### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Physical Chemistry I CH 431

NCS

GPA 3.54

### View Full Document

## 34

## 0

## Popular in Course

## Popular in Chemistry

This 55 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 431 at North Carolina State University taught by Staff in Fall. Since its upload, it has received 34 views. For similar materials see /class/224004/ch-431-north-carolina-state-university in Chemistry at North Carolina State University.

## Reviews for Physical Chemistry I

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/15/15

Chemistry 431 Lecture 2 Properties of Gases NC State University 8262008 Putting the results together When we combine of microscopic view of pressure with the kinetic theory of gases result we find the ideal gas law PV nRT This approach applies to a monatomic gas like neon or argon What about internal motions of molecules RT is a natural energy scale We can rewrite the ideal gas law in terms ofthe molar volume l7 Vn The ideal gas law has the form 13 39 The molarvolume at standard T and P V RT s 31 JmoleK29s K 0 0244 m3 24 4 L 1013 x 105Nm2 This is the volume of one mole of gas We could also write the unit as Lmol Microscopic variables Monatomic gases translation Pressure and temperature can be described solely in terms of the ballistic motion of the gas Diatomic gases translation vibration rotation 643 HO Center ofmass Quantized energy levels The constant h known as Planck s constant gives the scale for quantized energy levels h 6626 x 1034 Js Energy hv Translation particle in a box Vibration harmonic oscillator Rotation rigid rotator The energy levels for each ofthese is obtained by solution ofthe Schrodinger equation The energy level spacing The constant h known as Planck s constant gives the scale for quantized energy levels h 6626 X1034 J ra nslation We will see how to obtain these in the second half of the course 8262008 Levels are thermally populated Vibration Rotation Translation Key points regarding the microscopic view Translational energy levels are so densely spaced that these can be treated using classical methods We can treat particles as ideal even though they have vibrations and rotations The dynamics of the gas are not affected We will see that the heat capacity of the gas is affected by the internal degrees of freedom Key points regarding the microscopic view The kinetic energy of a large number of individual particles is proportional to the temperature of the system As the system heats up we can picture the molecules moving more rapidly Pressure results from the net momentum transfer between the particles and wall of the container Pressure of a dense fluid For a dense fluid or a liquid such as water we can think of the pressure arising from the weight of the column of fluid above the point where the measurement is made The force is due to the mass of water m kg accelerated by gravity 9 98 ms2 szmzm zm P A A Ah v pgh where p is the density p mN The dependence of atomspheric pressure on altitude We can think ofthe atmosphere is a fluid but it is not dense Moreover unlike water the density of the atmosphere decreases with altitude Thus at high elevations both the pressure and the density are decreased To obtain the dependence of pressure on height h above the earth s surface we use the ideal gas law to define the density of an ideal gas The dependence of atomspheric pressure on altitude The density of an ideal gas is p mN nMN MPRT The de endence of ressure on elevation 39Sap pg dh dh We need to collect variables of integration on the same side of the equation drank er 8262008 The barometric pressure formula Then we integrate assuming PD1 at h0 p 1 RT n P h quot E quotW P Puexp orP exp atm lsotherms We can plot the pressure as a function of the volume as shown below Each ofthe curves on the plot has a constant temperature Przssum p Vuim v Partial pressure For a y gas in a mixture of gases the partial pressure is de ned as PJ xJP where xJ is the mole 39action of component and P is the total pressure The mole fraction is de ned as n X 1 Zn i Chemistry 431 Lecture 3 The Schrc39jdinger Equation The Particle in a Box part 1 Orthogonality Postulates of Quantum Mechanics NC State University Derivation of the Schrodinger Equation The Schrodinger equation is a wave equation Just as you might imagine the solution of such an equation in free space is a wave Mathematically we can express a wave as a sine or cosine function These functions are oscillating functions We will derive the wave equation in free space starting with one of its solutions sinx Before we begin it is important to realize that bound states may provide different solutions of the wave equation than those we find for free space Bound states include rotational and vibrational states as well as atomic wave functions These are important cases that will be treated once we have fundamental understanding of the origin of the wave equation or Schrodinger equation The derivative The derivative of a function is the instantaneous rate of change The derivative of a function is the slope We can demonstrate the derivative graphically We consider the function fx sinx shown below The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinO the slope is 1 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinn4 the slope is 12 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinnZ the slope is 0 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin37c4 the slope is 12 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin37c4 the slope is 12 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinn the slope is 1 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin5n4 the slope is 12 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin5n4 the slope is 12 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope We see from of the black squares slopes that the derivative of sinx is cosx The derivative of sinx 10 39 00 10 i sinx cosx dX The derivative of cosx cl d X cosx SlnX 10 I I 00 05 10 quotI I I The second derivative of sinx cl sinx sinx The second derivative of sinx cl2 d XZ SInX SInX 1 0 I I I I I 00 05 1 0 quotI I I Sinx is an eigenfunction 2 If we define 2 as an operator G then we have X cl2 SinX SinX dX2 which can be written as G sinX sinX This is a simple example of an operator equation that is closely related to the Schrodinger equation Sinkx is also an eigenfunction We can make the problem more general by including a constant k This constant is called a wavevector It determines the period of the sin function Now we must take the derivative of the sin function and also the function kx inside the parentheses chain rule d smkx dX k coskx cl2 Sinkx k25nkx dX2 Here we call the value k2 the eigenvalue Sinkx is an eigenfunction of the Schrodinger equation The example we are using here can easily be expressed as the Schrodinger equation for wave in space We only have to add a constant 2 2 2 2 1i 2m dX2 2m In this equation h is Planck s constant divided by 27 and m is the mass of the particle that is traveling through space The eigenfunction is still Sinkx but the eigenvalue in this equation is actually the energy The Schrodinger equation Based on these considerations we can write a compact form for the Schrodinger equation HLP ELF in2 CI2 H Energy operator Hamiltonian 2m dX2 hzk2 E Energy eigenvalue Energy 2m LP Sin Wavefunction The momentum The momentum is related to the kinetic energy Classically The kinetic energy is 1 E mv2 The momentum is p mv So the classical relationship is p2 E 2m If we compare this to the quantum mechanical energy hzk2 we see that p hk 2m The general solution to the Schrodinger equation in free space The preceding considerations are true in free space Since a cosine function has the same form as a sine function but is shifted in phase the general solution is a linear combination of cosine and sine functions LP AsinkX Bcoskx Wavefunction The coefficients A and B are arbitrary in free space However if the wave equation is solved in the presence of a potential then there will be boundary conditions The particle in a box problem Imagine that a particle is confined to a region of space The only motion possible is translation The particle has only kinetic energy While this problem seems artificial at first glance it works very well to describe translational motion in quantum mechanics 0000 0 Allowed Region 39 The solution to the Schrodinger equation with boundary conditions Suppose a particle is confined to a space of length L On either side there is a potential that is infinitely large The particle has zero probability of being found at the boundary or outside the boundary 0000 0 Allowed Region 39 The solution to the Schrodinger equation with boundary conditions The boundary condition is that the wave function will be zero atx Oand atx L L110 AsinkO Bcosk0 O From this condition we see that B must be zero This condition does not specify A or k The second condition is L11L AsinkL O or kL arcsinO From this condition we see that kL In The conditions so far do not say anything about A Thus the solution f h 39 ort e bound state IS LPnoo AsmmnXL Note that n is a quantum number The probability interpretation The wave function is related to the probability for finding a particle in a given region of space The relationship is given by Pfwwv If we integrate the square of the wave function over a given volume we find the probability that the particle is in that volume In order for this to be true the integral over all space must be one 1 wwv If this equation holds then we say that the wave function is normalized The normalized bound state wave function For the wave function we have been considering all space is from O to L So the normalization constant A can be determined from the integral L L L 2 1 PZdX A2 sin 7ZX2dx AZI sinn 7ZX dX The solution to the integral is available on the downloadable MAPLE worksheet The solution is just L2 Thus we have 24 2 2 1 A2A LA tL As you can see the socalled normalization constant has been determined Normalization What is the normalization constant for the wave function expax over the range from O to infinity POW W Normalization What is the normalization constant for the wave function expax over the range from O to infinity POW W 1 r LIJZU X 0 Azexp ax20 X 4sz exp 2ax0 X A2i A 2 a 2a The appearance of the wave functions Energy arb units The appearance of the wave functions Note that the wave functions have nodes ie the locations where they cross zero The number of nodes is n1 where n is the quantum number for the wave function The appearance 0 L of nodes is a general feature of X solutions of the wave equation in bound states By bound states we mean states that are in a potential such as the particle trapped in a box with infinite potential walls We will see nodes in the vibrational and rotational wave functions and in the solutions to the hydrogen atom and all atoms Note that the wave functions are orthogonal to one another This means that the integrated product of any two of these functions is zero Energy arb units The energy levels The energy levels are n2h2 8mL2 The probability of finding the particle in a given region of space Using the normalized wave function LP sinn 7ZX one can calculate the probability of finding the particle in any region of space Since the wave function is normalized the probability P is a number between 0 and 1 For example What is the probability that the particle is between 02L and 04L This is found by integrating over this region using the normalized wave function see MAPLE worksheet P LP20 sinn 7ZX20 X O25 The appearance of the probability T2 Energy arb units The probability of finding the particle in a given region of space Using the normalized wave function LP sinn 7ZX one can calculate the probability of finding the particle in any region of space Since the wave function is normalized the probability P is a number between 0 and 1 For example What is the probability that the particle is between 02L and 04L This is found by integrating over this region using the normalized wave function see MAPLE worksheet P LP20 sinn 7ZX20 X O25 Solutions of the Schrodinger equation are orthogonal If the wavefunctions have different quantum numbers Then their overlap is zero We can call the integral Of the product of two wavefunctions an overlap We write L I Tana X 8m 0 Where firm is called the Kronecker delta It has the property 8 O min m 1 if m n For the particleinabox the orthogonality is written L sinsinn 7zxdx 8m 0 Postulates of quantum mechanics are assumptions found to be consistent with observation The first postulate states that the state of a system can be represented by a wavefunction L11q1 q2 q3n t The qi are coordinates of the particles in the system and t is time The wavefunction can also be timeindependent or stationary wq1 q2 q3n Corollary An acceptable wavefunction must be continuous and have a continuous first derivative Since the wavefunction is a solution of the Schrodinger equation it must be differentiable The wavefunction must also be singlevalued Postulate 2 The probability of finding a particle in a region of space is given by Pa a LIf illah Assumptions for the Born interpretation 1 L11LP is real LP is Hermitian 2 The wavefunction is normalized 3 We integrate over all relevant space Normalization is needed so that probabilities are meaningful Normalization means that the integral of the square of the wavefunction probability density over all space is equal to one IJIJdI 1 all Space The significance of this equation is that the probability of finding the particle somewhere in the universe is one Postulate 3 Every physical observable is associated with a linear Hermitian operator Observables are energy momentum position dipole moment etc operator 15 gt observable P The fact that the operator is Hermitian ensures that the observable will be real Postulate 4 The average value of a physical property can be calculated by fw wm P Normalization Postulate 4 The calculation of a physical observable can be written as an eigenvalue equa on New This is an operator equation that returns the same wavefunction multiplied by the constant P P is an eigenvalue An eigenvalue is a number The form of the operators is Position q q Momentum 15 mi 81 Time i t Energy H mi of Postulate 5 It is impossible to specify with arbitrary precision both the position and momentum of a particle This postulate is known as the Heisenberg Uncertainty Principle It applies not only to the pair of variables position and momentum but also to energy and time or any two conjugate variables Conjugate variables are Fourier transforms of one another Conjugate variables do not commute Commutator in quantum mechanics A commutator is an operation that compares the order of operation for two operators 1m m x19 Since the momentum p involves a derivative the order of application affects the result The way to see the result of the commutator is to apply it to a test function fx prx mfx xf x while xpfx mxf x Therefore px ih We say that position and momentum do not Commute 9 12008 Chemistry 431 Lecture 7 Enthalpy NC State University Motivation The enthalpy change AH is the change in energy at constant pressure Whe a c ange takes place in a system t at is open to the atmosphere the volume ofthe system changes butthe pressure remains constant In any chemical reaction that involve the creation or consumption of molecules in the vapor or gas phase there is a wum term associated with the creation or consumption ofthe gas Molar Enthalpy Enthalpy can be expressed as a molar quantity H n We can also express the relationship between enthalpy and internal energy in terms ofmolar quan quot H U PV For an ideal or perfect gas this becomes H U RT Usually when we write AH for a chemical or physical change we refer to a molar quantity for which the units are kJmol Enthalpy for reactions involving gases If equivalents of gas are produced or consumed in a chemical reaction the result is a change in pressurevolume work This is re ected in the enthalpy as follows AH AU PAV at const T and P which can be rewritten for an ideal gas HAUAnRT at const TandP The number ofmoles n is the number ofmoles created or absorbed during the chemical reaction For example CH2CH2Q H29 CHaCHa9 An 1 We arrive at this value from the formula An npmducts 39 reactants 1 39 2 391 The temperature dependence ofthe enthalpy change Based on the discussion the heat capacity from the last lecture we can write the temperature dependence of the enthalpy change as AH CFAT Note that we can use tabulated values of enthalpy at 298 K and calculate the value of the enthalpy at any temperature of interest We will see how to use this when we consider the enthalpy change of chemical reactions the standard ge The basic physics of all temperature de endence is contained in the above equation or more frequently in the equation below as molar quantity AHquot CHAT Another view of the heat capacity At this point it is worth noting that the expressions for the heat capacity at constant volume and constant pressure can be related to the temperature dependence of U and H respectively AHCFAT AUCVAT AH AU 0 AT BTL CV AT 67 The heat capacity is the rate of change of the energy with temperature The partial derivative is formal way of saying this The heat capacity is also a function of temperature We have treated the heat capacity as a constant up to this point That is a valid approximation under ma circumstances but only over a limited range of temperature In the general case the temperature dependence ofthe enthalpy can be described as AHIZCFTdT can bT 71 The parameters a b and c are given in Tables Actually this expression is readily integrated in the general case to give AHa7 T T Tf Enthalpy of physical change A physical change is when one state ofmatter changes into another state of matter ofthe same substance The difference between physical and chemical changes is not always clear however phase transitions are obviously physical changes Fusion AHM AHW Vaporization Solid ltgt Liquid ltgt Gas Freezing AHfreeze AHcond Condensation Sublimation AHSU So ltgt Gas Vapor Deposition AH van deg Properties of Enthalpy as a State Function The fact that enthalpy is a state function is useful for the additivity of enthalpies Clearly the enthalpy of forward and reverse processes must e re a ed by AforwamH ArevelseH so that the phase changes are related by Moreover it should not matter how the system is transformed from the solid phase to the gas phase The two processes of fusion melting and vaporization have the same net enthalpy as sublimation Question Which is statement is false A AWH gt o B Amii lt o c AMSH gt 0 DA Hlt0 vap Addivity of Enthalpies Because the enthalpy is a state function the same magnitude must be obtained for direct conversion from solid to gas s for the indirect conversion solid to liquid and then liquid to gas AsubH AfusH AvapH Of course these enthalpies must be measured at the me temperature Otherwise an appropriate correction would need to be applied as described in the section on the temperature dependence ofthe enthalpy Question Which statement is true A AWH AMSH AvapH B AvapH AsubH AmSH C Amer Asubn AvapH D AvapH AsubH AMSH 9 12008 Chemical Change In a chemical change the identity of substances is altered during t e course ofa reaction One example is the hydrogenation of et ene CH2CH2g H2g gt CHBCH3g AH 137 kJ The negative value ofAH signi es that the enthalpy of the system decreases by 137 kJ and if the reaction takes place at constant pressure 137 kJ of heat is released into the surroundings when 1 mol of CH2CH2 combines with 1 mol of H2 at 25 ElC Standard Enthalpy Changes The reaction enthalpy depends on conditions eg T and P It is convenient to report and tabulate information under a standard set of conditions Corrections can be made using heat capacity for variations in the temperature Corrections can also be made or variations in the pressure When we write AH39in a thermochemical equation we always mean the change in enthalpy that occurs when the reactants change into the products in their respective standard states Standard Reaction Enthalpy The standard reaction enthalpy AH39 is the difference between the standard molar enthalpies ofthe reactants and products with each term weighted by the stoichiometric coef cient A IB E vaproducts E vHZUeactants The standard state is for reactants and products at 1 bar of pressure The unit of energy used is kJmol The temperature is not part ofthe standard state and it is possible to speak ofthe standard state of oxygen gas at 100 K 20 K etc It is conventional to report values at 298 K and unless otherwise speci ed all data will be reported at that temperature Enthalpies of Ionization The molar enthalpy of ionization is the enthalpy that accompanies the removal of an electron from a gas phase atom or ion Hg gtHg e g AH 1312 kJ For ions that are in higher charge states we must consider successive ionizations to reach that charge state For example for Mg we have AH 733 kJ AH 1451 kJ Mgg Mgg 69 M919 gtM92g 69 enthlalpy change is 218 J for the reaction Mgg M9219 2619 We shall show that these are additive so that the overall 9 k Electron Gain Enthalpy The reverse of ionization is electron gain The corresponding enthalpy is called the electron gain enthalpy For example Cg e g 39CI39g AH 349 kJ The sign can vary for electron gain Sometimes electron gain is endothermic The combination ofionization and electron gain enthalpy can be used to determine the enthalpy offormation of salts Other types ofprocesses that are related include molecular dissociation reactions Enthalpies of Combustion Standard enthalpies of combustion refer to the complete combination with oxygen to carbon dioxide and water For example for methane we have CH4g 2029 gtCOZg 2HZOI AEH 890 kJ Enthalpies of combustion are commonly measured in a bomb calorimeter a constant volume device Thus AUn is measured To convert from AUW to AHn we need to use the relationship AHm Aum AvgasRT The quantity Avgas is the change in the stoichiometric coef cients of the gas phase species We see in the above express that Avgas 2 Note that H20 is a liquid 9 12008 Question Fill in the missing stoichiometric coef cients for the combustion reaction OSH12Q X029 gt YCOz9 ZH20 lt N N II N II II m N 074 449m Question Fill in the missing stoichiometric coef cients for the combustion reaction 05H12g 3029 5Coz9 5H20 ttmoo ltltltlt Qu esti o n Determine Avgas for the reaction as written 05H12g 3029 395Coz9 5H20 Qu esti o n Determine Avgas for the reaction as written 05H12g 3029 quot50029 5H20 AHm Aum AvgasRT The quantity Avgas is the change in the stoichiometric coef cients of the gas phase species We see in the above express that Avgas 2 Note that H20 is a liquid Qu esti o n What is the work term for expansion against the atmosphere 05H12g 3029 395Coz9 5H20 A AvgasRT B Aum AvgasRT 0 AU AvgasRT gas Qu esti o n What is the work term for expansion against the atmosphere 05H12g 3029 quot50029 5H20 A AvgasRT B Aum AvgasRT 0 AU AvgasRT gas Hess s Law We o en need a value ofAlF that isnot in the thermochemical tables We can use the fact that AH is a state funct39o o advantage by using sums and differences of known quantities to obtain the unknown We have already seen a simple example of this using the sum of AH offusion and AH of vaporization to obtain AH of sublimation Hess s law is a formal statement of this property The standard enthalpy of a reaction is the sum of the standard enthalpies of the reactions into which the overall reaction may be divided Question Consider the reactions Hg gt Hg e39g AH 1312 kJ Cg 69 C39g AH 349 kJ Clg Hg Cl39g Hg AH 963 kJ Which statement is true about the charge transfer from H to CI to form H and Cl39 A AH 963 kJ H 1661 kJ C A 1312 kJ D AH 349 kJ Question Consider the reactions AH 1312 kJ H9 e39Q Cl39g AH 349 kJ Hg C9 69 gt What further information do you need to calculate the enthalpy for the reaction H2 Cl2 2Haq 2Cl39aq A AH of ionization and AH of electron capture B AH of formation AH of dissociation and AH of solvation C AH ofionization and AH of solvation D AH of dissociation and AH of solvation 9 12008 Question Consider the reactions 9 Hg 59 H AH 1312 kJ Clg 59 019 AH 9kJ Which statement is true about the charge transfer from H to CI to form H and Cl39 A AH 963 kJ B AH 1661 kJ C AH 1312 kJ D AH 349 kJ Question Consider the reactions Hg gt Hg e39g AH 1312 kJ Cg 59 C39g AH 349 kJ What further information do you need to calculate the enthalpy for the reaction H2 Cl2 2Htaq 2Cl39aq A AH ofionization and AH of electron captu e B AH offormation AH of dissociation and AH of solvation C AH ofionization and AH of solvation D AH ofdissociation and AH of solvation Application of Hess s Law We can use the property known as Hess s law to obtain a standard enthalpy of combustion for propene from the two reactions CaHeQ H29 CaHHQ AH 424 kJ C3H8g 502g 393C02g 4H20l AH 2220 kJ If we add these two reactions we get C3H6g H2g 502g gt3C02g 4H20l AH 2344 kJ d now we can subtract H2t12029 gt H200 I AH39 286 kJ to o C3H6g 92029 30029 3H20I AH 2o5a kJ Variation In Reaction Enthalpy With Temperature Since standard enthalpies are tabulated at 298 Kwe need to determine the value of the entropy at the temperature of the reaction using heat capacity data Although we have seen this procedure in the general case the calculation for chemical reactions is easier if you start by calculating the heat capacity difference between reactants and products App E vcxproducts E vCpUeacfanfs and then substitute this into the expression AHBT2 AHBT1 IT2ACdT Ti lfthe heat capacities are all constant ofthe temperature range thequot AMT AWE mom 9 12008 Standard Enthalpies of Formation The standard enthalpy offormation AfH39is the enthalpy for formation ofa substance 39om its elements in their standard states The reference state of an element is its most stable form at the temperature of interest The enthalpy of formation ofthe elements is zero For example let s examine the formation ofwater H2g 12 029 H200 AH 286 kJ Therefore we say that AfH39HZO l 286 kJmol Although AfH 39for elements in their reference states is zero AfH39is not zero for formation of an element in a different phase Cs graphite 39 Cs diamond Ari l 1895 kJmol Question Consider the formation of carbon dioxide at 298 K 05 029 C029 How would you nd the heat of formation of oxygen A Look up AfH for Cs and subtract it from that of CO2 B Look it up the standard thermodynamic tables C The heat offormation of O2 is zero by de nition D It is equal to the standard bond energy of two oxygen atoms Question Consider the formation of carbon dioxide at 298 K 05 029 0029 How would you nd the heat offormation of oxygen A Look up AfH for Cs and subtract it from that of CO2 B Look it up the standard thermodynamic tables C The heat of formation of O2 is zero by de nition D It is equal to the standard bond energy of two oxygen atoms Question Consider the formation of carbon dioxide at 150 K 05 029 0029 How would you nd the heat of formation of C02 Apply a correction to the enthalpy 39om A Hess s law B the van der Waal s equation of state C ideal gas law D none ofthe above Question Consider the formation of carbon dioxide at 150 K 05 029 0029 How would you nd the heat of formation of C02 Apply a correction to the enthalpy 39om A Hess s law B the van der Waal s equation of state C ideal gas law D none of the above

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.