ANSWERS AND EXPLANATIONS TO PRACTICE EXAM I FREE RESPONSE QUESTIONS

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1 ANSWERS AND EXPLANATIONS TO PRACTICE EXAM I REE RESPONSE QUESTIONS. Mandatory Calculation - Equilibrium 0 points total NOCl (g) NO (g) + Cl (g) (a) (i) Initial pressure (atm) before decomposition: Given grams, temperature and volume think Ideal Gas Law PV= nrt irst, convert grams to moles and Celsius to Kelvin. mol NOCl 95.3 g NOCl =.46 moles g NOCl 35ºC + 73 = 308K Substitute and solve. nrt (.46 mol) (.08 L atm) (308K) P = = = 3.08atm V (.0 L) (mol K) (ii) ind concentration of NOCl. mol.46 mol = =. M liter.0 L (b) Calculate Kc if NOCl is 34.7% decomposed at equilibrium. Calculate the amount of NOCl present at equilibrium. (34.7%) (.46 mol) =.507 moles decomposed (this = x) NOCl left = =.95 moles Calculation of new molarities. Reaction NOCl (g) NO (g) + Cl (g) Initial Concentration.46 mol 0 0 Change -x +x +x Equilibrium Concentration.46 x.95 x.507 x.54 New Molarities.95 mol =.079 M.0 L.507 mol =.043M.0 L.54 mol =.0 M.0 L Set up the Kc expression. [NO] [Cl ] K c = [NOCl] Substitute and solve.

2 [.043] [.0] 3 Kc = = 6. 0 [.079] (pt) (c) Calculate Kp for the reaction. Set up the expression, substitute and solve. Kp = Kc (RT) n n = number of moles of gaseous product number of moles of gaseous reactant = 3 = Kp = (6. x 0-3 ) [(.08 L atm/mol K) (308K)] =.5 (d) Is reaction endothermic or exothermic? Justify. Must state and explain for the two points. The reaction must be endothermic. When the temperature increases from 35ºC to 50ºC, the number of moles of product greatly increases. With energy as a reactant, as in an endothermic reaction, equilibrium would shift to products in response to an increase in temperature. Energy + NOCl (g) W NO (g) + Cl (g) <explanation>. Kinetics 9 points total A + B ÿ C Experiment Initial [A] (mol. L B ) Initial [B] (mol. L B ) Initial Rate of Increase of [C], (mol. L B. sec B ) x 0 B x 0 B x 0 B4 (a) Write the rate law for the reaction. Rate = k [A] m [B] n irst calculate the order with respect to each reactant. Observe the data and notice that in experiment # and #3 the concentration of [A] is held constant. While [A] is constant, [B] doubles and the rate is quadrupled. When the concentration is doubled and the initial rate is quadrupled, the reaction is second order with respect to that component. This reaction is second order with respect to [B]. Notice that [B] is never held constant. Algebra will always work to solve problems of this type. Compare two rates, plug in the given values for concentrations and initial rates and solve. Showing the math always justifies the answer. Any two sets of data can be used.

3 m rate k [A] [B] = m rate k [A] [B] -4 m k (.040) (.80) -5 = m k (.00) (.0) m 6 = 6 m = *the k s cancel since they are constant m = 0 In other words, altering the concentration of A has no effect on the initial rate of the reaction. The rate law expression: rate = k [B] (b) Calculate the value of the rate constant and specify units. Substitute into the equation and solve. Any set of data may be used. rate = k[b] rate k = [B] for units mol/l s 3 k = =.5 0 L/mol s (.0 mol/l) (c) Calculate the initial rate of decrease of [A] in experiment 3. This is just stoichiometry from the reaction given at the beginning of the problem. A decreases at the same rate that C increases since they have a : mole ratio. ( mol/l s of C) mol A = mol/l s mol C Note: The negative sign is given to indicate that the rate is decreasing. (pt) (d) Which mechanism is consistent? Justify. In order to earn the points here, prove two things: The mechanism adds up to the balanced equation and the rate law written from the slow step matches the rate law calculated. I. A + B X (fast) X C +Y (slow) A + Y C (fast) II. B X A + X Y + B Y + A C (slow) (fast) (fast)

4 Both mechanisms add up to give the overall balanced equation. Mechanism I would give the following rate law: rate = k[x] (from slow step); [X] is an intermediate and its formation depends upon the fast step before. Using substitution, the rate law becomes: rate = k [A] [B] This does not match our calculated rate law. Mechanism II gives the following rate law: rate = k[b] calculated rate law, so this is the best choice. This does match our (e) Which substances are intermediates? Intermediates are those substances that are produced in one elementary reaction and used in a later reaction. The intermediates in the above reaction mechanism are X and Y. 3. Stoichiometry and Colligative Properties 9 points total (a) Determine the empirical formula from the data given. The empirical formula is the smallest whole number ratio of the elements in a compound. The first step in solving this problem is to assume that you have 00.grams of your compound and convert everything into moles. mol C g C = mol C.0 g C mol H 0.5 g H = mol H.0 g H mol O 9.95 g O =.46mol O 6.00 g O The second step is to calculate the mole ratio between elements. This will give subscripts. The moles of oxygen are the smallest so divide everything by the moles of oxygen mol C = 4.46 mol O mol H = 6.46 molo.46mol O =.46 molo

5 The empirical formula is: C 4 H 6 O (b) Calculate the molar mass from the freezing point data given. for calculating molality Molar mass = grams/ mole The problem gives the number of grams but moles must be found. Begin with the freezing point formula. T f = K f m i i = since the compound is a nonelectrolyte Calculate the temperature change: Normal freezing temperature: 79.8 ºC reezing point with solute added: 89. ºC T f = = 90.6 ºC Solve the equation for molality: o ΔT f 90.6 C m = m = o =.7 m K f 40.0 C/m for calculating molar mass ind the number of moles from the molality equation. m = mol solute kg solvent mol solute = m kg solvent =.7 mol/kg.045 kg =.0 moles solute (c) 8.75 g solute Solve for molar mass: molar mass = = 80.g/mol.0 mol solute Calculate mole fraction for camphor. Calculate the moles of each component. mol C 4 H 6 O 8.75 g C 4 H 6 O =.0 mol 80.g mol C 0 H 6 O g C 0 H 6 O =.96mol 5.00 g Calculate the mole fraction for camphor. moles camphor Χ camphor = total moles.96 = =

6 (d) Calculate the boiling point. Calculate the boiling point change. Substitute and solve. Δ T b = K b m i o = (5.6 C/m) (.7 m) o =.7 C Add the boiling point change to the normal boiling point. New boiling point: 07.4 º C +.7 º C = 0. º C (e) Calculate the vapor pressure of camphor above the solution. Vapor pressure is lowered for a nonvolatile solute according to Raoult s law. P solvent = (X solute ) (P solvent ) Calculate the mole fraction of the solute =.56 Calculate the change in pressure by substituting in to Raoult s Law. P solvent = (.56) (745 mm Hg) = 9 mm Hg Calculate the new vapor pressure. 745 mm Hg 9 mm Hg = 554 mm Hg Alternatively, the problem could be solved using the following: P solution = (Χ solvent ) (P º solvent) P solution = (0.744) (745 mm Hg) = 554 mm Hg

7 4. Net Ionic Equations choose 5 of 8 (3 points for each equation for a total of 5 points) or each reaction: point is awarded for correct reactants and points for correct products. All substances that ionize (strong acids, strong bases and soluble salts) must be written in ionic form. All spectator ions must be cancelled. point is awarded for balancing the reaction point is awarded for a correct answer to the descriptive chemistry question You can practice these to your heart s content at : (a) Dilute hydrochloric acid is added to a solution of potassium carbonate. (i) H + + CO 3 H O + CO (ii) Describe a test to positively identify the gaseous product. A glowing splint will be extinguished in the presence of the carbon dioxide gas. (b) Water is added to a sample of solid magnesium nitride. (i) 6 H O + Mg 3 N 3 Mg(OH) + NH 3 (ii) Describe any change in color or odor. The smell of ammonia will be noticeable as the reaction proceeds. (c) Ethanol is burned in oxygen. (i) C H 5 OH + 3 O CO + 3 H O (ii) Identify a possible source of the activation energy for this reaction. A lit match, spark or other ignition source could ignite this mixture. 5. Atomic Structure and Bonding---9 points total

In an essay of this type, be sure to read the prompt and address whatever is asked. In this essay it is important to discuss the bonding and/or draw Lewis structures to demonstrate understanding.

O O Cl O H O O Cl O H O Explain. Perchloric acid has four oxygen atoms compared to only three found in chloric acid.

8 In an essay of this type, be sure to read the prompt and address whatever is asked. In this essay it is important to discuss the bonding and/or draw Lewis structures to demonstrate understanding. When two substances are referred to in the question, be sure and refer to both substances in the explanation. (a) Perchloric acid is stronger than chloric acid. Explain. Draw each of the substances. O O Cl O H O O Cl O H O Explain. Perchloric acid has four oxygen atoms compared to only three found in chloric acid. Since oxygen is so highly electronegative it causes the net pull away from hydrogen to be greater with each additional oxygen, making it easier for water molecules to surround and remove the hydrogen of perchloric acid compared to that of chloric acid. An acid is considered to be strong if it ionizes close to 00% in solution. The less energy needed to remove the hydrogen, the stronger the acid. (b) C 4 H 0 is a liquid and CH 4 is a gas. Draw each of the substances. Both butane and methane are covalently bonded and are nonpolar molecules. The predominant type of intermolecular forces holding the molecules together are London Dispersion forces. (LDs) Butane, C 4 H 0, is a larger molecule with many more electrons than methane, CH 4. The more electrons in a substance, the greater chance for polarizability. Thus, butane is more polarizable and is held by LD more tightly than methane. The greater the intermolecular force, the more condensed the state of matter. (c) P 3 is polar and P 5 is not. Draw each of the substances.

P 5 P P P 3 P 5 Central Atom Central Atom Total Electron Pair: 4 Total Electron Pair: 5 Bonding Pair: Non-Bonding Pair: 3 P Bonding Pair: Non-Bonding Pair: 5 0 P Molecular Geometry: Trigonal

The one unshared electron pair on the central atom pulls with greater force on phosphorous than do the shared pairs of electrons with fluorine.

9 P 5 P P P 3 P 5 Central Atom Central Atom Total Electron Pair: 4 Total Electron Pair: 5 Bonding Pair: Non-Bonding Pair: 3 P Bonding Pair: Non-Bonding Pair: 5 0 P Molecular Geometry: Trigonal pyramidal Molecular Geometry: Trigonal bipyramidal P 3 has a trigonal pyramidal shape. The one unshared electron pair on the central atom pulls with greater force on phosphorous than do the shared pairs of electrons with fluorine. Because of this unequal pull on the central atom, the molecule has a net dipole moment and is polar. P 5 is a trigonal bipyramidal structure. The five shared pairs of electrons pull equally on the central atom causing the molecule to have no net dipole moment and thus nonpolar. (d) Bond lengths in the carbonate ion are all identical but are shorter than a single bond. Draw all of the resonance structures. The carbonate ion appears to have one double bond and two single bonds. Single bonds are longer than double bonds. None of the resonance structures above actually exist. The true structure is a composite of the three diagrams. Therefore, all of the bonds are about and /3 of a bond ---not single and not double.

10 6. Laboratory Question 9 points total (a) Steps necessary to carry out the experiment. When writing steps in an experiment, a brief list of the essential steps are a must. Steps for this experiment might include:. Mass out an alloy sample on the analytical balance.. Place the alloy into a beaker and add excess nitric acid. Be sure to perform this step under the fume hood. 3. Stir the beaker and its contents with a stirring rod to be sure that all metal is in solution. 4. Precipitate the silver ions by adding an excess of the sodium chloride solution. Stir with the stirring rod as sodium chloride is added. Continue this process until no more white precipitate forms. 5. Mass a piece of filter paper and set up the vacuum filtration system. 6. Empty the entire contents of the beaker into the Buchner funnel. Rinse the beaker with small amounts of distilled water to remove any residue. Empty this residue into the Buchner funnel. 7. Collect the solid and the filter paper and place in a drying oven. 8. Mass the solid and the filter paper. 9. Clean up the lab area. No two students will write the same procedure. The most essential steps that will earn points would include: React the alloy with nitric acid. Precipitate the silver with sodium chloride. Dry the sample of silver chloride. (pt) (b) List measurements: Mass of the alloy Mass of the filter paper Mass of the filter paper with the solid AgCl (one point for any two) (pt) (c) Calculations necessary to find % silver. mass of Ag from AgCl = mass of AgCl mass Ag % Ag = 00 total mass of alloy MM of Ag MM of AgCl (one point for each equation) (d) Would NaBr be appropriate? Explain.

11 (pt) NaBr would be a good substitute. The purpose of NaCl was to precipitate the silver ions and sodium bromide would be able to do this. Silver forms insoluble salts with chloride, bromide and iodide. Ag + + Br - AgBr 7. Extra Practice Essay ---Electrochemistry 8 points total (pt) (pt) (a) Predict sign of Eº and explain. The sign of Eº must be positive. The prompt gives a K value of.5 0 which means that the products are favored at equilibrium. Since the reaction proceeds as written, the voltage should be positive. (b) Identify reducing agent. Cd changes oxidation states from 0 to + thus, it is oxidized. Whatever species is oxidized is known as the reducing agent. (c) At a higher temperature, how would the cell potential change? Explain questions such as this with mathematical formulas if at all possible. There are two equations that allow this prediction. Gº = - neº and Gº = - RT lnk Therefore, - neº = - RT lnk Simplifying to solve for Eº: o RT lnk E = n The voltage should increase with an increase in temperature, all other factors held constant since voltage and temperature are directly proportional to each other. (d) Changing concentration of cadmium ions has what effect? Explain mathematically if possible. o.059 E = E lnq n (pt) + [Cd ].0 Q = = + [Cu ].0 The voltage will increase with a decrease in the concentration of cadmium. The ln of a number less than one will be negative. A negative multiplied by a negative will yield a positive a number, which will add to the standard voltage. (e) Cell potential at equilibrium. By definition, cell potential at equilibrium is zero

12 8. Thermodynamics 8 points total (pt) (pt) (pt) (a) Predict entropy change. Explain. S will be negative. Two moles of gaseous reactants form one mole of solid so the system becomes more ordered. (b) Predict free energy change at equilibrium. Explain By definition, Gº = 0 at equilibrium. (c) Temperature is increased G changes sign. Explain. The forward reaction is spontaneous at low temperatures, so G is negative. The Gibb s equation is useful in explaining this. Gº = Hº - T Sº Because S is negative, H must also be negative if the reaction is spontaneous. Therefore, if the temperature increases so much that the T S term is greater than the H term, the free energy value becomes positive and the reaction becomes nonspontaneous. (d) Write an expression for calculating the specific temperature referred to in part c. Begin with Gibb s equation: Gº = Hº - T Sº At equilibrium Gº = 0 and the equation becomes: 0 = Hº - T Sº Rearrange to solve for T: o ΔH T = o ΔS (e) What effect does adding more NH 4 Cl have on the value of the equilibrium constant? Explain. If the system is at equilibrium when more ammonium chloride is added, the equilibrium constant will remain the same. A solid is considered to have constant concentration and therefore, does not appear in the equilibrium expression.

ANSWERS AND EXPLANATIONS TO PRACTICE EXAM II FREE RESPONSE QUESTIONS

ANSWERS AND EXPLANATIONS TO PRACTICE EXAM II FREE RESPONSE QUESTIONS 1. Mandatory Calculation - Equilibrium 10 points total Ag 2 S (s) W 2 Ag (aq) S 2- (aq) a) Write the K sp and calculate its value. (solids