Practice Exam Solutions

Practice Exam Solutions - 1) Balance the following equation...

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1) Balance the following equation in an acidic solution: CH 3 OH(aq) + Ce 4+ (aq) & CO 2 (aq) + Ce 3+ H 2 O + CH 3 OH + 6Ce & 6Ce 3+ + 6H + + CO 2 2) Balance the following equation in a basic solution: H 2 CO(aq) + Ag(NH 3 ) 2 + (aq) & HCO 3 - (aq) + Ag(s) + NH 3 (aq) 5OH - + H 2 CO + 4Ag(NH 3 ) 2 + & 4Ag + HCO 3 - + 3H 2 O + 8NH 3 3) Why is a salt bridge or porous disk necessary for the flow of electrons in a galvanic cell? A salt bridge or porous disk allows charge balance (it prevents charge build-up). 4) For the following unbalanced equation: Cu 2+ (aq) + Mg(s) ? Mg + Cu(s) Draw a spontaneous galvanic cell labeling the anode, cathode, electron flow direction, + and – signs, electrodes, and solutions. Assume the concentrations are 1M and T=298K. In addition, write the balanced equation, cell diagram, overall cell potential, equilibrium constant, ?G°, and specify how many electrons are transferred. Cu + Mg & Mg + Cu x ° = +2.7 ¦ Mg ¦ Cu ¦ Cu (note I drew a porous disk, it could also be a salt bridge¦ ) ?G° = -nF x ° = -521 kJ/mol Solve for K by lnK = (nF x °)/(RT) I don’t have a calculator with me, but everything is known except K. 2 electrons are transferred per mole as seen by n = 2 from the total reaction. Cu 2+ 2+ Cu Mg Anode (-) e - e - Cathode(+)
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5) Identify the anode, cathode, path of electron flow, balanced equation, and give the cell potential with respect to your overall balanced equation. Ag + Ag + (1.0M) & + (0.01M) Cell potential: x = x ° -[(RT)/(nF)]lnQ = 0.12V x ° = 0, n = 1, Q = 0.01/1.0 = 0.1 6) Why does a catalyst increase the rate of a chemical reaction? Catalysts decrease the activation energy making it easier for the reaction to proceed. 7) H 2 and O 2 react explosively to form water, but a mixture of the two can be stable in a balloon. Why don’t they react under normal conditions? Under normal condition there isn’t enough energy to overcome the activation barrier. 8) Given the following reaction: 2I - (aq) + S 2 O 8 2- (aq) & I 2 (aq) + 2SO 4 2- (aq) The following experimental data was obtained: [I - ] [S 2 O 8 2- ] Rate Rate constant 0.080 0.040 12.5x10 -6 0.0038 0.040 0.040 6.25x10 0.0039 0.080 0.020 6.25x10 -6 0.0039 0.032 0.040 5.00x10 0.0039 0.060 0.030 7.00x10 0.0039 Where the rate is defined as the disappearance of S 2 O 8 (aq). Determine the rate law and calculate a rate constant for each experiment and determine the average rate constant. Overall rate = k[I - ][S 2 O 8 ] Ave rate constant = 0.0039 Remember to hold one constant to get rid of a variable. (ex. 1&3 to hold I - constant) [Ag + ]= 1M + ]= 0.01M Ag + ]= 1M + ]= 0.01M Ag Ag anode cathode e _ e _
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9) A certain 1 st order reaction is 45% complete in 65 seconds. What are the rate constant and ½ life for this process? 9.20 x 10 -3 s -1 75 sec Remember that 55% is left! Watch out for this type of wording so you don’t get confused. 10) The activation energy for the reaction: NO 2 (g) + CO(g) & NO(g) + CO 2 (g) is 125 kJ/mol, and ?E is -216 kJ/mol. What is the activation energy for the reverse reaction?
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This note was uploaded on 08/27/2009 for the course CHEM 6 taught by Professor Hale during the Spring '09 term at CSU Northridge.

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Practice Exam Solutions - 1) Balance the following equation...

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