Chem 1a Midterm 3 Practice Set

 

Chemistry 1A, Spring 2009 Midterm Exam III Practice Set and Solutions Author’s Note: This practice set for the third midterm is significantly shorter than the previous one in  terms of the number of questions. Unfortunately, I had greater time restrictions this  time around due to other midterms and other obligations. I spent the majority of my  limited time writing up expanded solutions for not only my questions, but also all the  multiple choice questions on the past midterms posted on bSpace. I felt that  understanding how to approach problems would be more important than just  memorizing the methods and equations for every type of problem, especially with the  conceptually heavy chapters that were covered this time. It’s also a minor excuse on  my part since writing decent problems actually take longer than writing solutions.  Hopefully you’ll find this practice set helpful for the the midterm. And as always, I  hope that you’ll all perform brilliantly Monday evening. ^_^ ~Stephen Kok

Informal Student Review Sessions If you haven’t been to any of the ones I held last midterm, these are basically Q&A driven reviews  where I’ll basically  answer questions and problems to clear up concepts that may still be a bit  fuzzy. For the 2nd review session, it is still tentative since apparently the buildings on campus will  be locked down due to Easter Sunday. If I actually cannot get a classroom that day, it will be held in  the Clark Kerr Campus “library” in Building 12, directions will be given if needed.  #1 Saturday, April 11th 6:00pm‐9:00pm – 120 Latimer *Tentative* # 2 Sunday, April 12th 2:00pm‐5:00pm – TBA  bSpace Chatroom  

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Practice Problems 2-6 Multiple choice and Free response Expanded Solutions 7-13 Practice Problems Expanded Solutions 14-27 Midterm III 2007 Expanded Solutions 28-33 Midterm III 2008                 *Any questions, complaints, etc.  regarding this practice set should sent to me.  [email protected]* ©2009 Stephen Kok UC Berkeley Undergraduate

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Practice Prob blems   

For probblems 1-3, usse the follow wing informaation.  Four ideentical flaskss are filled with w one molle of each off the followinng gases at 25˚C 2 and seaaled: 2 Oxygen (O2), Nitrogen N (N ), Krypton (Kr), ( and Chhlorine (Cl2) 1) In whiich flask do the particless have the leaast collisions per secondd with the glaass walls? B) Nitrogen-fille A Oxygen-fiilled A) N ed C Krypton-fi C) filled D) Chlorine-fille C ed E All of them E) m have the same collisioons per seconnd 2) Order the flasks by y decreasingg relative kinnetic energy B) Kr>Cl A N2>O2>Cll2>Kr A) K N2 2>O2>N B N2>Cl2>O B) O2>Kr D) Kr>O K > 2 2>Cl2>N E O2=N2=Krr=Cl2 E) 3) Calcullate the root mean square speed of eaach of the gaas samples. N2 O2

Cl2

Kr

4) A flaskk of 3.25L of o fluorine gaas at 34˚C annd 1.4 atm iss heated untiil its pressurre is at 12.6 atm. How has the root meean squared velocity v of thhe gas particcles changedd? A Doubled A) B) Tripled T C Quadrupleed C) D) More M than quuadrupled E Not enoug E) gh informatioon Below is a Maxwell Distributionn of speeds of gaseous paarticles. Usinng this graphh, answer thee next two questions. A fracction

vrms B C

D speeed

5) Given that Distribution A is Bromine B gas (Br2) at 35°C C, which of the distributtions would best b representt Argon (Ar)) at 35°C? A Distributio A) on A B) Distribution D B C Distributio C) on C D) Distribution D D E None of th E) he above

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6) Given that distribution D is Helium (He) at 255°C, which of the following temperatures would best define distribution B, if the gas of interest is still Helium? A) 16°C B) 63.8°C C) -240°C D) -141°C E) None of the above 7) Which of the following ways would reduce the internal energy of an open system? A) Add heat to the system B) Do work on the system C) Add matter to the system D) Remove matter to the system E) None of the above 8) Which of the following ways would reduce the internal energy of a closed system? A) Do work on the system B) Add heat to the system C) Have the system do work D) Add matter to the system E) None of the above 9) Which of the following ways would reduce the internal energy of an isolated system? A) Add heat to the system B) Do work on the system C) Remove heat from the system D) Have the system do work E) None of the above 10) A reaction occurs in a cylinder containing a pistol which is displaced 5.4L after the expansion of gas at 0.2kPa. Find the work done by the system. A) +1080J B) -1080J C) +1.08J D) -1.08 E) 0 11) In another cylinder, the system does 238kJ of work on a piston while 57kJ of heat is added to the system, following the reaction, 83kJ of work is done on the system to recompress the gas while removing 35kJ of heat. What is the change in internal energy? A) -299kJ B) +133kJ C) -133kJ D) +343kJ E) -343kJ 12) A gaseous mixture in a cylinder at 2.81 atm had an increase of 572J in its internal energy following a procedure involving work. What was the procedure, and how much did its volume change by? A) Expansion, +203.6L B) Compression, -203.6L C) Expansion, +2.01L D) Compression, -2.01L E) Not enough information 13) Find the temperature change of 24.5g of Ar(g) at 1.21atm when 357J are added to the system at constant pressure. Assume that chlorine behaves as an ideal gas. A) +4735K B) +28.0K C) +2841K D) +46.7K

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Δ

6.01

The following information is provided for the next question    Δ

40.7

    

4.184      °

2.03    °

2.03 °

14) 8.2kJ of heat is supplied to a sample of 21g of ice at -25°C, calculate the final temperature of the sample. Show your work.

Consider the combustion of pentane as shown below for the next two questions 11 5 12               Δ ° 3537 15) Calculate the amount of heat released in the combustion of 2.76 mol in the presence of excess oxygen. Show your work

16) In a particular combustion of pentane, the heat released was measured to be -537kJ. If the combustion occurred with excess oxygen, what volume of pentane was combusted? Assume standard temperature and pressure. (0°C and 1 atm) Show your work.

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17) Determine the reaction enthalpy for the hydrogenation of ethyne to ethane (shown below) from the following information. Show your work. 2 , 1300 / Δ ° Δ ° , 150 / Δ ° , 286 /

Use the following information regarding bond enthalpies to answer the next two questions Bond Mean Bond Enthalpy Bond Mean Bond Enthalpy C-H 412 kJ/mol H-Cl 431 kJ/mol C-O 360 kJ/mol C≡C 837 kJ/mol Cl-Cl 242 kJ/mol C=C 518 kJ/mol C-Cl 338 kJ/mol C-C 348 kJ/mol 18) Using the bond enthalpies in the table above, estimate the enthalpy of reaction for the following reaction:

19) Using the bond enthalpies in the table above, estimate the enthalpy of reaction for the following reaction:  

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Info Table Δ ° ° 

CaCl2(aq) C2H2(g) CaC2(s) HCl(aq) -877.1

+226.73

59.8

167.16

59.8

200.94

69.96

56.5

20) Using the information above, calculate Δ ° for the following reaction at 80°C  2   

21) 6 2 The reaction above has Δ ° and Δ ° which the reaction would be spontaneous.

3 Find the range of temperature at

22) Write the equation for the equilibrium constant K for the previous reaction and then calculate its value at 50°C. Determine the partial pressure of HF given that the pressure of BF3 is 3.54 bar under the same conditions

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Expanded Solutions  Practice Problems

1) (C) “Collisions per second” is basically another way of asking about the average velocity of the particles. Intuitively, the faster the particles are going, the more collisions there will be with the flasks. Velocity is proportional to the temperature of the gas, but in this instance, all the gases are at the same temperature. However, velocity is also inversely proportional to the molar mass of the particles. Thus, the gas particles that are the smallest will have the highest average velocity, while the largest gas particles will have the lowest average velocity, hence the least collisions per second, which in this case would be the huge krypton atoms. 2) (E) Relative kinetic energy can be easily confused with the velocity of the particles since many people directly relate kinetic energy with how fast the particles are moving. This is partially correct, however, if you look at the formula for the kinetic energy per mole of gaseous particles , where R is the gas constant in Joules per Kelvin per mole and T is the temperature in Kelvin, there doesn’t seem to be a velocity variable! However, to those that are also taking physics, you might also be wondering why there isn’t a variable for mass. The beautiful part of this formula is that the mass and velocity essentially cancel each other out, relating the kinetic energy only to the temperature of the gas (as long as it’s considered ideal). This is because the velocity of the larger particles is slower, but their mass is larger, thus their large mass essentially balances out the slower velocity. This holds true for the smaller particles moving at higher velocities are well. I will not go over the derivation here, but understand that although different gas particles may travel at different velocities, their average kinetic energy is the same for the same temperature because of the mass-velocity relationship. Thus, the flasks would all have the same relative kinetic energy because they’re all at the same temperature. 3) 3

3

8.31 · 0.028 8.31 · 0.071

298 / 298 /

515  /

323  /

3

3

8.31 · 0.032 8.31 · 0.084

298 / 298 /

482  /

297  /

Straight forward calculations, but you have to make sure your numbers are in the correct form. First, make sure that your temperature is in Kelvin not Celsius, you generally don’t even want to use temperatures in Celsius unless a specific heat is given in Celsius rather than Kelvin. Next, make sure that your molar mass (M) is in kilograms/mol rather than grams/mol. This is a simple but huge mistake that I generally encounter with others and myself sometimes. If you have a ridiculously small velocity, use your best judgment to decide whether you have made a unit error somewhere in the calculation. 4) A very simple problem that has a simple trap that could easily be missed. First off, we start with the ideal gas formula . Since the volume and moles of gas isn’t changing, we’ll simply ignore than along with the gas constant. Pressure is directly proportional to temperature in this case, which would mean an increase by a factor of nine in the pressure would ©2009 Stephen Kok UC Berkeley Undergraduate

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result in an increase by the same factor of the temperature. Now we’re see how this affects the root mean squared velocity.

The molar mass isn’t going to change in this problem

just like the gas constant, so we’ll focus only on the temperature. If the temperature increases to 9x its original amount, the square root over the temperature means that the velocity only increases by a factor of 3, tripled, not nine. Make sure that you check the formulas before jumping to conclusions and problems like these will be free points. 5) (B) If you’re given a Maxwell distribution of speeds, most likely you’ll encounter problems such as these. You’ll be comparing distributions by using the root-mean squares velocity formula 3 where R is the gas constant in J/mol, T is the temperature in Kelvin, and M is the molar mass in kilograms. Looking at the formula, the only variables are T and M, so all distributions are basically related to each other by a factor of

. You’ll be using this relationship to solve this

problem. This first question asks about the distribution of bromine versus argon. Note that the temperature does not change, but the molar mass of the compounds do. Bromine has a molar mass of 160g/mol or 0.16kg/mol, while Argon has a molar mass of 40g/mol = 0.04kg/mol. If we compare using the relationship

, we notice that Argon would have twice as much velocity as

Bromine since its molar mass is only ¼ as large, which through the square root becomes ½. Since it’s in the denominator, we take its reciprocal and see that the velocity is raised by a factor of two. Comparing distribution A to the rest of the distributions, B is a vrms twice as fast compared to A, so that’s the correct answer. 6) (C) In this problem, we encounter the relationship again, but this time the molar mass is staying constant while the temperature is changing. Comparing distribution D to distribution B, we find that B has a fourth of the vrms compared to D, so that means the new temperature would be one sixteenth of the original due to the square root. Here is a key point, the equation has temperature in terms of Kelvin (which is pretty much standard for all calculations with minor exceptions such as calorimetry problems), so first convert 255°C into Kevlin and then perform the operations. 255 273 528 528 33 16 33 273 240° 7) (D) An open system allows the system to do work and exchange heat and matter. In this case, removing matter from this system would decrease the internal energy of the system, since matter itself contains its own internal energy. Like removing money from your bank account, the “internal energy” of your account has decrease in terms of monetary values ^^. 8) (C) A closed system allows the system to do work and exchange heat, but not matter, hence “closed” system. The internal energy may be decreased by having the system do work, all the ©2009 Stephen Kok UC Berkeley Undergraduate

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others would increase the internal energy. This is similar to your own energy being consumed in work in terms of exercising. (However, understand that the human body is an open system) 9) (E) An isolated system is simply an “isolated” system in which matter, heat nor work could be exchanged or done. This means that the internal energy of the system stays constant because the surrounding conditions cannot affect it. The closest example to an isolated system is a sealed thermos, in which work cannot be done by or on it, matter cannot be added because it’s sealed, and heat generally isn’t exchanged. 10) (B) Straight forward calculation of work, don’t mess up your signs!! Δ w-work in Joules, Pex-external pressure in Pascals, ∆V-change in volume 200 5.4 1080 In case you get atm instead of Pa, convert your final answer by multiplying by 101.325J/L*atm 11) (C) Keep track of your signs! Work done by the system decreases its internal energy (negative), work done on the system increases its internal energy (positive), heat added to the system increases its internal energy, and heat removed from the system decreases its internal energy. 57 35 22   238 83 155 Δ 22 155 133 12) Another straight forward calculation, make sure you take into account the units! 572 5.645 ·   101.325 / · 5.645 · Δ Δ 2.81 Δ 2.01 , 13) (B) This is probably not a problem you’ll encounter on the midterm due to the difficulty of the concept. However, since it appeared on the homework, it’s still considered fair game. In this problem, you’ll be using the following formulas:                     Δ You have , the heat provided to the system, but you’ll have to solve for the heat capacity first. The heat capacity is defined as such, for every degree of motion possible for the particle, the heat capacity increases by a factor of ½R. Translational motion, the motion of the particle in the three-dimensional world gives the particles three degrees of motion. For molecules, you’ll also have to consider the rotation of the particle which could provide an additional two to three extra degrees of motion. Now we have a heat capacity of , however, this is only the heat capacity for atoms at constant volume. Luckily, the heat capacity for constant pressure is related by the , so the heat capacity for argon gas would be . Unfortunately, we’re not equation completely done, as if you remember, heat capacity is varies according to the quantity of matter.

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In this case, we’re using the gas constant R in J/mol, so the gas constant must be multiplied by the moles of argon gas present in this reaction. 5 2

24.5 40

Now, finally using the initial formula: 5 2

24.5 40

357 Δ

Δ

357

Δ

28

24.5 40

5 2

14) When doing calculations involving samples that could possibly undergo phase change, its best to break the problem into parts. In this case, we’ll start by determining the amount of heat necessary to heat the sample of ice to 0°C. Δ 2.03 0 25° 21 °

1066 We have more than enough heat for such a process. Energy remaining: 8200 Now we’ll calculate the energy consumed for the sample to undergo fusion. Δ 21 6.01 7.012 7012 18

1066

7134

The energy remaining now is 7134 7012 122 . It seems quite probable that the sample of water will not be heated to the boiling temperature, so we’ll use the remaining heat directly to calculate the final temperature. Δ 4.184 122 21 0 21

122 4.184

1.39

15) This is a simple calculation given that you understand that the enthalpy of combustion is the heat released for 1 mole of pentane and 11 moles of oxygen. Given that oxygen will not limit the reaction, the heat released for a particular quantity of pentane is simply the enthalpy of combustion multiplied by the quantity in moles. 3537 Δ 2.76 9762

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16) In this problem, we’re basically going in reverse, solving for the amount of pentane that was combusted. We start by taking the ratio of the heat released compared to the heat released by a mole of pentane with excess oxygen. 537 0.152    3537 Note that if you wanted to find the amount of carbon dioxide or water produced in this reaction, you would simply multiply this ratio by their stoichiometric coefficient. Since this problem wants this quantity in terms of volume, we’ll use the ideal gas formula. 0.152 8.06 10 1 3.34

0

273

17) This problem has Hess’ Law marked all over it. You should first start by writing the balanced equations for the combustions of the gases. (Note that fractional coefficients are allowed in order to preserve the enthalpy of combustion for 1 mole of the hydrocarbon) 5 2                  Δ ° 1300 2 7 2 3                Δ ° 1560 2 1                                            Δ ° 286 2 Manipulating the equations to get the reaction: 5 2                    Δ ° 1300 2 7 2 3                  Δ ° 1560 2 2 2                                                  Δ ° 2 286 2 312 Δ ° 1300 1560 2 286 18) For bond enthalpy type problems, it may help you considerably to draw a rough sketch of the molecule in order to help you decide which bonds are broken and which bonds are formed. In this particular reaction, a C-H bond and a Cl-Cl bond are broken apart, while a C-Cl bond and a H-Cl bond are formed. Remember that breaking bonds require energy while forming bonds releases energy. Thus Δ Δ Δ Δ Δ 412 242 338 431 Δ 115 / 19) Same as above, except with a different reaction. Δ Δ Δ Δ C C O H C C ©2009 Stephen Kok UC Berkeley Undergraduate

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C

O

C

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Δ

612

Δ

463

348 360 45kJ/mol

412

20) These types of problems have a high chance of appearing on the midterm since they cover all three calculations of state functions in one run. The problem itself is generally not difficult, just make sure you don’t make any calculation mistakes. , Δ ° 2Δ ° Δ ° Δ ° , Δ ° , , Δ ° 256.3 / Δ ° ° , ° , ° , 2 ° , Δ ° 77.8 / Important detail: Note that your calculated entropy is in Joules rather than kilojoules like your enthalpy, make sure you convert your entropy into kilojoules before using the Gibb’s free energy relation formula. Remember also to convert temperature to Kelvin. Δ ° Δ ° Δ ° 77.8 Δ ° 256.2 80 273 1000 Δ ° 283.7kJ/mol 21) If you recall from lecture, depending on the sign of the enthalpy and entropy of a reaction, it can be spontaneous or non-spontaneous at particular temperatures. In this case, we’ll be using the Gibb’s free energy formula relating the enthalpy and entropy of a reaction: Δ ° Δ ° Δ ° Note that for a reaction to be spontaneous, Δ ° 0. In this case of this reaction, we have a negative enthalpy along with a negative entropy, which means that the reaction is only spontaneous at certain temperatures. To find the boundary at which the reaction becomes spontaneous or non-spontaneous, we set Δ ° 0 and solve for T, the temperature in Kelvin. 0 232 379 232 612 379 / 1000 If T<612K, Δ ° 0, hence the reaction is spontaneous for temperatures less than 612K. 22) The equilibrium constant equation is written as shown:

Note that and do not appear in the equation since pure solids and liquids have a value of 1. The stoichiometric coefficients of the molecules determine how many times its partial pressure value is multipled by itself rather than being a simple multiple as when determining the other state functions such as enthalpy and entropy. To calculate the equilibrium constant with this formula, we would require the partial pressure of BF3 and the partial pressure of HF, but since we have neither piece of information, we’ll have to use a different equation. In the previous problem, we were given information regarding the enthalpy and entropy of the reaction. With those pieces of information, we can calculate Gibb’s free energy, which relates to the equilibrium constant in the following way: Δ ° First off, we’ll calculate Δ ° ©2009 Stephen Kok UC Berkeley Undergraduate

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Δ °

Δ °

Δ °

232

Δ ° 110 Rearranging the first equation to solve for K

50

273

379 1000

110000 °

6.14 10 Remember to convert your Gibb’s free energy in Joules in order to match the units of the gas constant or else you’ll end up with the wrong K value. Now that we have the calculated value of K, we can solve for the partial pressure of HF using the equilibrium constant equation.  3.54 6.14 10 3.54 0.00165  6.14 10 This is an extremely small partial pressure, however, if you look at the K value, you see that its extremely large. An extremely large K value generally indicates that that the reaction is productfavored, and in this case, nearly no reactants would be left, hence explaining the tiny partial pressure.

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Expanded Solutions  Midterm III 2007 1) A mixture of 1.00 g of H2 and 1.00 g of He exerts a pressure of 0.480 atm. What is the partial pressure of each gas in the mixture? A) PH2 = 0.240 atm PHe= 0.240 atm B) PH2 = 0.320 atm PHe= 0.160 atm C) PH2 = 0.160 atm PHe= 0.320 atm D) PH2 = 0.384 atm PHe= 0.096 atm E) PH2 = 0.096 atm PHe= 0.384 atm (B) This is a question that actually should have appeared on the last midterm since this deals with Dalton’s partial pressures. First calculate how many moles of gas you have 1 1 0.5                          0.25 4 2 Find the total amount of moles and then calculate the proportion of the gases in the mixture       0.5 0.25 0.75  0.5 2 0.25 1 %                      % 0.75 3 0.75 3 Now, since partial pressures is determined by the proportion of gas particles there are for each compound, we multiply the proportions by the total pressure. 2 1 0.480 0.320                     0.160 3 3 0.480 2) Consider the balanced reaction for the combustion of octanol: C8H18O (g) + 12 O2 (g) → 8 CO2 (g) + 9 H2O (g) What volume of carbon dioxide gas will be produced from 5.00 g of octanol and a stoichiometric amount of oxygen if the reaction is carried out at 28oC and 0.984 atm? A) 0.308 L B) 6.89 L C) 7.72 L D) 8.69 L E) 0.966 L (C) This is a limiting reaction-type problem. From the information, we know that by “stoichiometric amount of oxygen”, they mean that we have just enough oxygen to bring the reaction to completion given the amount of octanol. So… solving for the moles of octanol 5 0.0385    130 If this reaction goes to completion, then for every one mole of C8H18O that reacts, 8 moles of carbon dioxide is produced. However, we have less than 1 mol, so by proportion 8   0.0385   0.3077     1      Converting the moles of CO2 into volume by using the ideal gas equation 0.3077 8.2057 10 28 273.15 7.72 0.984

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3) Which change(s) in a sample of ideal gas would have the same effect on the pressure as increasing the velocity of the particles? Mark all that apply. A) Increase in temperature. B) Increase in volume. C) Decrease in pressure D) Increase in the number of moles of gas. E) None of these. (A/D) In this problem, we’re relating pressure to different situations applied on the system. Increasing the velocity of the particles will definitely increase the pressure of the system since there will be more collisions. So we’re finding the situations that would also increase pressure. Increasing the temperature would increase the velocity of the particles by the root mean square velocity formula essentially giving the same effect. Increasing the volume would decrease the pressure since the particles now have more space to occupy and would collide into the walls less frequently. Decreasing pressure… would decrease pressure, no explanation needed there. Increasing the number of moles of gas would mean there are now more particles occupying the same space, which would also increase the pressure exerted on the walls. It’s like starting out with a full lecture room and then doubling the number of students in there, which would increase the pressure in the room, not to mention the tension o.O… 4) What is the pressure required to compress a 1.00 L sample of air at 1.00 atm to 0.400 L at 25°C? A) 1.0 atm B) 1.5 atm C) 2.0 atm D) 2.5 atm E) 3.0 atm (D) Extremely simple calculation using a derived formula from the combined gas law    

   

 

1.00 1.00 0.400 2.5 Note that temperature isn’t used in this calculation because it doesn’t change. For the next three questions, consider four flasks filled with 1.0 mol of different gasses: Flask A: O2 at 760 torr and 10oC. Flask B: N2 at 900 torr and 80oC. Flask C: He at 800 torr and 10oC. Flask D: He at 500 torr and 80oC. 5) Which flask(s) will have molecules with the greatest average kinetic energy? (B/D) Whenever something refers to kinetic energy of gas particles, you should immediately . Note that mass and pressure are not taken into account, only refer to the formula temperature determines the average kinetic energy. Thus, gas particles at the same temperature would have the same average kinetic energy, which in this case, flasks B and D both have the highest temperature and thus the highest average kinetic energy.

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6) Which flask(s) will have molecules with the greatest root mean square velocity? (D) The room mean square velocity formula is

. There are two variables in this

equation, T, temperature in Kelvin, and M, the molar mass of the compound in kilograms. Basically, the highest the temperature and the lower the molar mass, the higher the velocity. In this case, we have both flasks B and D at the same high temperature, but since D contains helium which has a smaller molar mass compared to nitrogen gas (4g vs. 28g), the particles in D would have a highest velocity. 7) Which flask(s) will have the greatest number of collisions per second with the walls of the container? (B) Collisions per second is basically another way of asking which flask has the highest pressure. From the pressure values, we see that flask B is at 900 Torr, which is the highest pressure value thus it has the greatest number of collisions per second. For the next four questions consider a system which releases 546 kJ of heat while its internal energy drops by 125 kJ. 8) What can you infer about the process? Mark all that apply. A) Work is involved. B) The entropy of the system decreases. C) It is spontaneous. D) The first law of thermodynamics does not apply. (A/B) From the information, we are told that heat is released and the internal energy of the system decreases, but the numbers don’t match up. This means there’s a different factor, work, that’s involved in this process. Since the internal energy drops, this also means that entropy decreases since the loss of internal energy would mean that the particles are becoming less chaotic and becoming more ordered. Almost like how you would feel after a sugar rush. ^_^ 9) Based on the information given, which relationship(s) should be used to calculate the amount of work involved? Mark all that apply. A) w = -P ΔV B) ΔE = q + w C) qsystem = qsurroundings D) w = -RTlnK E) S = kBlnW (B) Simple enough. Since work is involved, C is removed. Pressure and volume isn’t given so A is removed. Temperature and the equilibrium constant are not given so D is removed along with E. So the answer is B, clear enough.

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10) How much work is done (if any) in kJ? (D) Δ and

A) -125 B) -421 C) +125 D) +421 E) 0 . In this case we have the values Δ 125 since the internal energy drops 546 since heat is released by the system. Solving for w, we get +421kJ.

11) What best characterizes the work done? A) Work is done on the system. B) Work is done by the system. C) Work is done both on and by the system. D) Cannot determine based on the information provided. (A) Since work is positive, work is done on the system. If this is hard to remember, refer back to the formula Δ . If work is done by the system, internal energy decreases since its consumed in the work, which makes it negative. If work is done on the system, internal energy increases since energy is being added by the work which makes it positive. 12) A system undergoes a change that can be broken down into two steps: 1) the system does 70 J of work while absorbing 62 J of heat, and 2) the system releases 29 J of heat while 94 J of work is being done on it. What is the overall energy change of the system? A) +57 J B) -57 J C) +9 J D) -9 J E) +255 J (A) Basic adding and subtracting of heat and work, make sure your signs are correct 29 62 33 70 94 24 Δ 57 13) Consider the chemical reaction that occurs in a hydrogen fuel cell: 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = -572 kJ What is the heat change (in kJ) when 186 g of oxygen reacts with excess hydrogen? A) -572 0 B) +5720 C) +3320 D) -3320 E) -1660 (D) Limiting reaction by oxygen. First off, solve for the number of moles of oxygen 186 5.8125      32 / Since this reaction produces -572kJ for each mole of oxygen, we multiply the enthalpy of reaction by the moles of oxygen we have (since the reaction goes to completion with excess hydrogen) 572 3325~ 3320 5.8125    

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14) Consider the following reactions carried out at a constant temperature and pressure. In which of these will the amount of work done be positive? Hint: only the coefficients of the reactants are given. All species are gas unless otherwise noted. A) The synthesis reaction of 2 mol SO2 and 1 mol O2 to give SO3. B) The synthesis reaction of 4 mol NO to give O2 and N2. C) The decomposition reaction of 2 mol CCl2O to give CO and Cl2. D) The combustion of 2 mol CH3OH. E) The conversion of 2 mol of solid I2 to gaseous I2. (A) This is a trickier problem. First off, we have to define when work is positive. If the value of work is positive, this means that work is done on the system, which would increase its internal energy. All the following reactions involve gaseous compounds, so to apply work to the systems would mean that there is a compression of gas. To determine which reaction would decrease the number of moles of gas particles, we’ll have to balance all of them. 2 2 2                 2

3

2

4

From balance reactions, we can see that all of the reactions except the first would produce more moles of gas, since the moles of reactants is less than the moles of products. The first reaction has 3 moles of gas going into 2 moles of gas, so this means that there is an overall compression occurring which gives the positive work the question is asking about. 15) Consider the reaction for which ΔHo~200 kJ/mol and ΔSo~100 J/mol CaO (s) + CO2 (g) → CaCO3 (g) Under which condition(s) will this reaction be spontaneous? A) At high temperatures. B) At low temperatures. C) The reaction will always be spontaneous. D) The reaction will never be spontaneous. (A) This problem deals with the formula relating Gibb’s free energy Δ and the entropy and enthalpy of the system, which is Δ ° Δ ° Δ °, where T is the temperature in Kelvin. So in this problem, we’re given information about the enthalpy Δ ° and the entropy ΔS°, so putting it into the equation. Remember that entropy is given in Joules per mole rather than kiloJoules. Δ ° 200 0.1 Now we discuss what Δ would imply when it is a positive value versus a negative value. A positive Δ ° would mean that the reaction would not be spontaneous, while a negative Δ ° would imply a spontaneous reaction (though in later chapters we refined this definition to be “reactant-favored” and “product-favored” respectively). So if we want the reaction to be spontaneous, we would want Δ ° to be negative. This is the case only if the temperature, T, is larger than 2000K. Hence, the condition that would make this reaction spontaneous would be at high temperatures.

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16) A 46.2 g sample of Cu is heated to 95.4°C and then placed in a calorimeter with 75.0 g of water at 19.6°C. The final temperature of the water is 21.8°C. What is the heat capacity of the Cu in kJ/ °C·g if the calorimeter does not absorb any heat? A) 0.203 B) -0.203 C) 0.0770 D) -0.0770 E) 4.18 (A) This is a basic calorimetry problem, easier too since we don’t have to take into account the heat absorbed by the calorimeter as in the lab. Since heat must be conserved in an isolated system, we’ll use the following equation: Δ Δ We know all the values except for the specific heat of copper , where we are solving for, so rearranging the equation we get: Δ Δ 4.184 75 21.8° 19.6° ° 46.2 21.8° 95.4° 0.203 / ° Remember to check your units in these problems since sometimes the heat capacity is given in kilojoules rather than Joules, but temperature is generally kept in Celsius instead of Kelvin. For this problem, I think they meant Joules as their unit instead of kilojoules since the specific heat of water is 4.184J/g° rather than 4.184kJ/g° as given on their equation list. 17) Along which isotherm(s) will a phase change be observed? Mark all that apply. A) A B) B C) C D) D E) none of these (A/C) With respect to the phase diagram, an isotherm is a line in which the temperature stays constant throughout the particular line, hence the prefix of “iso” which basically means the same or constant. Since the temperature increases as you move from left to right, there will be an isotherm only when a line is pointed upward or downward since that would indicate only a change in pressure and not temperature. D is disregarded in this case because there is no “phase change” that occurs, which is visually represented as the intersection of a phase boundary with the isotherm. 18) Along which line will there be a positive change in entropy? Mark all that apply. A) A B) B C) C D) D E) none of these (B) An increase in entropy, or simply put, an increase in chaos always occurs when moving from any lower energy phase to a higher energy phase. Examples would include melting (solid to liquid), vaporizing (liquid to gas), or subliming (solid to gas). The reverse phase changes would indicate a negative change in entropy, moving towards a more ordered state. A represents deposition (gas to solid), a negative entropy; B represents melting (solid to liquid), a positive entropy; C represents condensation (gas to liquid), a negative entropy; and D doesn’t represent any phase change, but there is still an increase in pressure. An increase in pressure generally

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means a negative entropy as well since pressure generally decreases the volume of the content and would force it to become more ordered. 19) Which of the following represents a plot of the vapor pressure vs. T? (A) This one is pretty much you know it or you don’t. Vapor pressure is directly proportional with the vaporization and condensation of compounds, which occurs on the phase boundary between the liquid and gas phases. Thus, the same line represents the vapor pressure, and hence the answer is A. For the next five questions, consider the gas phase reaction N2 + 3 H2 ↔ 2 NH3 with an equilibrium constant of K = 0.278 at 800 K. The entropies of formation (in kJ/mol) are: H2=130.6, N2=191.5, & NH3=192.3. 20) What is the standard entropy change associated with this reaction (J/mol)? A) –198.7 B) –129.8 C) -99.4 D) 129.8 E) 198.7 (A) We’re given the standard entropies of formation for all the molecules so we can simply use the formula: Δ ° Σ ° Σ ° 191.5 130.6 192.3 3 Δ ° 2 Δ ° 198.7 / Remember to multiply by stoichoimetric coefficients if there are any in the balanced equation. 21) Which is the free energy change associated with this reaction in kJ/ mol? A) –84.0 B) +84.0 C) -8.51 D) +8.51 E) +4.26 (D) First we look at what values we’re given for this problem which are K, the equilibrium constant, T, the temperature at which the reaction occurs, and the recently calculated Δ °, the standard entropy of reaction. From the formula sheet, we find an equation that involves two of the three values for the calculation of Gibb’s free energy, excluding the entropy of reaction (which is common, since it’ll prevent students from getting both equations wrong due to the miscalculation of the first). The formula is Δ ° 8.3145 800 ln  0.278 Δ ° Δ °

8514.94

8515 /

However, the problem wants the answer in kilojoules per mole, so convert by dividing by 1000. 8515 8.515 / 1000

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In this problem, keeping your signs straight is crucial in getting the correct answer since many calculation type problems will have similar values with only a simple change in sign. Make sure when using this formula that your temperature is in Kelvin, though it was already given in this problem. 22) Analysis of a reaction mixture yielded the following partial pressures: N2 = 0.417 atm, H2= 0.524 atm and NH3= 0.122 atm. What is the reaction quotient for the reaction? A) 0.116 B) 0.248 C) 2.90 D) 1.91 E) 0.881 (B) The reaction quotient, Q, is given by the following formula:                    . Since this is specified as a “gas phase reaction” in the problem, we won’t have to worry about partial pressures of pure solids and liquids (which would be set at 1). The partial pressure of each molecule is taken to the power of its coefficient, rather than simply multiplied. So throwing in the values of the partial pressures: 0.122 0.248   0.417 0.524 23) Which best describes the condition of the reaction mixture in the previous question? A) The reaction mixture is at equilibrium. B) The reaction mixture is not at equilibrium – it will proceed toward reactants. C) The reaction mixture is not at equilibrium – it will proceed toward products. D) Can’t tell. (C) This one is linked to the previous problem regarding the reaction quotient since we’ll be comparing the equilibrium constant with the quotient value. From the problem we have K=0.278 and Q=0.248 and the formula for Q is:    Since Q≠K, the reaction is not at equilibrium, so now the question is which way the reaction would proceed. Since Q
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24) If equilibrium is reached for this reaction, and the volume of the container is then reduced by a factor of two. Which best describes how the reaction mixture will proceed? A) Toward products. B) Toward reactants. C) No change. (A) Now that equilibrium was reached, the volume is reduced by a half (factor of two), which would mean the pressure of the system increases due to less space for the gas particles. Looking at the formula for the reaction quotient to see if the system is at equilibrium:   We assume that since volume decreases by a half, partial pressures are doubled since pressure and volume are inversely proportional. (Remember ?) So…changing the quotient 2 2

4 2

2

4 4

2

2

2 So the reaction is no longer at equilibrium since its value is different due to the decrease of volume. Since the reactants’ partial pressure value (bottom) is now larger, the reaction quotient would be smaller, and if the reaction quotient is smaller, than there must be an increase in products to increase the partial pressure of the top of the quotient. Another way to look at this is with le Chatelier principle in terms of the reaction equation: N2 + 3 H2 ↔ 2 NH3 All the molecules are gases, so we can compare the moles and thus partial pressures of the gases. On the reactant side, there are four moles of gas, while there are only two moles of gas on the product side. If there is a decrease in volume, and thus an increase in pressure, the reaction would move in the direction that would “save” more space. Since two moles worth of partial pressure would take up less volume than four moles worth of partial pressures, the reaction would move in the product direction. Consider the following three reactions for the next two questions: 1) N2 (g) + O2 (g)→2 NO (g) K1 2) N2 (g) + 2 O2 (g) → 2 NO2 (g) K2 3) 2 NO (g) + O2 (g) → 2 NO2 (g) K3 25) What is an expression for the reaction quotient for reaction 1? (E) Simple enough, since all the reactions are in the gas phase, it’s the sum of the partial pressures of the products divided by the sum of the partial pressures of the reactants, all taken to the power of their stoichiometric coefficient.

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26) What is an expression for K3? (A) This problem involves the manipulation of the reactions that may remind you of Hess’ law. In this case, you’ll be manipulating the first and second reactions in order to determine the third’s equilibrium constant in terms of the first and second’s.         2  2 2            2   2              We want NO on the left side of the equation so we “flip” it, but instead of making the equilibrium constant negative (not possible), we’ll use its reciprocal. 1 2      2 2            Cancelling out compounds that appear on both sides we get the third reaction. 2              2   Now the other difference is instead of adding the equilibrium constants, we would multiply them to get The mathematical proof of what went on: “Flipping the equation” 2

   

2 1

“Multiplying the constants”

1

27) What is the effect on reaction 3 if the partial pressure of NO is increased at equilibrium? A) an decrease in the partial pressure of O2 B) a shift toward reactants C) a shift toward products D) a change in the K1 E) none of these (A/C) This problem can be solved by observing the reaction quotient of the third reaction:

If the partial pressure of NO is increased from equilibrium, then the quotient would become smaller due to the larger figure on the denominator. To return to equilibrium, the partial pressure of NO2 must increase, which explains the reaction shift towards products. Also, another way to return to equilibrium is to decrease the partial pressure of O2 which would help counteract the ©2009 Stephen Kok UC Berkeley Undergraduate

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effects the increased partial pressure of NO. Both of these work simultaneously to return the reaction back to dynamic equilibrium. This problem may also be partially solved by le Chatelier’s principle, but you’ll only be able to see the shift towards the reactants and might miss the decrease in partial pressure that can also occur. By the way, how does a change in the third reaction affect the equilibrium constant in the first? It doesn’t. ^_^ 28) Given that the reaction is endothermic, what is the effect of increasing the temperature of reaction 1 at equilibrium? Mark all that apply. A) An decrease in the partial pressure of O2. B) A shift toward reactants. C) A shift toward products. D) A change in K1. E) None of these. (A/C/D) In this problem, we’re going to look at the chemical equation for the reaction.     2   Since the reaction is endothermic, it means that it requires energy to occur, thus heat is a reactant in this process. Through le Chatelier’s principle, if we increase the temperature of the reaction, thus increasing the heat, the reaction would move to the product side. A easier way to think of le Chatelier’s principle is a balanced see-saw where the reactants are on one side while the products are on the other side at equilibrium. If we add heat, which “increases the weight” on the reactant side, then to rebalance the see-saw, we would need more products, which means the reaction would shift towards the products. ^^ The equilibrium constant K, is dependent on the temperature at which the reaction takes place. If the temperature increases for an endothermic reaction, then the equilibrium constant is also increased to the temperature. Thus there is a change in K1. Now to determine whether there are changes in partial pressure, we’ll look at the reaction quotient

Since K1 increases, then there is an increase in the partial pressure in NO, which was stated before by le Chatelier’s principle, but there can also be a drop in partial pressure of the reactants N2 and O2 in order to compensate for the increase of K1. 29) What is the reaction quotient after the volume is halved for reaction 1 at equilibrium? A) Q = K1 B) Q = 2K1 C) Q = ½ K1 D) Q = ¼ K1 E) Q = 4K1 (A) This is a particular interesting problem that shows that changing the conditions of a reaction might not affect its dynamic equilibrium. The reaction quotient for the first reaction:

If the volume is halved, the pressure of the overall system would double in response. (By ) Thus all the partial pressures would also double. 2 2 ©2009 Stephen Kok UC Berkeley Undergraduate

4 2

4

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The multiples cancelled each other out which means that the equilibrium is not disturbed when the volume is reduced by a half. Thus 30) Which of the following is the correct plot of ΔG° vs. T for the evaporation of water? (C) This problem can be looked at in several ways. You could use the simpler definition of Gibb’s free energy which is that if Δ ° 0 then the reaction is spontaneous and if Δ ° 0 then it’s not spontaneous. We know that water evaporates as temperature increases, which means that it is “spontaneous” for high temperatures. If it is spontaneous, then Δ ° 0 at those temperature, which is the case for graph C, where Gibb’s free energy becomes negative after a particular temperature (boiling temperature). The first graph is incorrect because if the free energy is always negative, then water evaporates spontaneously at all temperatures; the second is incorrect because it is the exact reverse; and the last graph is incorrect because it implies that water never evaporates since the free energy is never negative. 31) The equilibrium position is quantified by the equilibrium constant, K. Which of the following generalized statements regarding K is false? Mark all that apply. A) When K is larger than 1, the equilibrium position favors the formation of products. B) When K is larger than 1, the value of ΔGo for the reaction is always negative. C) When K is equal to 1, the number of moles of reactants is always equal to the number of moles of product. D) When K is larger than 1, the forward reaction is always occurring faster than the reverse reaction is. E) When K is smaller than 1, the value of ΔH for the reaction is always negative. (C/D/E) This problem involves exploring the relationships between the equilibrium constant K, Gibb’s free energy, enthalpy, and entropy. In the statement A, it is referring the refined definition of Gibb’s free energy where a positive would imply a reactant-favored reaction while a negative value would imply a product-favored reaction. The relationship between K and Gibb’s free energy is given by the formula: Δ ° If K is bigger than 1, then ln(K)>0, which would make the value of free energy negative due to the negative sign in front. If it’s negative, then the equilibrium position does favor the products, which also answer B. Statement C only applies to some reactions and generally doesn’t work for most reaction as like the following                       For instance, if K=1 in this reaction, then [C]=[B]=[A]=1. So there are 2 moles of reactants, but only 1 mole of product. This is a very common reaction structure and would contradict the statement.

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32) In order to determine the identity of a compound with a molecular formula of CxHy you carry out an experiment with a Dumas bulb. In a bath of boiling water, you determine that a 2.0 L Dumas bulb will hold exactly 1.829 g of gaseous unknown. If the atmospheric pressure is 1.0 atm, what is the identity of your unknown? A) CH4 B) C2H2 C) C2H4 D) C2H6 E) C3H6 (C) Simple enough, using the ideal gas formula, we solve for the number of moles that are held by the Dumas bulb. Note that temperature is 100°C or 373.15K since it is in boiling water. 8.206 10 1.0 2.0 100 273 0.06534 

1.829 0.06534 The only compound with a molar mass of 28g/mol is C2H4.  

~

28

33) Recall the experiment that you did in lab to determine ΔH for the reaction between magnesium oxide and hydrochloric acid. Which of the following would result in ΔH value that is more exothermic than the actual value? A) The top of the calorimeter is not on tightly. B) The student accidentally put some ice into the calorimeter when determining the calorimeter constant, Kcal. C) The heat capacity of the solution is actually smaller than the heat capacity of pure water. D) None of the above will result in a low enthalpy of reaction. (C) We can pretty look just look at the situations and determine if they would result in a more exothermic value. The first statement would imply that heat leaked out of the system during the reaction, this would probably decrease the calculated value since there was heat that was not accounted for, unless heat leaked “into” the calorimeter which would only be possible if the room was at a higher temperature than the reaction. The second state in regarding the Kcal would probably result in a lower calculated value as well. If you remember, the equation used to solve for the reaction was: Δ Δ Δ If ice was accidently put into the calorimeter when determining the calorimeter constant, then the temperature drop would have been more than expected. Calculation of the constant: Δ

1

Δ

Δ

Δ Δ Δ Looking at the final calculation, if the change in temperature was larger, than the Kcal would be smaller and in the original enthalpy calculation, a smaller Kcal would give a smaller enthalpy, hence less exothermic. Statement C deals with the assumption regarding hydrochloric acid in which we used the specific heat of water as an estimation. Looking at the formula again: Δ Δ Δ

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If is smaller than we had assumed, then the enthalpy we calculated would have been more negative than it was supposed to be, hence more exothermic than the actual value. 34) Potassium thiocynate (KSCN) is a very soluble salt. Iron thiocyanate is a red complex ion that forms from the combination of iron (III) which is yellow and thiocynate ions which are transparent: Fe3+ + SCN- ↔ FeSCN2+ The equilibrium constant for the formation of iron thiocynate is moderate. What will happen to the color of the solution as potassium thiocynate is introduced into a test tube containing equal amounts of iron and thiocyanate? A) The color will not change. B) The solution will become redder. C) The solution will become more transparent. D) The solution will become more yellow. E) Cannot determine the color change based on the information provided (B) A very simple problem as long as you can apply le Chatelier’s principle. The potassium thiocynate (KSCN) would disassociate in solution creating K+ and SCN- ions. Due to the addition of SCN- ions, the concentration of the reactants is now higher than the concentration of the products. Assuming that it was in dynamic equilibrium before KSCN was added, the reaction would be tipped in the direction the product side to counterbalance the increase of reactants. If more FeSCN2+ is created, then the solution would become more red.

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Expanded Solutions  Midterm III 2008 1) Which of the following gases would you expect to behave least like an ideal gas? A) He B) CO2 C) H2O D) Xe E) Br2 (C) Simple enough, all the molecules are in gaseous form at standard temperature and pressure except water. Water also has extensive hydrogen bonding that also affects gaseous behavior. 2) Using the diagram below, which of these is a plot of the vapor pressure of water vs. temperature? (B) Vapor pressure is present only on the phase boundary between liquid and gas, and since its directly proportional as well, it is the same shape as the boundary line. 3) A system absorbs 716 kJ of heat and does 598 kJ of work on the surroundings. What is the change in internal energy of the system? A) 1314 kJ B) 118 kJ C) -118 kJ D) -1314 kJ E) 598 kJ (B) Make sure your signs are correct! q  Δ

716 598 716

598

118

4) When breathing, the average human lung expands by 0.50 L. If this expansion occurs against an external pressure of 1.0 atm, how much work is done by the system? (A) The formula that relates expansion with work is pressure. So for this problem: 1.0 0.5 However, you have to make sure and convert to Joules 101.3 0.5  

Δ where P is the external 0.5 51

5) For which of the following is ΔS° expected to be positive? A) H2O(g) → H2O(l) B) C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) C) 2CH3OH (g) + 3O2(g) → 2CO2(g) + 4H2O(l) D) 2Mg(s) + O2(g) → 2MgO(s) E) 2H2(g) + O2(g) → 2H2O(l) (B) Δ ° is a measure of entropy, or simply chaos. A positive value indicates that the reaction is heading towards disorder while a negative value indicates that the reaction is becoming more ordered. In terms of phase change, the most ordered phase is the solid state, and the most ©2009 Stephen Kok UC Berkeley Undergraduate

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disordered phase is the gaseous phase. So any reaction that goes from a lower state of matter to a higher state of matter will have a positive entropy. All the reactions are becoming more ordered except the second reaction where a solid and gaseous mixture combusts into a gaseous and liquid compound. Overall, you’ll be observing whether the ending phase is more or less disordered from the original phase. 6) Which phases are present at point ‘a’? A) Supercritical Fluid B) Solid and Liquid C) Liquid and Gas D) Solid and Gas E) Solid, Liquid and Gas (B) Since the point is on the boundary between the solid and liquid phases, then the phases present at that point is as stated. The boundary indicates where both phases coexist. Just as a note, a supercritical fluid is simply a compound that is beyond the critical point, which means that it is a “gas” that is so dense that it literally acts as a fluid. 7) At sea level, what is the approximate temperature at which CO2 sublimes (solid to gas phase change)? A) 0°C B) 30°C C) -78°C D) -55°C E) -115°C (C) At sea level, the atmosphere is approximately 1 atmosphere. Sublimation is the term used to describe the phase change from the solid state directly into the gaseous phase, which is indicated on the diagram by the boundary between the solid and gas phase. At intersection of the 1 atmosphere level with that specific boundary line occurs at (1,-78°C) 8) What does point ‘b’ represent in the phase diagram? A) Triple Point B) Condensation C) Vaporization D) Critical Point E) Sublimation (D) The only “points” that would exist on a phase diagram is the triple point and the critical point. In this case, point B is on the critical point in which the characteristics of the substance can no longer be defined solely as gaseous or fluid, it takes on characteristic of both phases becoming a supercritical fluid (technically a gas). The other point, the triple point is when all the phases coexist, and this is defined as the intersection of the phase boundary lines.

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9) The decomposition of H2O2 is exothermic: H2O2(g) → H2O + ½ O2 Average Bond Enthalpies (kJ mol-1) O-H 463 O-O 146 O=O 497 Using the bond energy data above, what is the molecular structure of H2O2? (C) To start out with, break all the bonds on the product side. You have 2 (O-H) bonds from water and ½ (O=O) bond from oxygen gas. This gives you a total of 2 463  ½ 497 1174.5   Since the formation of bonds is exothermic and the breaking of bonds is endothermic, to have an overall exothermic reaction would require that the energy put into breaking the bonds is less than the energy that’s released in forming the bonds. Δ q   2 497 2 463 1423   497 2 463 1423     146 2 463 1072 Only the third structure requires less energy than is formed and thus the correct answer. The enthalpy of reaction in this case would be Δ 1072 1174.5 102.5 (Note that (B) doesn’t even exist in terms of Lewis structure ^_^) 10) Which of the following cannot occur at the same time? A) ΔSsys > 0 and ΔSsurr < 0 B) ΔSsys < 0 and ΔSsurr < 0 C) ΔSsys > 0 and ΔSsurr > 0 D) ΔSsys < 0 and ΔSsurr > 0 E) All of these can occur at the same time. (B) This basically tests whether you understand what entropy is and what rules apply to it. The universe is constantly heading towards chaos (sound familiar?), that means the universe’s entropy is always either zero or positive. The “universe” that we have to take into account is sum Δ 0. of the change of entropy of the surroundings and the system. That means, Δ So if the system’s entropy is negative (becoming more ordered), the surrounding’s entropy must make up for the decrease in entropy by becoming more chaotic (positive entropy) to balance out the universe; an example is ice forming over a lake if the temperature is low enough, the phase change from a liquid to a solid signifies a negative entropy since the water is becoming more ordered, the surroundings respond by absorbing the heat that’s released from the water due to the phase change and has a positive entropy since there is more thermal disorder. This also works the other way around where the system becomes more disordered, while the surroundings become more ordered. However, one situation that can never occur is the system and surroundings

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having negative entropy, which means they are both becoming more ordered, thus having a negative sum and that’s answer B. 11) A reaction will occur spontaneously at standard conditions if: A) ΔH > 0 B) ΔS > 0 C) ΔG > 0 D) K > 1 E) ΔH < 0 (D) Let’s go through all the answers. For (A), a positive enthalpy means that the reaction is absorbing heat, which in some cases can be spontaneous, though with enthalpy alone, you cannot predict reactivity. For (B), you have a positive entropy for your system, that means it’s becoming more disordered like the universe wants, however, it’s not enough to information. For those of you who actually read the book, you’ll might remember at some point there was a section that said a positive entropy indicates that a reaction is spontaneous, however this positive entropy , which is the sum of the entropy of the system and the entropy of refers to the total entropy Δ the surroundings. Δ alone generally means the entropy of the system, and since you don’t know how the entropy of the surroundings will respond, you don’t know if the reaction is spontaneous. Answer (C) deals with Gibb’s free energy, which if you remember, a negative value would indicate that the reaction is spontaneous while a positive value would indicate that it’s not. So since the Gibb’s free energy is positive, the reaction is not spontaneous. Jumping to answer (E), a negative enthalpy means that the reaction is exothermic, but that isn’t enough to prove that a reaction is spontaneous; you can base such assumptions only on Gibb’s free energy values or the equilibrium constant K. Now for answer (D) which has K>1. If you can’t remember that a positive K indicates a spontaneous reaction, then you can look at the relationship between Gibb’s free energy Δ and the equilibrium constant Δ in the equation: Δ ° If K=1, ln(1)=0 and hence Δ 0, which indicates equilibrium. If K>1, then lnK>1 and Δ 1, which indicates a spontaneous reaction. And the last condition if K<1, then lnK<1 and Δ 0, which indicates a non-spontaneous reaction. Keep in mind that both values are directly linked and should give the same information regarding the direction of a reaction. 12) Consider the following exothermic reaction at equilibrium in a sealed container: N2(g) + 3H2(g) → 2NH3(g) Which of the following conditions will result in the reaction favoring products? A) High temperature B) Low Pressure C) Thermal Insulation D) High Pressure E) None of these (D) First off, this is an exothermic reaction, which means that heat is released from this reaction, hence heat is a product. By le Chatelier’s principle, if we increase the temperature of an exothermic reaction, the reaction would proceed in the reactant’s direction. An easy way to remember this is the see-saw example again, where if we increase the temperature and thus heat, we would increase the “load” on the product side, and thus to balance out the see-saw, we’ll have to have more reactants, thus reactant favored. Thermal insulation will pretty much produce the same result since this is an exothermic reaction, the heat released will stay isolated within the system, and since there’s an increase in heat, the reaction will be pushed to the reactant side ©2009 Stephen Kok UC Berkeley Undergraduate

College of Engineering Bioengineering – Premedical

General Chemistry 1A Spring 2009 Professor Nitsche

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again. Now let’s explore how pressure affects this reaction. Generally reactions will involve only gaseous compounds, but if there are solids and liquids around, you can generally ignore them since pressure has little to no effect on their volume. In this reaction, we have four moles worth of gas on the reactant side and two moles worth of gas on the product side. Now if we decide to decrease the pressure, the reaction would like to proceed in the direction where the gas can have the greater expansion since they’re under less pressure. Thus a lower pressure would push the reaction towards the reactant side which has four moles of gas versus two moles of gas on the product side. Now if we decide to increase the pressure, the reaction would proceed in the direction that would reduce the stress of the extra pressure, which in other words, would proceed in the direction that would decrease the number of particles floating around in order to decrease pressure. This would force the reaction to the product side in this case since the products only have 2 moles worth of gas which would take reduce the number of particles in the system and thus reduce the overall pressure. 13) Which of the following correctly represents the Ksp expression for this reaction: K2CO3(s) 2K+(aq) + CO32-(aq) A) [K+] [CO32-]/[K2CO3] B) [K2CO3]/ [K+] [CO32-] C) [K+]2 [CO32-]/[K2CO3] D) [K2CO3]/ [K+]2 [CO32-] E) [K+]2 [CO32-] (E) When writing the formula for the equilibrium constant K, or more commonly the reaction quotient Q, you have to make sure that you’re not including the concentrations or partial pressures of pure solids or liquids. The only ones that appear in the quotient are those in the gaseous form or the aqueous form. Also, a common mistake carried over from finding a reaction’s enthalpy, entropy, or free energy is that the coefficient isn’t multiplied, but rather, the concentration or partial pressure is taken the power of that coefficient. For example a particular compound in a reaction is 3X, the corresponding value in the quotient would be not 3 . The similar factor that carries over though is that the products “go first” which in this case, they appear in the numerator, and the reactants are in the denominator. So for this reaction: Note that

was left out of the quotient because it was a pure solid (s).

14) A 30.0 g bar of iron (Fe) is heated to 106.0°C and then placed in a calorimeter with 100.0 g of 20.8°C water. The final temperature of the water is 23.4°C. What is the heat capacity of the Fe in J g-1 K-1? A) 439 B) -439 C) -0.439 D) 0.439 E) -39.9 (D) A simple calculation, but make sure you don’t end up with a negative heat capacity, because that’s impossible ^_^. The heat that’s released from the iron is absorbed by the water, so following the rules regarding enthalpies, q is negative for iron and positive for water. Δ

Δ

Solving for

©2009 Stephen Kok UC Berkeley Undergraduate

College of Engineering Bioengineering – Premedical

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Δ Δ 4.184 100 23.4° 20.8° ° 30 23.4°C 106.0°C 0.439 ° 15) In the presence an acid solution, chromate and dichromate exists in equilibrium with a moderate equilibrium constant. 2  2 The chromate ion ( ) is yellow in color, while the dichromate ( ) anion is orange. If we add potassium chromate (K2CrO4), a very soluble salt, what would happen to the color of the solution as it regains equilibrium? A) The color will not change. B) The solution will become more orange. C) The solution will become more yellow. D) The solution will become transparent. E) The color change cannot be determined from the data given. (B) This problem involves a reaction in dynamic equilibrium. This means that in response to a change that affects its equilibrium, the reaction would proceed in the direction that opposes the changes until it reaches equilibrium again. In this case, we have the addition of potassium chromate, which is stated to be very soluble. This means that in the presence of a solution, the compound would disassociate into its ions, K+ and CrO42-. The potassium ion does not play a role in this reaction so we can ignore that one, but the chromate ion is a reactant in the chromatedichromate reaction complex. With the addition of more CrO42-, the concentration of the reactant side becomes higher, and would thus push the reaction towards the product side as expected with le Chatelier’s principle (the see-saw principle). If the reaction becomes product-favored, then the quantity of dichromate ions and water will increase while the quantity of hydrogen ions and chromate ions decreases since they’re used up in the reaction. This would decrease the yellow color produced by the chromate ions and increase the orange color produced by the dichromate ions. Thus, the reaction will become more orange. Be careful though, not all conditions will affect an equilibrium, sometimes there is no change.

©2009 Stephen Kok UC Berkeley Undergraduate

College of Engineering Bioengineering – Premedical

General Chemistry 1A Spring 2009 Professor Nitsche

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