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# General Chemistry 2 week 2, test 2 Chem 202

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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 67 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 DISCLAIMER: FROM HERE ON OUT IN CHEM 202 YOU WILL BE DOING LOTS OF ALGEBRA. IF YOU ARE RUSTY, PLEASE TALK TO YOUR PROFESSOR OR SEE A TUTOR. YOU WILL STRUGGLE IF YOU DON’T UNDERSTAND ALGEBRA WELL!!!! Chapters 49-51 Chapter 49 49.1 - At Any Rate- Learning to set up rate problems Last chapter we covered speed. Now we will cover rate. Rate = k [A] [B] g This is called the rate law o Rate law is equal to the prior speed relationship we covered in chapter 48 o Symbols [A] and [B] represent actual concentrations during the reaction not changes o Symbols f an g are the order of the reaction o K is called the rate constant, and remains constant as long as temperature does -1 -1 o K units can vary between per second (s ) or per minute (min ) o Units for rate in the end must be concentration per time Orders: 1 = time -1 rate = k [A] nd -1 -1 2 2 = conc time rate = k [A] Lets try an example: - - - CH 3(aq) + Cl (aq) CH Cl3aq) + F (aq Rate = k [CH F3 [Cl ] 46.2- An Integrated Approach and Solving Rate Problems It is important for your exam you know how to actually solve rate problems; we only focus for on 1 and 2 ndorders. Here are a few from lecture Dr. Kuta gave. Reaction: H O(2) 2H (g) + 2 (g) 2 -1 -This is a first order reaction with k = 0.0114 s . Calculate the speed of the reaction when the initial concentration of H O = 2.02 M Rate = k [A] Rate = k [H O]2= 0.0114 1 x 2.04 M = 0.0233 m/s s Reaction: 2 C H (g) C H (g) 2 4 4 8 -This is a second order reaction. Were going to find the speed of ethylene -1 producing 1-butene when k = 0.042 M at 452C and the initial concentration of ethylene = 0.00722 Rate = k [A] 2 Rate = k [C H2] 4remember the 2 exponent for the leading coefficient) 2 -6 Rate = 0.042 1/m x s (0.00722 M) = 2.19 x 10 M/s The rate for C H 4s8equal to the rate we just found for C H , 2 4 remember speed relationships equation! -6 -6 So now -2 x 2.19 x 10 m/s = -4.38 x 10 m/s for the speed of disappearance of the reactant. We use -2, because it is disappearing and 2 is the leading coefficient. If you want to find the concentration after a certain amount of minutes follow this: 1 (Coefficient, if any) x (k) x (minutes given) + or 2 kt + concentration 1 substance -Next, it is super important you know how to solve percent remaining and percent completion of a reaction. Don’t worry its super easy, you just plug and chug. concentrationat timet Percent remaining = x 100% intialconcentrationatt=0 All you have to do is plug in the values you just solved for. Remember to find concentration after a certain time is right above. initial−conc.aftertimet Percent completion = initialconc x100 46.3 - Half-life (two halves do not make a whole) o Half life - This is the time period which is needed for one half of the amount of a reactant to react. So the time it takes for 50% of your reactant to do its thing. o First half-life ends at (1/2) . When 50% remains and there is a 50% completion o Second half-life ends at (1/2) , when 25% remains and there is a 75% completion Learn how to solve these for your test!!! ln2 Half-life equation First order: 1/2= ak Second order: t 1/2= A¿ o ak¿ 1 ¿ This is also plug and chug, I’ll show an example using the 2 C H 2g)4 C4H 8g) equation. 1 T = l =1600min Remember the 2 is there cause 1/2 2x0.042 x0.00722mol/l mol it’s the coefficient 49.4 - k and T o Know that k increases as T increases o The relationship between them is given as k = A e -Ea/RT Example: CO(g) + NO (g) 2 CO (g) +2NO(g) Ea= 141 kJ A = 8.9 x 10 (l/mol)/s 10 -0.0340 10 K22 c= 8.9 x 10 x e = 8.60 x 10 o We can also find k at different temperatures within the reaction Set the appropriate k to the appropriate temp. Ea Ln k 1 ln A - RT Both of these re-arrange to: ln 1 k2= −Ea (1 − 1 ) k R T T 1 2 1 Ea Ln k 2 ln A - RT 2 k 2 o We can use this equation to find any variable within it. Finding ln k 1 is very simple, just plug your values in. Temperature is simple too but ill show an example from the book. K2/k1= 100, Ea= 141,000 J, T = 251C (298k) find T 2 ln 100 = - 141,000 J 1 1 ( − ) 8.314 J /K T 2 298K 1 1 1 -0.00027154 = − 0.0030841 = Basically its T 2 298K T 2 solved algebraically T2 = 324K Chapter 50 50.1 - Getting into reverse Reactions can go forward and reverse. We will still be using rate laws o Rate fwd k fwd= rate rev k rev kfwd kfwd o K = k rev= krev K o H = E - E You can also rearrange this equation to find the a,fwd a,rev other variables o Know that the energy barrier to go backwards is greater than to go forward Ex: Br(g) + H 2g) HBr(g) + H(g) rate fwd = 720 (L/mol)/s rate rev = 5.6 x 10 (L/mol)/s Rate = k [Br] [H ] rate = k [HBr] [H] kfw= [HBr ]H] fwd fwd 2 rev rev krev [Br][H ] 2 720/s −7 9 =1.3x10 5.6x10 /s 50.2 - Reaction Energy Diagram o Useful for activation energies and reaction enthalpies We can apply the H = E - E equation to these diagrams. Basically plug a,fwd a,rev and chug. 50.3- Catalysts o Catalyst- a component in the system which speeds up the overall reaction by reacting or interacting with a reactant or intermediate o Chemical equations stay same, mechanisms will change o Does not change thermodynamic parameters, but kinetic parameters o For a catalyst, you basically add up the steps. Its super simple, but don’t worry about them for the exam, just know what they do o Stabilizer-agent whose purpose is to stabilize something from doing an undesirable reaction (opposite a catalyst) Chapter 51 51.1 - More Q and K We will be focusing on a new equation derived from G = G + RT lnQ Q o G = RT ln K Recall what all of these variables mean Q o Q = K at equilibrium and Q/K = 1 so we have G = RT ln K = RT ln (1) = 0 o Q is where the system is at some point o K is where the system is heading o Q decreases as K increases, vise versa 51.3 - Three Relationships for K and Q First relationship o When you reverse an equation you will invert K K-1 1/K 1 Second relationship o When you add an equation, you multiply the K’s (and Q’s) to get the new K for the sum of the equation K sum= K 1 K 2 Q sum= Q 1 Q 2 Third relationship o When K in an equation is multiplied K 3 K 2n Q 3 Q 2n These are important because we can use these different equations when we need to manipulate equations and we don’t have to go through G every single time. 51.4 - Examples Ways we can use “K” in equations o If system is in equilibrium, measure amounts of all reagents present, then calculate K which allows us to find G o If we know K we can measure amount of some of the reagents Example 1 from text: This equation is at equilibrium at 298 K H 2g) + I2(s) 2 HI(g) Two parts to this example: Part A: During one run, at equilibrium 0.0689 atm H 2 0.00691 mol I 2 0.140 atm HI 2 2 HI¿ 0.140¿ Calculate K K = ¿ K = ¿ P¿ ¿ ¿ ¿ Part B: Another run lead to these measurements 0.0418 atm H , 2.00771 mol I 2 0.132 atm HI Is this system in equilibrium? Remember to be in equilibrium Q = K 2 2 HI ¿ 0.132¿ So Q = ¿ = ¿ = 0.417 P¿ ¿ ¿ ¿ So is Q = to the K value we found earlier? Nope! So this system is not in equilibrium, its going in reverse since Q is greater than K, the exergonic direction. Example 2: Here is one where we know K In equilibrium: N (2) + 3 H (g2 2 NH (g) K3= 36.6, P Nitro = 0.217 atm, P Ammon. = 0.261 atm What is the pressure of Hydrogen? N H 3 2 ¿ 3 We know that K = H 2 soooooo what’s next? P ( ) ¿ 2 P ¿ ¿ 2 0.261¿ ¿ 2 H ¿ 3 0.261¿ 1/3 36.6 = 2 Re-arrange P(H ) = 2 (¿¿ 0.217 (36.6)) = 0.205 (0.217)P¿ ¿ ¿ ¿ You can figure out the total pressure by just adding the pressures all together.

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