Chemistry 2. week 1 test 2
Chemistry 2. week 1 test 2 Chem 202
U of L
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Popular in Chemistry
This 8 page Class Notes was uploaded by Catherine Montz on Thursday February 25, 2016. The Class Notes belongs to Chem 202 at University of Louisville taught by Dr. Kuta in Spring 2016. Since its upload, it has received 109 views. For similar materials see General Chemistry 2 in Chemistry at University of Louisville.
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Date Created: 02/25/16
General Chemistry 202 Professor Kuta Test 2 3 Chapters covered weekly Chapters 4648 Chapter 46 46.1 Stoichiometry A more in depth review is on noblereaction.org chem 201 text, chapter 19. Below is a quick example for quick review. o Know that with stoichiometry we can calculate G as well as H Example problem: How many kJ’s can you get from the combustion of 1.000g propane at standard conditions? o Okay, kJ’s are not used for enthalpy, but for free energy. (hint: combustion) o Write an equation for this reaction. C 3 8 ) + 5 O (2 3 CO (g)2+ 4 H O(g)2 H 2 is a gas in this equation, due to the types of chemicals in the equation, and we are calculating free energy. o Now we simply set up the equation for stoichiometry 1.0 g C H x molC 3H 8 x −2,073.19kJ = 47.02 kJ 3 8 44.09gC 3H 8 molC 3H8 46.2 Temperature effects Know your different variations o Variation 1 always exergonic and temperature will not change that o Variation 2 always endergonic and temperature cannot change that o Variation 3 this is exothermic G = ()H T()S S is negative, making T and G more positive o Variation 4 this is endothermic G = (+)H T(+)S Here is an example problem using variation 4’s equation: Calculate G at 20.00C (293.15 K) and at 130.00C (403.15 K) if H = 15.0 kJ and S = 45.0 J/K o This is easy as pie, simply plug….. and chug. 293.15 15.0 kJ (293.15 K)(45.0 J/K) = 15.0 kJ 13.1 kJ = 1.9 kJ AZ 398.15 15.0 kJ (398.15 K)(45.0 J/K) = 15.0 kJ 17.9 kJ = 2.9 kJ AZ Simply reverse variation 4 by making 15.0 kJ, 15.0 kJ and making 45.0 J/K, 45.0 J/K to apply this example to variation 3. Give it a try for practice Answers: 1.9 kJ and 2.9 kJ o Temperature can determine whether a reaction will happen or not and the direction (only for variations 3 and 4) o If G switches since between two temperatures, at some point G goes through zero, which is the temperature of equilibrium at standard conditions o We can determine T (temperature), using the example above G = H TS 15.0 kJ = T (0.0450 kJ/K) 15.0/0.0450 = 333.3 K This system is at equilibrium at 333.3 K REMEMBER to always set G = 0 o All phase changes for single compounds are either variation 3 or 4 o You will always multiply by 1, if one molecule is present, if two are present multiply by 2 Ex: H 2= 1(293.15) 1(293.15) for one molecule 2H O= 2(293.15) 2(293.15) 2 46.3 Standard Conditions Know what standard conditions are Standard pressure = 1 atm Concentration = 1M Temperature = 25 C Important items to know: o Energy and entropy will always be affected by temperature o Pressure will always have impact on thermodynamic properties of gases, but not on liquids or solid o symbolizes standard conditions. o We compare the final state to initial state (under standard conditions) o Sign of G is used to determine direction of net change, whether or not a reaction can happen in one direction or the other. 46.4 General Conditions Not all conditions will be standard and these will be called general conditions (just G). o The connection is G = G + RT lnQ R= 8.134 J/(mol K) T= temperature Q= reaction quotient (ratio of given thermodynamic conditions of all components in the balanced equation, or activities) o Activities is its actual given condition relative to standard conditions y z a(Y) a(Z) o Activity is written as a( ) so Q will be Q= a(A) a(B) b (n= coefficient) o a(A) = actualconditionof A actual condition being what is given standard conditionof A actual Solids : a(A) = = 1 standard So all solid compounds are entered into Q as 1. You can leave out all solids because their activities equal 1. actual Liquids : a(Z) = = 1 standard You may also leave out liquids because they too equal 1. actual P ∈atm Gases : a(B) = actual = B Remember to drop units, because standard oneatm activities do not have units actual actualY∈M Solutes: a(Y) = = drop units standard oneM Examples to clear things up: A) O 2g) O (a2) [O 2(aq)] Q = P(O g ) 2 B) NaCl(s) NaCl(aq) Must present dissociation of ions + NaCl(s) Na (aq) + Cl(aq) +¿ aq ) N a ¿ ¿ + Q = −¿ a(¿) = [Na (aq)][Cl(aq)] C l ¿ ¿ ¿ Make sense? More examples for you to work will be in the study guide. 46.5 Getting into Q We will need to use the equation above G = G + RT lnQ Example taken from book: You have a solution which contains 0.663 M C H OH(aq) at 4 9 25.00 C. You add C H OH4l)9to the system. What is G for dissolving under these conditions? Can any of the added C H OH(l) dissolve? G = .017 kJ. 4 9 soln First off: write the equation! C 4 O9(l) C H OH(4q9 Know Q = 0.663 M from above G = G + RT lnQ = 0.17 kJ + (8.314 J/K)(298.15 K) ln(0.663) G = 0.17 kJ + (1020 J) = 0.17 kJ + (1.02 kJ) = 1.19 kJ Since G is negative, the process is allowed and yes, more of the C H OH can 4is9olve. o G is usually in kJ and RT is usually in J, so you have to convert J to kJ If you have the book, I would recommend trying example 5, pg. 481. I want you to try it on your own so you can understand the concept. If you struggle, don’t worry. I will be posting the exact problem in the study guide for you to work. It is important you try these on your own to get a complete understanding for your exam. 46.6 Keeping the Perspective This is simply a summary of the chapter on what we have learned, but I will break it down o We know if something can or cannot happen now o We have seen how these processes break down with free energy relationships o How to calculate Q, T, G, and G Chapter 47 47.1 Equilibrium Aspects Let’s start by remembering some equilibrium conditions (CASE 3) o S univ= 0, Gsys 0 o S univstays the same o G ssyss the same Ultimately we can make a connection and apply to Q, so G = 0 = G + RT lnQ or G= RT lnQ These values are all constants except lnQ, so we will now add lnK G= RT lnK K = equilibrium constant or quotient G /RT We can determine K by this equation K = e Know that K is one specific condition of Q, specific to the equilibrium condition Know variations of Q o Q = reaction quotient for the given conditions o Standard: Q = 1 o Equilibrium: Q = K Here is an example where we know K 2NO(g) + 2Cl (g2 2Cl O(g2 + N (g) 2 equilibrium is at 298.15 K K= 0.000213, what is G? Use G= RT lnK So, G = (8.314 J/K)(298.15K) ln(0.000213) = 21.0 kJ 2 Cl 2¿ P(N ) 2 ¿ Cl ¿ 2 22 K = NO¿ P¿ The ratio of products over reactants has the numerical value of K = P ¿ P ¿ ¿ 0.000213 Example: Calculate EVP for water at 20.00C from G vap vap8.56 kJ −∆G° = −(8.56kJ) o Remember K = e G /RT RT J Change 8.56 to J so (8.314 K )(293.15K) 8,560 J Which equals 3.5122 so plug this into the (K=) equation K = e3.51= 0.0298 so EVP at 20.00C = 0.0298 K = P(H O(g)) 2 Let’s do one more regarding solubility, I think this one is very important; Example: Calculate solubility (in M) of CO in water at 25.00C when the pressure of CO above 2 2 the solution is 4.0 atm. o CO (g2 CO (aq)2 [C O2 aq )] o K = P(CO 2(g)) o [CO (a2)] = K x P(CO (g)2 G /RT 8.41/(8.314)(298.15) o K = e Gsoln 8.41 kJ K = e = 0.9966 o [CO (a2)] = K x P(CO (g)2 = K x 4.0 so this equals 3.98 47.2 Drive o Driving force Free energy which would be available from the exergonic process or minimum payment into the system which would be required for an endergonic process. o Nature’s drive towards maximum S univ o Doesn’t last forever, greatest driving force is when G is most negative, it gets weaker and is lost, G becomes 0. Decreases as reaction proceeds due to changing Q value o Natural drive for every process is G = 0 because this is the point of equilibrium Balance associated with equilibrium approaches o By speeds of the opposing processes. At equilibrium these speeds are equal and balanced, this is kinetic description o Enthalpy and entropy; the systems enthalpy and entropy are in balance at equilibrium, this is the thermodynamic approach. Know that a system in equilibrium is still in motion and this is referred to as dynamic, because there is motion, but everything is BALANCED. Chapter 48 48.1 Kinetics, Will it happen? o Called the rate of reaction o Factors effect a reaction and those are called mechanisms 48.2 Two Examples o Mechanisms depend on who’s reacting with whom o Ex: reaction in water between CH F (a3) + Cl(aq) CH Cl(aq) 3 F(aq) (exergonic) o Ex: CH (g) + Cl (g) CH Cl(g) + HCl(g) 4 2 3 Discussed in recitation, these may not be of importance for the exam however I would suggest reading it for your own benefit. 48.3 It’s Elementary o Most reactions occur in multiple steps (composite mechanisms) 1 step = simple o Overall rate depends on the rates of individual steps of the reaction o Each step called elementary step conveys specific action on a particletoparticle basis of who is doing what to whom o Only includes chemical units happening in that reaction at that time o Overall equation tells us net, overall outcome of all steps done (sum of steps) o Need to know whether it is an overall or elementary equation Elementary = molecularity (unimolecular, bimolecular, trimolecular) Elementary can only have 1, 2, or 3 particles shown on the left Overall = four or more particles on the left Overall = equations with fractional coefficients 48.4 Speed Relationships o Speed is how fast it goes, how fast reactants are disappearing or how fast products are appearing ∆[A] ∆[Z ] o Speed rate = ∆t = ∆t change in concentration over a period of time o A (reactants) is negative and Z (products) is positive Example with equation of coefficients of 1 CH 3(aq) + Cl(aq) CH Cl(a3) + F(aq) [CH F3 = [Cl] = [CH Cl] 3 [F] −¿¿ −¿ ¿ ∆[C H F] Cl ∆[C H Cl] F speed rate = 3 = ¿ = 3 = ¿ ∆t ∆¿ ∆t ∆¿ ¿ ¿ 1 ∆[A] 1 ∆[B] 1 ∆[Y ] 1 ∆[Z] o Speed = a ∆t = b ∆t = y ∆ t = z ∆t Coefficients represent lower case letters o Only gases and solutes will be worked via speed relationships o Speed will depend on how the coefficients are written in balanced equations
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