 3 (1 mark) iii. Rate Law = k NO] [ O ] (1 mark) [ iv. Using experiment 1 Rate k [ NO] [ O 6 7 *10 (0.001) (0.00) 3500mol ] dm 6 s 1 14