Formulas and Constants (you may remove this page) NA = 6.0x0 3 mol - h = 6.66x0-34 J s c =.99x0 m s - e =.60x0-9 C me = 9.09x0-3 kg Å = x0-0 m = 00 pm atm = 760 torr R = 0.006 L atm K - mol - =.34 J K - mol - ~ ν = λ = hc λ = hcν~ [gas(aq)]eq = KH (Pgas)eq vpsolution = Xsolvent vppure solvent Tf = Kf m Tb = Kb m Π V = nrt Π = MRT ln [A] [A]o = kt [A] = [A]o ekt t½ = ln /k = 0.693/k t½ = [A]ok [A] = kt [A]o = a,f a,r ln k k = a R T T ln k k = a R T T k = A e a/rt IA A IIA A 3 IIIB 3B 4 IVB 4B 5 VB 5B 6 VIB 6B 7 VIIB 7B 9 VIII 0 IB B IIB B 3 IIIA 3A 4 IVA 4A 5 VA 5A 6 VIA 6A 7 VIIA VIIIA 7A A H.00 3 4 Li Be 6.94 9.0 Na Mg.99 4.3 9 0 K Ca 39.0 40.0 37 3 Rb Sr 5.47 7.6 55 56 Cs Ba 3.9 37.3 Sc 44.96 39 Y.9 57 La* 3.9 Ti 47. 40 Zr 9. 7 Hf 7.5 3 V 50.94 4 Nb 9.9 73 Ta 0.9 4 Cr 5.00 4 Mo 95.94 74 W 3.9 5 Mn 54.94 43 Tc (9) 75 Re 6. 6 Fe 55.5 44 Ru 0. 76 Os 90. 7 Co 5.47 45 Rh 0.9 77 Ir 90. Ni 5.69 46 Pd 06.4 7 Pt 95. 9 Cu 63.55 47 Ag 07.9 79 Au 97.0 30 Zn 65.39 4 Cd.4 0 Hg 00.5 5 B 0. 3 Al 6.9 3 Ga 69.7 49 In 4. Tl 04.4 6 C.0 4 Si.09 3 Ge 7.59 50 Sn.7 Pb 07. 7 N 4.0 5 P 30.97 33 As 74.9 5 Sb. 3 Bi 09.0 O 6.00 6 S 3.07 34 Se 7.96 5 Te 7.6 4 Po (0) 9 F 9.00 7 Cl 35.45 35 Br 79.90 53 I 6.9 5 At (0) He 4.003 0 Ne 0. Ar 39.95 36 Kr 3.0 54 Xe 3.3 6 Rn ()
CH45 Chemistry 45 Practice Test Prof. Shattuck Name R =.34 J mol - K - = 0.006 L atm mol - K - T(0 C) = 73.5 K Part : Answer of the following 0 questions. If you answer more than cross out the one you wish not to be graded, otherwise only the first will be graded. points each.. In the following pairs, which substance has the higher boiling point: a. CH6 or C6H4? more electrons b. HS or HO? hydrogen-bonding c. CO or NO? NO has a dipole moment. Which of these statements about benzene is true? A. All carbon atoms in benzene are sp 3 hybridized. B. Benzene contains only π-bonds between C atoms. _* C. The bond order of each C C bond in benzene is.5. D. Benzene is an example of a molecule that displays ionic bonding.. All of these statements are false. 3. From the following information calculate the Keq for reaction 3. Reaction Ag NH3 = Ag(NH3) K =.7x0 3 M - Reaction Ag(NH3) NH3 = Ag(NH3) K =.x0 7 M - Reaction 3 Ag NH3 = Ag(NH3) Keq = K K =.7x0 0 4. Give the hybridization for NH3 sp 3 and for SO _sp. 5. For the reaction NO CO NO CO the experimental rate law is: rate = k [NO] The following mechanism has been proposed: NO NO3 NO NO3 CO NO CO a. Which step is the rate determining step? Slow step is the rate detemrinining step step b. Is there a reactive intermediate in this reaction? If so what is it? NO3 (not a reactant or product) 6. For a first order reaction, the concentration of the reactant dropped from 0.00 M to 0.00 M in 6.00 minutes. How long does it take for the concentration to drop from 0.00 M to 0.000 M? Answer: The first time interval is for [A] = [A]o/, that is one half-time, t½= 0.693/k = 6.00 min. The rate constant is then k = 0.55 min -. The second time interval is: ln([a]/[a]o) = ln(0.000 M/0.00 M) = -.996 = (0.55 min - ) t or t = 5.9 min
CH45 3 7. The rate law for a third order reaction is [A] [A]o = kt To make a straight line plot to verify third order behavior, a. what do you plot on the vertical axis? /[A]. b. what do you plot on the horizontal axis? t (time). c. what is the slope equal? slope = k. The osmotic pressure of a solution of a protein in water is.54 torr at 5.0. The solution contained 0.700 g of protein per liter of solution. Calculate the molar mass of the protein. Answer: In atmospheres, Π =.54 torr/760 torr/atm =.06x0-3 atm. From Π = MRT the molar concentration is M = Π/RT =.06x0-3 atm/0.006 L atm K - mol - /9.5 K =.x0-5 mol L -. The molar mass is then MM = 0.700 g/.x0-5 mol =.45x0 3 g mol -. 9. Which molecule has a stronger bond, CF or F? xplain your answer for credit. CF F σ * u (p z) σ * u (p z) π * g (p x) π * g (p y) π * g (p x) π * g (p y) p C πu (p x) σg (p z) πu (p y) p F p F πu (p x) p σg (p F z) πu (p y) σ * u (s) σ * u (s) s C s N σg (s) s C s N σg (s) CF C CF F F FF F Bond order: BO = 3 = ½ BO = 6 = stronger bond in CF (greater bond order) 0. Determine the rate law for the reaction NO (g) Br(g) NOBr(g) from the following initial rate study (you don t need to get a numerical value for k, just leave it as k ):
CH45 4 [NO] (mol L - ) [Br] (mol L - ) rate (L mol - sec - ).00.00.00x0-6.00.00.00x0-6.00.00 4.00x0-6 Answer: xperiments &: when [NO] doubles the rate doubles the order with respect to NO is one. xperiments &3: when [Br] doubles the rate quadruples the order with respect to Br is two: rate = k [NO][Br] If it were requested: For experiment, to get the rate constant rate = k [NO][Br] or:.00x0-6 l mol - s - = k (.00 mol L - )(.00 mol L - ) giving k =.00x0-6 L 3 mol -3 s - Part. Answer 3 of the following 5 questions. If you answer more than 3 cross out the ones you wish not to be graded, otherwise only the first 3 will be graded. points each.. (a.) Draw an energy level diagram for the molecular orbitals for the CN - ion. (b.) Label each orbital with the type of orbital, sigma, or pi and bonding or anti-bonding. (c.) Fill the levels with the proper number of electrons, and (d.) calculate the bond order. CN σ * u (p z) π * g (p x) π * g (p y) p C πu (p x) σg (p z) πu (p y) p N σ * u (s) s C s N σg (s) CN C CN N Bond order: BO = = 3. NOBr decomposes according to NOBr (g) NO (g) ½ Br (g) With Kp = 0.5. If.0 atm of NOBr, 0. atm of NO, and 0.4 atm of Br are mixed, will any reaction occur? If a net reaction is observed, will NO be formed or consumed?
CH45 5 Answer: Q = PBr½ PNO PNOBr reverse direction. = (0.4 atm)½ (0. atm) atm = 0.506 with Q > Kp the reaction runs in the 3. What is the molarity of 50.0% by weight NaOH solution? The density of the solution is.53 g ml -. (Molar Mass(NaOH) = 40.0 g mol - ) Answer: Assume 00 g of solution, which then contains 50.0 g of NaOH or.5 moles of NaOH. The volume of the solution is Vsoln = 00.0 g/dsoln = 00. g/.53 g ml - = 65.36 ml or 0.06536 L. The final molarity is c = (moles solute)/vsoln = 9. mol L - 4. The rate constant for the following reaction is.05x0-4 sec - at 76 K. The rate constant increases to 7.x0-4 sec - at 34 K. Calculate the activation energy for the reaction: C4HO C4H6 HO Answer: ln k = a k R giving T T ln 7.x0-4.05x0-4 a =.34 J K - mol - 34. K 76. K.05 = a/.34 J K - mol - (-7.3x0-5 ) a = 9. kj mol - 5. NOBr decomposes according to NOBr (g) NO (g) Br (g) With Kp =.5. If.00 atm of NOBr is placed in a constant volume container, calculate the equilibrium pressure of Br. Answer: NOBr (g) NO (g) Br (g) initial.00 atm 0 0 change x x x equil.000 x x x Kp = PNO PBr PNOBr = x x =.5 cross multiplying gives: x =.5.5 x or x.5 x.5 = 0 solving: x = b ± b 4ac a =.5 ± (.5) 4(-.5) =.5 ± 3.636 = 0.743 atm