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Chem 222 Study Guide for First Exam

by: Leslie Pike

Chem 222 Study Guide for First Exam Chem 222

Marketplace > Western Kentucky University > Chemistry > Chem 222 > Chem 222 Study Guide for First Exam
Leslie Pike
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These notes cover everything lectured on in Chapters 12 & 13. Solutions, colligative properties, van 't Hoff, equation kinematics, Arrhenius equation AND the derived Arrhenius equation
College Chemistry 2
Darwin Dahl
Study Guide
Chemistry, Chemistry 222, chem 222
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This 6 page Study Guide was uploaded by Leslie Pike on Wednesday February 17, 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 66 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.


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Date Created: 02/17/16
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 , ha2 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 x y Rate=k[A] [B] 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: 1/2[A] 02k. 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: t1/2n(2)/k. This half life does not change with respect to time. Second-order Reactions: For a second order equa2ion 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: ln k2 = E a( 1 − 1 ) (k1 R T 1 T 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 2s 2etected 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 +2O  22O 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: Rate for first step = k[NO] 2 Rate for second step = k[N O ]2O 2 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.


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