Chem 222 Final Study Guide
Chem 222 Final Study Guide Chem 222
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This 20 page Study Guide was uploaded by Leslie Pike on Thursday May 5, 2016. The Study Guide belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 57 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.
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Study Guide for Chem 222 Final General Tips: From what I’ve seen in the ACS study guide, the problems on the final will be much more straightforward than the ones on the midterms. The problems will include all of the information you need to work the problem (molar masses, equations, etc.) and will not include useless information designed to trick you. The problems will also be less complex. This should ideally make the test a little easier. Things to watch out for: Units. Energy must be in joules, pressure in atmospheres, temperature in Kelvins. Since each multiple choice question must have three plausible wrong answers, and since you’ve been given all the information (minimizing the amount of ways you can get a wrong answer), you can be certain that there will be a wrong answer where the procedure was done correctly but the units were not converted. Balancing coefficients. In expressions for Q and K, raise to the balancing coefficient. When calculating ΔG rxnH ,rxndΔS , mrxniply by the balancing coefficient. When calculating EcellIGNORE the balancing coefficient. It’s an easy mistake to make, and you can bet that the ACS exam will exploit that fact to get their three wrong answers. Time. The exam material itself will be easier, but you will have less time per problem. Don’t spend too much time on any one problem. All that being said, here is all of the stuff that was covered in Chem 222. Chem 222 things to know Chapter 12: Solutions There are seven fundamental SI units: • Mole is the unit of quantity. Mole has the value 6.02*1023 • Meter is the unit of length. • Kilogram is the unit of mass, NOT WEIGHT. Weight is a force, not a fundamental property, and it is measured in newtons, not in kilograms. • Second is the unit of time. • Candela is the unit of light intensity. • Kelvin is the unit of temperature. Celsius is NOT the unit of temperature. • Ampere is the unit of electric current. Examples of what are NOT fundamental SI units: • Liters (or any other volume measurement). Volume is measured in length * width * height; therefore, the unit for volume is the cubic meter, which is derived from the meter. • Energy. The SI unit for energy, the joule, is defined to be kilograms * meters2 / seconds2. Obviously, it is a derived unit. • Weight. Weight is measured in newtons. One newton is equal to kilogram * meter / seconds2. Weight IS NOT MASS. Mass is a scalar, weight is a vector quantity having both direction and magnitude. • Everything else (power, speed, pressure, etc.). Just because something has a nice, metric-sounding name does NOT mean that it is a fundamental SI unit. Liters, joules, pascals, newtons, watts, and the like are DERIVED units, not FUNDAMENTAL units. To sum up, there are seven fundamental SI units, and all other units are derived units. The mole is used by chemists because the mass of mole of atoms or molecules, in grams, is equal to the mass of one atom or one molecule in AMUs. The periodic table does double duty. For example, helium has a weight of 4 on the periodic table. This means four AMUs per one helium atom, and four grams per mole of helium. Moles are very important in stoichiometry because equations are written in mole-to-mole relationships. For solid substances, grams are converted to moles using the molar mass. For liquid solutions, multiply the molarity by the volume used (molarity is defined as moles per liter). For gases, use the ideal gas law, PV=nRT. (These things should be familiar to you from Chem 120. If not, I have a complete set of notes for Chem 120 available for you to review.) Percent yield is actual yield divided by predicted maximum yield. Reactions do not always go to completion; thus, you may have less product than stoichiometry predicts. A solution contains a solvent and a solute. Theoretically, the solute is dissolved and the solvent does the dissolving. Sometimes, such as in a mixture of two metals, the solvent-solute relationship is not exceptionally clear. In this case, the solvent is the substance present in the largest amount. Miscible substances can be mixed with one another and will not separate back out. The solubility of something is how much of that something will dissolve in a given amount of a given solvent at a given temperature. In a saturated solution, the dissolved solute and the precipitated solute are in equilibrium; the solvent is full. In a supersaturated solution, there is too much dissolved solute which has no way of escaping. Scratch the glass of the container, and the solute will precipitate out on the scratches. Entropy is the measure of the disorder of a system. Polar solvents (i.e. water) dissolve polar solutes because the polar water is attracted to the polar molecules (sugar, alcohol, some salts, etc.) and vice versa. Oil and water will not mix because oil is a nonpolar hydrocarbon and is less dense than water. Oil is more attracted to itself than to water and vice versa; therefore, the two liquids will not mix. Not all salts dissolve in water. If the lattice energy is higher than the ion- dipole attraction (the charged ions’ attraction to the polar water molecules), the crystal stays together. Molecular geometry determines whether or not a molecule is polar. Carbon tetrachloride has four polar bonds, but these four directions cancel each other out, resulting in a net polarity of zero. (Polarity was covered in Chem 120 and should be familiar to you; if not, I have materials that explain it.) Henry’s Law of Gas Solubility: InitialsolubilityFinal solubility Initial pressure Final pressure The solubility of gas increases with increasing pressure and decreases with increasing temperature. The solubility of liquids and solids is not affected by pressure. As a general rule, the solubility of solids and liquids increases when temperature increases. Exceptions to this rule are some substances which release heat when they dissolve. According to Le Chatelier’s principle, adding more heat to the system would discourage the solute from dissolving. Le Chatelier’s Principle: When a system in equilibrium is shifted from its equilibrium, the system will try to offset the shift. Ex. The breakdown of dinitrogen tetraoxide to form two NO2 molecules. If the pressure is increased in the container, NO2 will combine to form dinitrogen tetraoxide, reducing the number of molecules in the container and offsetting the pressure increase. If the pressure is decreased, dinitrogen tetraoxide will break apart to form NO2, increasing the number of molecules in the container and offsetting the pressure increase. Mass percentage: Mass of solute divided by mass of solution. If you have a saline solution that is 20% by mass, in 100 grams of the solution, 20 grams will be salt and 80 grams will be water. Molality: Moles of solute divided by kilograms of solvent. The cool thing about molality is that it is not affected by temperature. Molarity (moles solute per liter solution) is affected by temperature. (This is why containers in the lab are more likely to be labeled in molals than in molars.) Mole fraction: Numbers of moles of given substance divided by total number of moles of everything. Van’t Hoff factor: Number of ions per formula unit. Ex. Calcium chloride, CaCl , h2s a van’t Hoff factor of 3, NaCl has a van’t Hoff factor of 2. Represented by a lowercase “i”. Colligative Properties: Property that depends on the concentration of the solute. The following are colligative properties: Vapor-pressure lowering: Δ P=i P A B Freezing point depression: Δ T=iK mf Boiling point elevation: Δ T=iK mf Osmotic pressure: π=iMRT Chapter 13: Kinematics Kinetics deal with how fast a reaction takes place. Thermodynamics deal with whether or not a reaction will occur. Kinetics and thermodynamics are not related to each other. The rate of reaction is the change in molarity with respect to time, dM/dt. For a reaction A 2B, the rate expression for the rate of reaction would be: Rate= -Δ[A]/Δt Rate= Δ[B]/2Δt Reactants have a negative sign in front of them, and the expression is divided by any coefficient that may be in front of a reactant or product. Four things affect rate: 1. Reactant concentration 2. Presence of a catalyst 3. Temperature 4. Surface area A rate law describes the rate of reaction in terms of the concentration of the reactants. RATE LAW IS DETERMINED EXPERIMENTALLY. A sample rate law for an equation: 2A + 3B 5C Rate=k[A] [B] y Notice that C, a product, is not a part of this rate law. Notice that the balancing coefficients in front of A and B do not necessarily correspond to X and Y. X and Y are usually small integer values, such as 0, 1, or 2. The order of a reaction is the sum of the exponents. Say X is 2 and Y is 1, the reaction above would be a third-order reaction. Note: the equations below WILL be given to you on the exam Zero-order Reactions: For a zero order equation A B + C, the rate expression is: Rate= -Δ[A]/Δt. The rate law is: rate=k. A is raised to the zero power because the sum of all exponents must equal 0 since this a zero-order reaction. Anything raised to the 0 power is 1; therefore, it disappears out of the equation. The equation for determining the concentration after a set amount of time for a zero-order reaction is: [A]t=[A] 0kt The expression for the half-life is: t 1/2] /2k0 This half-life decreases with respect to time. First-order Reactions: For a first order equation A B + C, the rate expression is: Rate= -Δ[A]/Δt. The rate law is: rate=k[A]. A is raised to the first power because the sum of all exponents must equal 1 since this a first-order reaction, and there is only one exponent so it must be 1. The equation for determining the concentration after a set amount of time for a first-order reaction is: Ln[A] tLn[A] -0t The expression for the half-life is: t 1/2(2)/k. This half life does not change with respect to time. Second-order Reactions: For a second order equation A B + C, the rate expression is: Rate= -Δ[A]/Δt. The rate law is: rate=k[A] . A is raised to the second power because the sum of all exponents must equal 2 since this a second-order reaction, and there is only one exponent so it must be 2. The equation for determining the concentration after a set amount of time for a second-order reaction is: 1/[A]t=1/[A] 0kt The expression for the half-life is: t1/2/k[A] .0This half-life increases with respect to time. Calculating the Rate Constant, k: K=Zfp, where Z=collision frequency F=fraction of collisions with minimum energy P= orientation factor Arrhenius equation: Ln(k)=-E /Ra + Ln(A), where A is the frequency factor and R=8.314J/(mol*K). IMPORTANT: R DOES NOT EQUAL 0.0821 LITER*ATMOSPHERE/ (KELVIN*MOLE)!!!! Use 8.314J/(mol*K) Also important: Temperature does NOT equal Celsius!!! Temperature is in Kelvins!!! Modified Arrhenius equation WHICH WILL NOT BE GIVEN TO YOU, SO MEMORIZE IT OR LEARN HOW TO DERIVE IT: k E ln 2 = a( 1 − 1 ) (k) R T T 1 1 2 Reaction Mechanics: Elementary steps are the steps taking place during the reaction. For example, in the reaction 2NO + O 2NO 2 The presence of N O 2s2detected during this reaction. Dinitrogen dioxide is produced during an elementary step and consumed during another elementary step; that’s why it doesn’t appear in the equation above. There are two elementary steps to the above reaction: Step 1: NO + NO N O 2 2 Step 2: N 2 2 O 2NO 2 Notice that dinitrogen dioxide is a product, then it is a reactant. It does not appear in the net reaction. Substances which are formed in an early elementary step and are consumed in a later elementary step are known as intermediates. Catalysts are not generally considered intermediates because they are not produced or consumed by the reaction, generally. Rate laws can be written for intermediates. These do not have to be determined experimentally; instead, the concentration is raised to the power of the balancing coefficient. Using the reaction above gives the results: 2 Rate for first step = k[NO] Rate for second step = k[N O ]2O2] 2 The rate of the slowest elementary reaction should agree with the experimentally determined rate for the overall reaction. A catalyst does not change whether or not a reaction will occur, but it speeds up the rate of the reaction by lowering the activation energy. Given the Arrhenius equation -(Ea)/(RT) k=Ae where k is the rate of the reaction and Ea is the activation energy, a decrease in Ea will result in a corresponding increase in k. This should be obvious from looking at the equation, because e is raised to a negative power. Solubility and Dissociation A system is at equilibrium when there is no observable change with time. The reaction going from the reactants to the products is equal to the reaction going from the products to the reactants. For some reactions, there is no equilibrium point, the reaction goes until all reactants are used up. Ex. 2H 2 O 2H O 2 For other reactions, they can go both ways: N 2 4 2NO 2 Equilibrium can be physical equilibrium, such as water or mercury in a closed container at equilibrium with its vapor. Equilibrium can be chemical, such as in the reaction shown above. For a reaction aA + bB cC + dD, the equilibrium constant is determined as: c d K= [C] [D] [ A] [B]b K is unitless because it is a ratio. K can be written as Kc, the concentration equilibrium constant. At a homogenous equilibrium, all reactants are in the same phase. For a heterogeneous equilibrium, the reactants are in different phases. For gases, there is a pressure equilibrium constant (the partial , Kp, in addition to the concentration equilibrium constant. One can convert between the two with the following equation: Kp=Kc(RT) Δ n Here, Δn is the change in the number of moles of gas, or the number of moles of gas on the right minus the number of moles of gas on the left. It can be a negative number. R is 0.0821 L*atm/(mol*K) Q can be found by taking the equation for K and plugging in the initial concentrations of the reactants and products. If Q is greacer than K, the reaction will go to the left because there are too many products. If Q is lesscthan K, the reaction will go to the right because there are too many reactants. If Q = K, c equilibrium is reached and nothing happens. To work an equilibrium problem, remember ICE: Initial, Change, and Equilibrium. Initial is your initial concentrations. Change is how they change (multiply by balancing coefficient, and use a positive or negative sign in front depending on whether the species is being produced or consumed). Equilibrium is the final concentrations (initial concentrations plus or minus however much they changed). Le Chatlier’s Principle: The system does the opposite of what you do to it. For example, take the following reversible reaction: N +3H 2NH , delta H > 0 2 2 3 All three of these are in the gas phase. Increasing the concentration of N will caus2 the system to want to decrease the concentration of N , so the 2eaction will go to the right. Taking N o2t of the system will cause the system to want to produce more N 2 so the reaction will go to the left. The reaction is exothermic, meaning that heat can be considered a product. Adding more heat to the system will cause the reaction to go to the left, absorbing heat. Removing heat will cause the system to go to the right, releasing heat. If the above system is in a closed container, decreasing the size of the container (or increasing the pressure) will mean that the molecules have less room. Therefore, to save space, N an2 H will2become NH . (Four m3les of gas on the left, two moles on the right.) Increasing the size of the container (or decreasing the pressure) will give the molecules more room, and NH will bre3k apart into N and H . 2 2 EASY MISTAKES TO MAKE: The stressor on the system MUST be something that will actually affect K. Adding more liquid water to a system, for instance, won’t change anything, because water is a solvent instead of a reactant. Increasing or decreasing the pressure or volume of a system which contains no reacting gases won’t change the equilibrium. The presence of a catalyst doesn’t affect K. Temperature affects K differently, depending on whether the reaction is exothermic or endothermic. If endothermic, treat heat as a reactant and solve the problem accordingly. If exothermic, treat heat as a product and solve the problem accordingly. Sample problem: The equilibrium constant for silver carbonate, Ag CO , is K =28 *3 sp -12 10 . Find the solubility of silver carbonate. First write the equation for the dissociation. Ag 2O 32Ag + CO+ 3 2- Then, write the expression for K : sp K spAg ] [CO ] 32- Notice that silver carbonate, being a solid, is NOT included in the expression! Notice that the products are raised to their balancing coefficients. Use the ICE principle: Initial: Ag = 0, CO = 3 Change: Ag = +2x, CO = +x 3 Equilibrium: Ag = 0+2x = 2x, CO = 0+x =3x Plug the values into the K expspssion: 8*10 -1= [2x] [x], x=1.26*10 M -4 -4 Therefore, the solubility for silver carbonate is x=1.26*10 M. Given sufficient silver carbonate in the water, it will dissolve until solubility concentration is reached. (Note: silver carbonate splits into three ions, so the concentration of silver ion will be twice the solubility number and the concentration of carbonate ion will be equal to the solubility number.) Determining whether or not a given concentration of ions will cause a precipitate: Calculate Q for the given concentrations of the two ions. (Remember, the expression for Q is the same as the expression for K, and you plug in your initial concentrations instead of your equilibrium concentrations.) If Q < K, you have a dilute solution and no precipitate will form. If Q = K, you have a saturated solution. If Q > K, you have more compound in the system than can be dissolved, and some will precipitate out. Common ion effect: the presence of an ion decreases the solubility of a compound containing that ion. Ex. An 0.1 M solution of silver nitrate will dissolve less silver chloride than a solution of pure water. When making the ICE table, be sure to write in the initial concentration of whatever ion is already present in the solution. Other than that, the problem is worked no differently than when you are dissolving a compound in pure water. Acids and Bases + An Arrhenius acid produces hydronium ion (H O ) in wat3r. An Arrhenius base produces hydroxide ion (OH ) in water. A Bronsted-Lowry acid is a proton donor. A Bronsted-Lowry base is a proton acceptor. - + An acid-base conjugate pair differs by one proton, i.e. HF/F , NH /NH . 4 3 NOT an acid/base conjugate pair: H SO /SO 2 Th4se t4o compounds do not differ by one proton. They differ by two protons. -14 The equilibrium constant for the auto-ionization of water is K =10w. The expression is Kw=[H ][OH ], where the concentrations of H and OH are equal. The pH is calculated as –log[H ] and is on a scale from 1 to 14, with 1 being the most acidic (highest concentration of hydrogen) and 14 being the most basic (highest concentration of hydroxide). Pure water has a pH of 7. The pOH is - calculated as –log[OH ]. The pH and pOH sum to 14, so it is easy to convert from one to the other. Strong acids dissociate completely in water and have K=infinity. The six strong acids are: HCl HBr HI HNO 3 HClO 4 H 2O 4 Strong bases dissociate completely in water and have K=infinity. Almost all bases are strong. Ammonia is a weak base. When given a problem, assume a strong base unless you are told otherwise. Sample problem: You mix 20 mL 0.1 M HCl with 15 mL 0.2 M HClO . What is4the pH of the resulting solution? Since pH only takes into account the hydrogen ion concentration, we will ignore the fact that two different acids are used. We don’t care about chloride and perchlorate ions because they don’t affect the pH. The first step is to calculate the number of moles of hydrogen ion we have. Each acid is a strong acid, so the number of moles of hydrogen will be equal to the number of moles of acid. 0.1 M HCl * 0.02 L = 0.002 mol HCl = 0.002 mol H + + 0.2 M HClO *40.015 L= 0.003 mol HClO = 0.004 mol H 0.002 + 0.003 = 0.005 mol H + The next step is to convert this to molarity. Our total volume of acid is .02 + .015 = . 035 L. .005mol/.035L = .357 M pH=-log[.357]=.845 Is our answer reasonable? We are mixing two strong acids, so our pH value should indicate an acidic solution. A pH value of .845 is very acidic. The pH of a weak acid is calculated using the same method used to determine the + concentration of sparingly soluble salts. The concentration of H is determined, and then the pH is determined by taking the negative log of this value. In using the above method, it is often useful to neglect x to save having to use the quadratic equation. X can be neglected when concentration/K > 100. The bigger Ka is, the stronger the acid is, and the less basic that its anion is. For all strong acids, the anion is neutral, but for weak acids, the anion is basic. Ex. Perchlorate and sulfate are neutral anions, but fluoride and cyanide are basic anions. If a salt contains a basic anion, the salt will form a basic solution when dissolved in water. Ex. Sodium acetate, potassium fluoride. Useful equations for pH problems: −¿ OH ¿ ¿ ¿ +¿¿ H ¿ ¿ pKa+pKb=14 Sample problem: You have a solution of 0.1 M HCl and 0.1 M HF. Calculate the pH. The HCl will dissociate completely, so that the concentration of the free hydrogen ions will be equal to the HCl concentration. The HF does not dissociate completely. It dissociates according to the following equation: HF H + F - According to Le Chatelier’s Principle, addition of an ion to one side of the equation + will push the reaction to the other side. The additional H caused by the dissociation of HCl will make HF even less likely to dissociate. Thus, the weak acid can be ignored in the calculation of the pH; only the concentration of the strong acid is needed. Take the negative logarithm of the concentration of the strong acid to find the pH. Acidic salts are the result of the neutralization of a strong acid and a weak base. Acidic salts can also result when the salt contains small, highly charged ions, like aluminum. The easier the free hydrogen atom is to remove, the stronger the acid. For a binary (two-atom) acid, strength increases down a column (HI is stronger than HBr which is stronger than HCl which is stronger than HF). The bigger atoms have more electron shells between themselves and the hydrogen proton; therefore, they do not attract it as strongly. For oxoacids (acids containing oxygen atoms), acid strength increases as the number of oxygen atoms involved increases. Acid strength also increases as the non-oxygen non-hydrogen atom increases in electronegativity. HClO is stronger than HBrO which is stronger than HIO. For polyprotic acids, acid strength decreases as the acid loses protons. How to choose a buffer: the pKa of the buffer should match the pH you desire to buffer at. Sample problem: You mix 5 ml of 0.1 M NaOH with 20 ml of 0.1 M of a weak acid, HA, with a Ka of 10 . What is the resulting pH of this solution? This is a limiting reagents problem. The reaction will go to completion because strong bases such as NaOH react until all is used up. Convert everything to mmol by multiplying molarity by mL. You will get 0.5 mmol NaOH and 2 mmol HA. Use an ICE table. Your limiting reagent is NaOH, as there are fewer moles of it present and species react in a 1:1 mole ratio. Your final concentration of HA is 2mmol-0.5mmol=1.5mmol. Your final concentration of NaA (which will dissolve in - solution and be present at A ) is 0.5 mmol. Now, ask yourself: Do I have a buffer in solution? In this case, the answer is yes, because both HA and A- are present in the solution. Since you have a buffer, use the Hendersen-Hasselbalch equation. −¿ A HA ¿ ¿ ¿ pH=pKa−log¿ Is your answer reasonable? You mixed a base and an acid together at the beginning, but you added more acid than base, so your solution will be acidic, so your pH value should be less than 7. Sample problem 2: You mix 20 ml of 0.1 M NaOH with 5 ml of a weak acid, HA, with the same Ka as before. What is the resulting pH of the solution? This is a limiting reagents problem. This reaction will go to completion because of your strong base. Convert everything to mmol by multiplying M by ml. You should get 2 mmol NaOH and 0.5 mmol HA. Use an ICE table. All of your HA will be used up in the reaction; you should end up with 1.5 mmol NaOH and 0.5 mmol A . - Now, ask yourself: Do I have a buffer in solution? In this case, the answer is no, because no HA is present, only A-. You CANNOT use Hendersen-Hasselbalch when you do not have a buffer in solution. What you DO have in solution are two anions, - - OH and A . One is the ion of a strong base and the other is the ion of the weak base. Ignore the weak base and calculate pH using the strong base. −¿ OH ¿ ¿ ¿ pOH=−log¿ pH=14−pOH=12.78 Now ask yourself: Is this answer reasonable? You mixed an acid and a base, and you had more base than acid, so your resulting solution will be basic, so your pH should be above 7. Titrations Titration can be used to determine the concentration of an acid or base. A titration curve is a plot of pH versus amount of titrant added. It is shaped like an “S” with the midpoint of the S being the point at which there is neither acid nor base present in solution, only salt and water. If both the acid and base are strong, your pH will be 7 at the equivalence point. If one is strong and the other is weak, your salt will be acidic or basic and will affect the pH. To tell if a salt is acidic or basic: Strong base strong acid = two neutral ions making a neutral salt. Strong base weak acid = a neutral cation and a basic anion (makes sense that an acid containing a basic anion won’t be as strong of an acid, right?) making a basic salt. Weak base strong acid = an acidic cation and a neutral anion (a base containing an acidic cation will be a weaker base—think aqueous ammonia, NH OH, whi4h + contains the acidic cation NH )4making an acidic salt. Weak base weak acid = if the acid is stronger, you get an acidic salt and vice versa. For example, strontium carbonate is a salt containing strontium and carbonate (well, duh). The strontium would have come from strontium hydroxide, a strong base. Carbonate would have come from carbonic acid, a weak acid (if they are not strong, acids are weak—again, duh). Neutral cation (strontium) and basic anion (carbonate) make a basic salt. To distinguish between base-acid titrations and acid-base titrations: If you have a base and you are titrating it with an acid, your pH will start out high and then decrease. If you have an acid and you are titrating it with a base, your pH will start out low and then increase. Chapter 18 Entropy A reaction is spontaneous if it happens of its own accord, without requiring any outside work. Iron rusting is a spontaneous reaction. Sugar dissolving in water is spontaneous. Spontaneous reactions can be exothermic or endothermic, but ALL spontaneous reactions result in a loss of energy. ΔG is negative. If ΔG is positive, that means that you have to add energy to the system to make the reaction proceed in that direction (i.e. the reaction is not spontaneous.) Spontaneous reactions usually have an increase in entropy, symbolized by ΔS. Entropy is the measure of the disorder of a system. When order increases, entropy decreases. When disorder increases, entropy increases. Do not confuse entropy with enthalpy. Enthalpy is the internal energy (or internal heat) of a system. Remember than enthalpy is heat, which you can measure with a thermometer. Enthalpy is symbolized ΔH. The entropy of a solid is less than the entropy of a liquid, which is much less than the entropy of a gas. A gas has very, very high entropy. If the change in entropy is positive, the system is becoming more disordered. If the change in entropy is negative, the system is becoming more ordered. A positive ΔS means that more disorder has been added to the system. A negative ΔS means that disorder has been lost (the system has become more ordered). Entropy can be calculated by: S=k∗ln ?(W) Here, k is a constant. W is the number of microstates of a system. As W increases, so does entropy. To illustrate: I have two pairs of shoes at school with me. I’m not the most organized person and my shoes tend to get strung out all over the room, but since I only have four shoes total, they don’t make much of a mess. I have four microstates (four shoes) and low entropy. My roommate has lots of shoes—let’s say 7 pairs. She is not the most organized person either, but her shoes make a considerable mess on account of there being so many of them. She has fourteen microstates (fourteen shoes) and high entropy. Basically, the more stuff you have, the bigger potential mess you can make, and the higher the entropy. Thus, bigger molecules will have higher entropies than smaller ones (bigger molecules have more protons, neutrons, and electrons, and thus more microstates). At absolute zero, the entropy is zero. This is because, at absolute zero, there is only one microstate (everything is completely still) and the natural log of one is zero. Adding more heat to a system will cause an increase in entropy (ΔS will be positive) because things move around more when they are heated, meaning you get more microstates. A phase change of a solid to a liquid or gas gives ΔS > 0 because liquids and gases are more disordered than solids, and vice versa. Any element in its elemental state has an enthalpy of formation (ΔH ) of 0. fhus, graphite, O 2 and other standard states have no ΔH and they can be ignored in calculations of ΔH . rxnse elements in their standard states DO have a ΔS, because ΔS and ΔH are not the same thing. ΔH is energy gained or lost (aka heat, thus the capital H), and ΔS is disorder gained or lost. You can remember that disorder is symbolized with a capital S because the word disorder contains the letter “s” and the word “heat” doesn’t. For a given reaction, increase in the number of moles of gas means a positive ΔS. Decrease in the number of moles of gas means a negative ΔS. When no gas is involved, ΔS will be small. (Remember this if we get a test question that asks which reaction has the least ΔS; the correct answer will be the reaction which involves no gas molecules.) ΔS 0 is the standard entropy change for a reaction that takes place at 1 atm and rxn 298 K. It can be calculated by the sum of the ΔS of the products, minus the sum of the ΔS of the reactants. Remember to multiply by the balancing coefficients. Gibbs Free Energy To determine whether or not a reaction is spontaneous at a given temperature, we calculate the Gibbs free energy for that reaction. Gibbs free energy is calculated with the following equation: ∆ G=∆H−T ∆S If the ΔG is negative, that means that energy is released by the reaction when it proceeds, and thus the reaction is spontaneous. Imagine a child on a sliding board. The child slides down spontaneously (without any outside work). The child loses energy in the act of sliding down (potential energy of the child at the top is transferred to the slide and the child’s shirt as heat). If ΔG is positive, that means that energy has to be added in order for the reaction to proceed. The child is not going to slide up the sliding board unless somebody is pushing him or her. If ΔG is zero, the reaction is at equilibrium and it does not proceed either way. A child sitting at the bottom of a sliding board won’t go up or down. 0 ΔG fis the standard free energy of formation of one mole of a material. It is calculated by summing the ΔG’s of one mole of all the elements that make up the material. Superscript zeros mean standard conditions, temperature 298 K and pressure 1 atm. Sample problem: Calculate ΔG for a given reaction. ΔS is 42 J/(K*mol) and ΔH is 0 35 kJ/mol. First step: figure out what you have, what you need, and what equation you can use to get from what you have to what you need. You have ΔS and ΔH, and you need ΔG. So, you can use the following equation: ∆G=∆H-T∆S This looks fine, except you weren’t given T. That’s okay, you don’t need to be given T. The superscript zeroes should clue you in to the fact that you are at standard conditions, which means that T is 298 K. So, you can just start plugging in numbers now, right? WRONG!!!! Before you plug anything in, make sure that it is in the proper units. ∆S was given to you with J, and ∆H with kJ. Convert kJ to J before plugging in your numbers. ∆G=∆H-T∆S=35000-298*42=22484 J/mol The equilibrium constant can be calculated from the Gibbs free energy with the following equation: 0 ∆ G =−RT∗ln(K) 0 Here, R=8.314 J/(mol*K). Common pitfall: be sure to convert C to K and kJ to J before you start plugging numbers in. If K is greater than 1, the natural log will be positive, ∆G will be negative, and the reaction proceeds spontaneously to the right. If K is less than 1, the natural log will be negative, ∆G will be positive, and the reaction will not proceed spontaneously to the right (in fact, it will proceed spontaneously to the left). Chapter 19: Electrochemistry Electrochemistry deals with redox reactions. Electrochemistry deals with one of two situations: a reaction making electricity, or electricity being used to force a reaction to go in a certain direction. Oxidation states Redox reactions involve a change in oxidation states. An oxidation state is the electrical charge on an atom or molecule. Free elements (Pd , O ) have an oxidation state of 0 because they have no (s) 2 (g) charge. Monatomic ions have an oxidation state equal to their superscript number 3+ (Fe has an oxidation state of +3) Alkali metals (in solution or in compounds) have an oxidation state of +1 Alkali earth metals (in solution or in compounds) have an oxidation state of +2 Fluorine (unless it is a gas) ALWAYS, ALWAYS, ALWAYS has an oxidation state of -1. All other halogens also have an oxidation state of -1, unless they are bonded to fluorine. Oxygen (unless it is a gas) has an oxidation state of -2. As peroxide, the oxygens have a charge of -1 apiece (for a total charge of -2). Hydrogen has an oxidation state of +1 when it is written at the beginning of the compound, and -1 when it is written at the end of the compound. All other elements have whatever charge it takes to balance out the charges of the above atoms. For instance, in K CrO , t2e p4tassium contributes a charge of +2 and the oxygen contributes a charge of -8. Chromium therefore must have a charge of +6 to make the molecule neutral. Oxidation states do not have to be whole numbers. Redox reaction balancing Balancing a redox reaction (acidic aqueous solution): Cr2O 72-+ Fe 2+ Cr 3++ Fe 3+ Split it into two half-reactions. 2- 3+ Cr2O 7 Cr Fe2+ Fe3+ You will have to balance both equations. Start with the easier reaction (in this case, the one with iron). First, mass balance. (Already done.) Then, charge balance. Charge balancing looks intimidating when you first try it, but actually it’s just simple arithmetic. The left side and the right side have to equal. You have +2 on the left and +3 on the right. To make the equation balanced, subtract 1 from 3. +2 = +3 -1 Ta-da, it’s that easy. You add one electron to the right side of the equation, and you’re done! Now, for the other half-reaction. Mass balance it first. (You are in aqueous solution so you can add all of the water molecules you need, since you have plenty of them floating around. You are in acidic solution so you can add all of the hydrogen ions you need, since you have plenty of them floating around.) 14H + Cr O 2 7- 2Cr 3++ 7H O2 Now, you need to charge balance it. This equation looks more intimidating than the previous one, but you can just break it down into smaller pieces. Hydrogen, oxygen, and water do not participate in the actual electron transfer—they can be ignored. This leaves us with chromium. The chromium in dichromate has a charge of +6, and the free chromium ion has a charge of +3. DO NOT NEGLECT THE FACT THAT THERE ARE TWO CHROMIUM ATOMS IN DICHROMATE, AND TWO FREE CHROMIUM IONS ON THE RIGHT SIDE OF THE EQUATION. Write the amount of charge on each side of the equation, and subtract where you need to subtract to make both sides equal (electrons) + 2(+6) = 2(+3) -6 +12 = +6 You have to add six electrons to the left. Check your work; one of the half-reactions should have electrons on the right, and the other should have them on the left. (This is why you do the easy one first, so you can check the harder one against the easy one.) The electrons will need to cancel. You have 6 electrons in the chromium equation, and 1 electron in the iron equation. Multiply the iron equation by 6 so that the electrons cancel, and add the two equations. + 2- 2+ 3+ 3+ 14H + Cr O 2 7 + 6 Fe 2Cr + 7H O2+ 6Fe This is how to balance a redox reaction in acidic aqueous solution. For a basic aqueous solution, do the same thing. Then, after you have the final equation, add hydroxide to both sides to cancel the H . For the above equation, this would be 14 hydroxide. You would have 14 water on the left, 7 water on the right, and 14 hydroxide on the right. Waters cancel each other, and you have 7 water on the left and 14 hydroxide on the right. Electrochemical cells There are two types of electrochemical cells: galvanic (voltaic) or electrolytic. In a galvanic cell, the reaction makes electricity. In an electrolytic cell, electricity makes the reaction. Galvanic cells have a positive EMF, and electrolytic cells have a negative EMF. Oxidation occurs at the anode, and reduction at the cathode. (The word “oxidation” and the word “anode” both start with vowels, and the word “reduction” and the word “cathode” both start with consonants.) Anodes shrink and cathodes grow. Oxidation is the loss of electrons (metal turns to aqueous ions), and reduction is the gain of electrons (aqueous ions turn to solid metal). When writing the shorthand notation for a battery, you start at the anode and end at the cathode. (A comes before C in the alphabet, so you do the anode first.) For the traditional zinc-copper battery: 2+ 2+ Zn (s) (aq)(x M)||Cu (aq)y M)|Cu (s) Here, x and y represent the concentrations of the zinc and copper ions, respectively. The standard EMF of the cell is calculated as: 0 0 0 E cell E cathode (-E anode The cathode reaction will be going in the forward direction as it is written in a table. The anode reaction will be going in the reverse direction as it is written in the table; hence, its EMF will have an opposite sign of whatever the table says (this is why it has a negative sign in front of it in the equation). Standard conditions are 298 K, 1 M concentration and 1 atm. To determine the EMF of a cell not at standard conditions: 0 E = E – (.0592/n)*log(Q) Calculate Q for your reaction, calculate E cellrom the E values of your cathode and anode, and determine the number of free electrons involved in the reaction. Then, plug everything in and solve. (I personally do not recommend solving for E cathodend E separately; this creates extra steps and extra potential for mistakes.) anode To convert from free energy to EMF: ΔG=-nFE F is the Faraday constant, 96,485 Coulombs/mole of electrons. Lowercase n is the number of free electrons involved in the reaction, E is cell EMF. Sample problem: Calculate ΔG for the following reaction at standard conditions: + 2+ 2Ag (aq)+ Cu (s)> 2Ag (s) Cu (aq) The standard reduction potential for silver is 0.799, and the standard reduction potential for copper is 0.337. First step: figure out what you have, what you need, and what relationship gets you from what you have to what you need. You have the E for your two reactions, and you need ΔG. ΔG is related to E by the following equation: ΔG=-nFE. E for the complete reaction is the sum of the E of the two half-reactions. One half reaction is going forward, and the other is going backward. In the equation at the top, silver is written as a reduction; thus, its reduction potential is the same as given in the table. DO NOT multiply by the balancing coefficient. Copper is being oxidized in the above reaction, not reduced, meaning that its E will have an opposite sign of the number given in the table. Add the two half-reactions to get the E for the full reaction: E = +0.799 + (-0.337) = 0.462V. cell Plug this value into the expression for ΔG (n is 2 because you have 2 electrons being transferred in the above reaction, and F is Faraday’s constant, 96485): ΔG=-nFE = -(2)(96485)(.462)= -89000 J In an electrolytic cell, an electric current drives a nonspontaneous reaction. Electrolytic cells have a negative E, this is how you distinguish an electrolytic cell from a galvanic cell. Positive E means spontaneous, negative E means nonspontaneous. (Notice that this is the opposite of what the sign means when you are talking about free energy.) In an electrolytic cell, electrons are added to the cathode and removed from the anode. Metal plates out on the cathode, because it is reduced from its ionic form to its solid form when it takes the extra electrodes from the cathode. Whatever anion the metal was bounded to donates its electrons to the anode. Sample problem: You are purifying copper by electrolysis. You need 2 kg of pure copper. You have a 10 amp current source you can plug your cathode and anode into, and an unlimited supply of impure copper. How long does it take for 2 kg of pure copper to plate out onto the cathode? This problem looks hairy, but trust me, it’s nothing you haven’t already done in Chem 120. This is a regular stoichiometry problem; you just need to remember that a copper ion is missing two electrons (two moles of electrons are needed per one mole of copper), and you need to learn two new relationships: 96,485 C =1 mole electrons An amp is a coulomb per second (C/s), making a 10 amp current 10 C/ 1 s. 2000 g copper * (mol copper/63.55 g copper) * (2 mol electrons/mol copper) * (96485 coulombs/mol electrons) * (second/10 coulombs) = 607301 seconds = 168 hours = 7 days. You will be waiting a whole week for that copper. Better find a stronger current source.
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