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Gen Chem II Study Guide

by: Brianna Hinton

Gen Chem II Study Guide CHEM 102

Brianna Hinton

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These notes cover what will be on exam 2.
General Descriptive Chemistry II
Dr. Jennifer Krumper
Study Guide
Enthalpy, entropy, Chemistry
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This 13 page Study Guide was uploaded by Brianna Hinton on Sunday October 9, 2016. The Study Guide belongs to CHEM 102 at University of North Carolina - Chapel Hill taught by Dr. Jennifer Krumper in Spring 2016. Since its upload, it has received 21 views. For similar materials see General Descriptive Chemistry II in Chemistry at University of North Carolina - Chapel Hill.


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Date Created: 10/09/16
General Chemistry II Chapter 16: Thermodynamics Warm-Up Assignment​ (Due 9/21/2016) 1. Which of the following is the best description of the f ​ irst law of thermodynamics? a. An object will maintain its current state of motion unless acted upon by an outside force. b. Thermal energy flows from the colder object to the warmer object. c. The entropy of an isolated system will continue to increase until the system reaches thermal equilibrium. d. The change in thermal energy of a system is equal to the energy transferred in or out of the system as work, heat, or both. 2. Place an arrow on each of the diagrams below to show the direction heat energy will naturally move when the substances are placed in contact or mixed together (U represents internal energy. Metal Pan, 800 g, 35 degrees Celsius ← ​ ​ Water, 1000 g, 50 degrees Celsius Glass cup, 200 g, 20 degrees Celsius​ → ​Ice cubes, 250 g, 0 degrees Celsius Argon gas, U= 1000 J, 30 degrees Celsius ← ​ ​Helium gas, U= 500 J, 40 degrees Celsius *​Remember that thermal energy flows from the warmer/hotter object to the colder object.* ​ 3. Which of the following statements regarding entropy is FALSE? a. Sugar poured into a container of water dissolves. The entropy of the sugar and water increases. b. A log in a fire burns to form ashes. The entropy of the log increases. c. A candle burns to heat the surrounding air. The entropy of the candle and surrounding air increases. d. Crude oil is refined to make gasoline. The entropy of the crude oil increases. 4. Predict how the entropy of the substance is affected in the following processes. Ar (l) → Ar (g): ​Entropy increases. O​2​g, 300 K) → O​ (g,2​00 K): ​Entropy increases. CO​ 2​, 57 degrees Celsius) → CO​ (s, -80 2​grees Celsius): E ​ ntropy decreases. 5. Without referring to a data table, rank the following compounds by standard molar entropy. Highest to Lowest: CH​ OH (g3​ O​ (g), N2​(g) 2​ 6. The change in entropy, DeltaS​ is relrxn,​to the change in the number of moles of gas molecules. Determine the change in the moles of gas for each of the following reactions and decide if the entropy increases, decreases, or has little or no change. KO​ 2​) → K(s) + O​ (g)2​ 1 mol, entropy increases H​2​g) + Cl2​(g) → 2HCl (g) 0 mol, has little or no change C​2​​2​g) + 3O​2​g) → 2CO​ (g2​+ 2H​ O(2​ -2 mol, entropy decreases NH​ (g) + HCl (g) → NH​ Cl (s) 3​ 4​ -2 mol, entropy decreases 7. Which of the following reactions have a positive DeltaS​ ? Checrxn​l that apply. a. 2A (g) + 2B(g) → 5C (g) b. A (g) + 2B (g) → 2C (g) c. A (s) + 2B (g) → C (g) d. A (s) + B(g) → 2C (g) Wednesday, September 21, 2016 System and Surroundings: ● The ​universe​ is everything ● The universe= ​system + surroundings ● The ​system​ includes the particles we want to study ● The s​ urroundings​ are everything else Energy: ● Energy can be transferred as work or as heat Work: energy used to cause an object that has mass to move. ● Heat: Energy used to cause the temperature of an object to rise. Energy Transfer The internal energy, U, of a system can be changed by heat flow and work. If heat flows into the system, q​ in​r work is done on the system, w​ , Delon​ >0. If heat flows out of the system, q​ , out​ or work is done by the system, w​ , itby​nternal energy decreases, DeltaU <0. Endothermic Reaction: q>0, the system gains energy in the form of that heat Exothermic Reaction: q<0 I drive from my house to campus and my parking spot every day. Which of the following variables describing this trip is/are state function(s)? A. Time in the car B. Mileage car is driven C. Elevation change D. All 3 are state functions E. None are state functions State Functions: Paths X and Y represent two different routes to the summit. Both have the same change in elevation (​altitude​ is a state function); it does not depend on the path, but they have different distances (distance walked is ​not​ a state function; it depends on the path). First Law of Thermodynamics​: ● Energy cannot be created or destroyed​ (Chapter 5). ● Therefore, the total energy of the universe is constant. ● Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa. Enthalpy​: ● Enthalpy,​ ​the total energy or heat content of a syste​ ​ is the sum of the internal energy plus the product of pressure and volume: H= E + PV Only when we are at CONSTANT PRESSURE: DeltaH= q So, at constant pressure, the change in enthalpy IS the heat gained or lost. Which of the following has the higher enthalpy (heat content): A. 1 mol CO​ 2​) at 298K B. 1 mol CO​ (g) at same T 2​ C. A and B have the same enthalpy Think about this as a chemical reaction. It is an ENDOthermic reaction, because you have to have heat as a reactant in order to break apart those intermolecular forces to convert it from a solid to a gas. Which of the following has the higher enthalpy (heat content): A. 2 mol H atoms B. 1 mol H​ 2 C. A and B have the same enthalpy Hydrogen is found in nature (diatomic). It is an ENDOthermic reaction. Think about this as a chemical reaction. H​ (g) → 2H (g) 2​ You have to have energy to break the covalent bonds between the hydrogen atoms. BREAKING A BOND IS ALWAYS AN ENDOTHERMIC PROCESS. Enthalpies of Reaction​: For a chemical reaction, the change in enthalpy, Delta H, is the enthalpy of the products minus the enthalpy of the reactants. Delta H= H​ productsreactants Enthalpy is a state function. Calculating Delta Hrxn from Delta Hf We can use Hess’s law. Product-or reactant-favored: ● Exothermic reactions (that release heat energy to the surroundings) are sometimes called “product-favored” reactions ● Conversely, endothermic reactions (that require the input of heat energy into the system) are called “reactant-favored.” Chapter 16: Thermodynamics Spontaneous processes disperse matter over the universe. Spontaneous processes are those that can proceed without any outside intervention. The gas in the left vessel will spontaneously effuse into the right vessel, but once the gas is in both vessels, it will not spontaneously localize to one vessel. Helium balloon popped, helium particles will not spontaneously undergo the random and chaotic motion of gas molecules, and they will fill the room. Spontaneous processes disperse energy. ex) Placing a hot object in contact with a cool object. Spontaneity​: Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. And they’re usually reversible. Spontaneity may depend on temperature. Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. ex) Take ice cube out of the freezer and put it on the table- it will melt. Take ice cube out of the freezer and put it in a colder freezer- it will remain as ice. Above 0, it is spontaneous for ice to melt. Below 0, it is nonspontaneous for ice to melt. How many of the following processes are spontaneous? I. Fragrance molecules spreading through a room II. Condensing water vapor at 110 degrees Celsius, 1 atm pressure III. Condensing water vapor at 95 degrees Celsius, 1 atm pressure IV. Evaporating liquid water at 95 degrees Celsius, 1 atm pressure Entropy​: Entropy (S) is a term coined by Rudolph Clausius in the 19th century. Clausius was convinced of the signficance of the ratio of heat delivered and the temperature at which it is delivered q/T Entropy can be thought of as a measure of the randomness of a system. DeltaS: q​ /T rev​ Like total energy, E, and enthalpy, H, entropy is a state function. Therefore, DeltaS= Sfinal-Sinitial For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred Predicting the sign on Delta S Entropy increases with the freedom of motion of molecules. Therefore, S(s)<S(l)<S(g) Generally, when a solid is dissolved in a solvent, entropy increases. Entropy changes in chemical reactions. In general, entropy increases when: ● Gases are formed from liquids and solids ● Liquids or solutions are formed from solids ● The number of gas molecules increases ● The number of moles increases Entropy and Molecular Formula: An increase in molecular complexity generally leads to increase in S. Which substance has the larger standard entropy, in each pair: A. 0.5 mol of N2 (g) at 298 K, 20-L volume B. 0.5 mol CH4 (g) at 298 K, 20-L volume Why? Methane with its 4 covalent bonds has more places to hold energy. Which substance has the larger standard entropy, in each pair: A. 0.5 mol of N​2​g) at 298 K, 20-L volume B. 0.5 mol of N​2 ​) at 298 K, 10-L volume Why? You have the same number of moles, but one is in a larger container, increasing the freedom of movement of molecules. Which substance has the larger standard entropy, in each pair: A. 100.0 g NaNO3 (s) at 298 K B. 100.0 g NaNO3 (aq) at 298 K Why? Look at the phases of the compound. Entropy is higher in aqueous than solid phase. What is the sign of DeltaSsys when LiCl (s) is formed from Li(s) and Cl2 (g)? Li(s) + Cl2 (g) → LiCl (s) A. Positive B. Negative C. Zero Why? Turning 1.5 moles of things to 1 mol in the products. Gas is also going to a solid, which decreases the entropy. Friday, September 23, 2016 Second Law of Thermodynamics The second law of thermodynamics states that: ● The entropy of the universe (DeltaSuniv) increases for spontaneous processes, and ● The entropy of the universe does not change for reversible processes (ex. Phase change at transition temperature) As a result of all spontaneous (irreversible) processes, the entropy of the universe increases. DeltaSuniv > 0: spontaneous DeltaSuniv <0: nonspontaneous (spontaneous in opposite direction) DeltaSuniv =0: reversible (system is at equilibrium) Entropy of Rusting Consider the oxidation (rusting) of iron to make iron (III) oxide, where the system is the chemical reaction that takes place. What can you say about the signs on DeltaSsys, DeltaSsurr, and DeltaSuniverse? (Positive, negative, 0) It may help to write out the reaction of iron rusting. 4Fe(s) + 3O2(g) → 2Fe2O3 DeltaSuniverse: Positive because the nail rusts DeltaSsurr: Positive DeltaSsys: Negative Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. Increasing the temperature​: The entropy of a pure crystalline substance above absolute zero is a positive value since it has random motion due to nonzero kinetic energy. It has 0 disorder because all the dipoles have positioned themselves in a way that maximizes the bonds between them. Standard Entropies, S(degree) Increase molecular complexity, increase the standard entropy. Entropy Change for surroundings Exothermic Reaction: DeltaHsys < 0 Heat is released to the surroundings The surrounding molecules gain disorder by gaining that heat energy, so they will move faster because of that gain of kinetic energy. DeltaSsurr > 0 Endothermic Reaction: DeltaHsys > 0 Heat absorbed from surroundings so the surroundings become more organized. DeltaSsurr < 0 What is the sign of DeltaSsys for the production of ammonia gas (NH3) from its elements, nitrogen and hydrogen? A. Positive B. Negative C. Zero D. Can’t tell without more information Using the table, estimate the entropy change associated with the production of one mole of ammonia gas from its elements, nitrogen and hydrogen. Substance Standard Molar Entropy, S (degree) [J/mol K] N2 (g) 192 H2 (g) 131 NH3 (g) 193 N2(g) + 3H2(g) = 2NH3(g) (2 x 193) - (192 + [3 x 131])= -199 You are making two moles of ammonia, though and the question only asks for the production of ONE mole of ammonia gas. This means that you must divide the -199 by 2, which will give you -99 J/K. A. -130 J/K B. -193 J/K C. -99 J/K D. -199 J/K In order to do this, you have to use the stoichiometry and the coefficients from the balanced chemical equation for the formation of ammonia gas. It is also always the products subtracted from the reactants. The math for this can be seen just above the available answer choices. What is the sign on DeltaSsurr for the production of ammonia gas (NH3) from its elements, nitrogen and hydrogen? N2(g) + 3H2(g) = 2NH3(g) Delta H= -91 kJ A. Positive B. Negative C. Zero D. Can’t tell without more information Consider the process by which 234 g benzene (78.1 g/mol) vaporizes reversibly at its normal boiling point of 80.1 degrees Celsius. The enthalpy of vaporization of benzene is 30.8 kJ/mol. What is the entropy change for the surroundings? Report your answer in J/K to units place. The sign is going to be negative. DeltaHsys: 30.8 kJ/mol (234g)(mol/78.1g)= 92.4 kJ -92.4 x 10^3 J/353= -262 J/K Consider the process by which 234 g benzene (78.1 g/mol) vaporizes reversibly at its normal boiling point of 80.1 degrees Celsius. What are the entropy changes for the system and universe for the same proess? DeltaSsurr: -262 J/K DeltaSsys: +262 J/K Sapling Chapter 16 Graded Homework Daily Quiz: 1. For the spontaneous reaction below, what are the signs for Delta G, Delta H and Delta S? Remember that Delta G= Delta H - T(Delta S) (-,-,-) 2. For the spontaneous reaction below, is the reaction enthalpy or entropy driven? A. Enthalpy B. Entropy C. Can’t tell without more information Delta H <0 Product Favored Delta S <0 Reactant Favored Determining Spontaneity Delta S​univ​Delta S​system​ -DeltaH​system​ -T(Delta S​universe​Delta H Gibbs Free Energy -TDeltaS​ universe ​efined as the Gibbs free energy, Delta G. When Delta S​ is positive, Delta G is negative. Therefore, when Delta G is negative, a process is universe​ spontaneous. Delta G= Delta H​ system​T Delta S​system Reactants are pure solids and liquids, if concentrated they are 1 M and if there are gases, they are at 1 atm. G= H-TS If DeltaG Delta S​universe ​ Delta G < 0 Spontaneous ​Exergonic Delta S​universe ​ Delta G > 0 Nonspontaneous ​Endergonic Delta S​ = 0 Delta G = 0 Reversible (at equilibrium) universe ​ Is Gibbs energy, G, a state function? G= H- TS A. Yes B. No C. Can’t tell without more information Gibbs energy is a state function because it is made up of state functions. Standard Free Energy Changes Analogous to standard enthalpies of formation are standard free energies of formation, Delta G​ f For a change at any temperature, DeltaG= DeltaH - TDeltaS How does DeltaG change with temperature? Delta H > 0 (endothermic) Delta H < 0 (exothermic) Delta S > 0 (increase in entropy) Delta G < 0 at high temperature Delta G < 0 at any temperature Delta G > 0 at low temperature Process is spontaneous at any Process is spontaneous at high temperature temperature Delta S < 0 (decrease in entropy) Delta G > 0 at any temperature Delta G < 0 at low temperature Process is nonspontaneous at Delta G > 0 at high temperature any temperature Process is spontaneous at low temperature Signs the same: that’s when Delta G is a function of temperature Spontaneous at high T: Delta H > 0 endo, Delta S > 0, entropy-driven Spontaneous at low T: Delta H < 0, Delta S <0, enthalpy-driven High temperature for one reaction may be a low temperature for another reaction. ex) Consider an ice cube removed from the freezer (-10 degrees Celsius) and placed on the countertop in a kitchen at t= 20 degrees Celsius. Answer the following questions for this phase change. a. What is the sign of Delta H (system)? ​positive b. Is this process exothermic or endothermic? ​endothermic c. What is the sign Delta H (surroundings)? ​negative d. Is the product more or less ordered than the reactant? ​less ordered e. What is the sign of Delta S (system)? ​positive f. Is the process spontaneous under these conditions? ​yes g. What is the sign of Delta S (universe)? ​positive h. What is the sign on the change in free energy, Delta G? ​negative i. Is the process driven forward by change in enthalpy or change in entropy? ​Change in entropy j. At what temperature (degrees Celsius) is the process at equilibrium? (Do this by setting Delta G equal to 0)​ degrees Celsius Delta G at equilibrium: 0 That means Delta H= T Delta S T= Delta H/ Delta S Units on T: Kelvin!! An exothermic reaction is always spontaneous if ΔS is positive, and only spontaneous at low temperatures if ΔS is negative. An endothermic reaction is spontaneous only at high temperatures if ΔS is positive and never spontaneous if ΔS is negative. In other words, the only time the spontaneity is temperature dependent is when ΔΗ and ΔS have the same sign. Sapling Warm Up Homework: Wednesday, September 28, 2016 Do DeltaS (sys) and DeltaS (surr) always have opposite signs? A. Yes B. No C. Maybe… Consider the combustion of glucose…. C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l) What are the signs on DeltaH (sys), Delta S (sys), Delta S (surr)? Create a three-character code where P=positive, N= negative, U= unknown. Combustion reactions are always exothermic. Delta H (sys)- negative Delta S (sys)- positive Delta S (surr)- positive What can you say about the sign on Delta G for this reaction? Is it positive, negative, or does it depend on temperature? -​Increasing in disorder, Delta S is positive, making Delta G more n ​ egative​. Free Energy and Equilibrium Delta G dot- system is in standard conditions, solids and liquids are pure substances Under any conditions, standard or nonstandard, the free energy change can be found this way: Delta G= Delta G (dot) + RT ln Q For a system at equilibrium, Delta G= 0 and Q=K: Delta G (dot)= -RT ln K Delta G is proportional with the number of moles reacting. Rearranging: K= e^ -Delta G (dot)/ RT K>1, Delta G <0, products are more abundant at equilibrium K <1, Delta G >0, reactants are more abundant at equilibrium K=1, Delta G=0, If Delta G (dot) is negative, K > 1 and products are favored over reactants at equilibrium. *Look at the reaction coordinate diagram.* If Delta G (dot) is positive, K < 1 and reactants are favored over products at equilibrium. *Look at the reaction coordinate diagram.* If Delta G (dot) is zero, K = 1. *Look at the reaction coordinate diagram.* This doesn’t happen very often. Gibbs Free Energy Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium. Ex) Consider the weak acid dissociation of hydrofluoric acid, HF: HF (aq) + H2O (l) <-> H3O+ (aq) + F- (aq) where Ka= 6.8 x 10^-4 Delta G (dot)= -RT ln K = +18.1 kJ/mol What is Delta G when [HF]= [H3O+]=[F-]= 0.50 1. Write out your Q expression. Q= [H3O+][F-]/[HF]= 0.50 2. Delta G= 18.1 kJ + 8.314J (298) ln (.50)(kJ/10^3)= 16.4 kJ/mol What point is this likely to be on the graph? Point A, B, or​ C Q= 0.50 Methanol, CH3OH, can be made by controlled oxidation of methane. CH4 (g) + 1/2 O2 (g) <-> CH3OH (g) First, predict the sign on DeltaS (rxn): A. Positive B. Negative (Delta S rxn < 0) C. Zero Given that Delta H (dot) = -146 kJ, calculate Delta G (dot) at 298K. Delta S (dot) = -49.1 J/k Delta G (dot)= -131 kJ Under standard conditions, is the reaction spontaneous at this temperature? A. Yes B. No What is the value of the equilibrium constant, K? (Expect K to be greater than 1 because Delta G (dot) is negative) 9 x 10^22 Graph A


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