Final Exam Review

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divide this number of moles by the TOTAL concentration of the reaction. {Your total volume is your initial volume PLUS the volume you added in step 3}). 5. Find the pH of the salt. Ex. Find the pH at the equivalence point of a titration of 100 mL 3M HA titrated with 2M Ba(OH) 2 (Ka of HA = .0005). 1. 2HA + Ba(OH) 2 Ba(A) 2 + H 2 O 2. 3 * .1L = .3 moles HA. 3. Notice it is not a 1:1 ratio. The change in Ba(OH) 2 is –x, while the change in HA is –2x. Therefore, HA gets used up twice as much. So if we add .15 moles of Ba(OH) 2 , we see that .15 moles - .15 moles (the titrant) = 0. Also, .3 moles – 2(.15) (for the analyte HA) = 0. Therefore, we need to add .15 moles of Ba(OH) 2 . Now solve for volume: V=mol/c or .15/2 which equals .075. We need 75mL of Ba(OH) 2 . 4. Since the change in number of moles in the salt is +x, not +2x (check the coe ffi cients), we know the moles of salt produced is going to be equal to . 15 mol. Divide that by our total volume, which is equal to 100 mL + 75 mL = .175 L. .15/.175 = .86M salt. 5. Now solve for the pH of the salt. (Note that the salt has formula Ba(A) 2 ). Take that in to account when solving for your dissociation concentration in step 2 of pH of salts. Thermodynamics: Δ E = q + w Δ H 0 = (n prod Δ H f 0 (prod)) -- (m react Δ H f 0 (react)) w = —P( Δ V) Δ S sys = (q rev )/(T) ( spont. when Δ S sys >q rev /T’)

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Δ H = Δ E + P( Δ V) Δ S 0 rxn = (n prod Δ S 0 (prod)) -- (m react Δ S 0 (react)) q = mC( Δ T) Δ G 0 = Δ H 0 – T Δ S 0 ( if not standard, Δ G = Δ H – T Δ S) q cal = C cal ( Δ T) G = H – TS q rxn = -q cal Δ G 0 rxn = (n prod Δ G 0 (prod)) -- (m react Δ G 0 (react)) Δ G= Δ G 0 + RT(lnQ) Δ G 0 rxn = --RT(lnK) at equilibrium Finding temperature at where a reaction becomes spontaneous: 1 Using the information given, determine if the reaction is spontaneous or not at the given temperature. This will be handy later. 2 We know that a given reaction is spontaneous when Δ G<0. Δ G becomes less than zero JUST after it is equal to zero. So therefore, we will solve for Δ G=0 (i.e. 0= Δ H—T Δ S). 3 Rearrange that equation to give you T= Δ H/ Δ S and solve for T. 4 Use your information from step 1 to find your appropriate symbol. (You found the equilibrium temp., and it has to be either higher or lower than that temp. Let’s say your initial temp was 100K and it was non- spontaneous. Your equil. temp was 400K. Since 100 was NOT spontaneous and it’s less than 100, you know the rxn is spontaneous when T>400.) Rate: For rate of one reactant: --1/a * ( Δ {A}/ Δ t)
For rate of one product: + 1/c * ( Δ {C}/ Δ t) Generic A+B C+D: --1/a * ( Δ {A}/ Δ t)= --1/b*( Δ {B}/ Δ t)=1/c*( Δ {C}/ Δ t)=1/d* ( Δ {D}/ Δ t)=Rate Rate Law: (for above rxn): Rate=k[A] m [B] n Integrated Rate Laws: Zero Order: [A] t =[A] 0 —kt First Order: ln[A] t =ln[A] 0 —kt 2 nd Order: (1/[A] t )=(1/[A] 0 )+kt Half-life: t 1/2 = (.693/k) Arrhenius: k = A e^(--E a /RT)

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