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For all phase changes the sign of Δ S and Δ H is always going to be the same

For all phase changes the sign of Δ S and Δ H is always going to be the same Sublimation, melting, vaporization = + Freezing, deposition, condensation = - Correct answer is Δ S > 0. Vapor pressure increases as IMF’s get weaker Ranking IMF:

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For all phase changes the sign of Δ S and Δ H is always going to be the same

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  1. For all phase changes the sign of Δ S and Δ H isalways going to be the same Sublimation, melting, vaporization = + Freezing, deposition, condensation = - Correct answer is ΔS > 0

  2. Vapor pressure increases as IMF’s get weaker Ranking IMF: Ionic strongest  H-bond dipole dipole London Dispersion Forces What does each have? BeO – Ionic Ethyl Amine( CH3CH3NH2) - H-Bonding Benzene - (C6H6) CsF- ionic C3H6O (acetone)- dipole-dipole N2- LDF We have two ionic salts so you rank them based on charge density BeO has much large charge density so therefore will have a lower vapor pressure Correct answer is #2

  3. So if the ΔH lattice is larger ( always a positive number) than ΔH hydration (always a negativenumber) then which is true? So if ΔH lattice is a large positive number and ΔH solution is a smaller negative number than ΔH solution is going to be positive and endothermic So salt can dissolve if temperature of the solution is increased.

  4. Triple point is intersection of three phases. This can be solid liquid gas or three different phases it doesn’t matter So the triple point here is at 200K and 50 atm. Answer choice #2

  5. So container is at 150 atm and at 25 degrees C. Convert to K which is a classic mistake and thus we are at 300K and 150 atm which is in the Liquid area of the graph which you will have to label. So we are at star one and when the container is opened the pressure is relieved (violently). The liquid would vaporize (something like a propane tank). L S 1 G 2

  6. Question type is : calculating delta H from heating across phase transitions. So we have ice steam and water all together We need to bring them all in the same phase to figure our how we can find what the equilibrium temp is. First lets see how much energy it takes to melt the 80 g of ice and get 160 g of water at 0 degrees C q=m ΔH fusion  (80g) (334 J/g) =26720 joules of heat Now we have 160 grams of water at 0 degrees C on one side of the equation On the other side we have 16 g of steam. Lets take this 26720 joules of heat and use it to cool down the 120 degree steam to hot water at 100 degrees C So first we cool down the steam to 100 degrees C steam. q=m CsteamΔT  (16g)(2.08 J /g C) ( 20 C)  665.6 J Now lets turn this 100 degrees C steam to 100 degrees C water q=m ΔH vap (100g) (2256 J/g) = 36160 J Now on the cooling down steam side we have 36160 + 665.6 J = 36825.6 J On the water side we have 26720 joules. Therefore if you take the difference between the water and steam you have 36825.6 J – 26720 J= 10105.6 J of heat that will heat up the water. q=m CwaterΔT = 10105.6 = (160g)(4.184 J/gC)(X) X =15.09 degrees C Finally we have the same phase now we just have to do an equilibrium problem. We have on one side 160 g water at 15.09 degrees C and on the other we have 16g water at 100 degrees C m CwaterΔT = m CwaterΔT (160 g)(4.184 J/g C)(T eq - 15.09 )=(16g)(4.184 J/gC)(T eq - 100) Solve for T eq= 176 grams water at 22.8 degrees C

  7. Solubility of gas is exothermic therefore it is inversely proportional to temperature. Also solubility of gas is directly proportional to pressure. N2 is a non-polar molecule and there fore a less polar solvent would help solubility So the answer is I and III

  8. Like dissolves like so octane which is non-polar will dissolve in non-polar media Therefore the most polar will be least soluable for gasoline Oil/ fuel are going to be non-polar Wax is also non-polar Sugar is polar with hydrogen bonds Salt is the most polar –ionic So the answer is salt

  9. Temperature and pressure has exponential relationship #2 statement would raise the boiling point #3-#4 are correct

  10. We need to figure out molar mass So pressure= Moles/liter * R *temp Lets turn 1500 Pa to atm 1 atm= 101325 pa 1500 pa = .014 atm So .014 atm = X/.2L *0.0821 * 300 K Solve for x = .000114 moles Moles= Grams/molar mass .000114= 1/X  X is molar mass X= 8771.93 choice 2

  11. We want grams of salt i =2 for NaCl Molality = moles /kg solvent T= 4 4= 2 * X/2 *.512 X=7.81 We have 7.81 moles of NaCl To get grams we multiply by molar mass of NaCl which is = 58.44 456 grams so answer #4

  12. Increase , entropy

  13. Reactants over products Use solutions and gasses Therefore the correct answer is #4

  14. Kc= [H2][I2]/[HI]2 [H2] at eq is .233 thus [I2] must also be .233 Kc =.012 Rearrange equation and solve for [HI] [HI]2 =[H2][I2]/Kc [HI]=([H2][I2]/Kc) ^(1/2) [HI] = [(.233)(.233)/(.12)] ^(1/2) answer choice 5

  15. Kc= 4 Initial concs are NO = 5 initial  Change -X  eq 5-X CO2 = 10 initial change -X eq 10-X NO2 = 2 initial change+X  eq 2+X CO = 0 initial change +X  X

  16. Kc= 11.7 SO2, O2, SO3 = .015 Kc= [SO3] 2/[SO2]2 [O2] (.015)^2/(.015)^2 (.015) 1/.015 = 66.666 So Q > K reaction will shift to the left So answer choice is #3

  17. Le Chatelier says if we add product reaction will shift to the left to reestablish eq

  18. So this is an endothermic reaction So as T goes up K will go up So answer choice one is correct

  19. If K is a very small number or less than 1  delta G will be a positive number If K is a large number greater or than 1  delta G will be negative If K is 1 then delta G is 0 So answer choice 5 is impossible

  20. It will never be O

  21. Kw gets larger as T increases… so answer I is wrong Kw is 10 ^-14 at room temp this is correct [H+]=Kw/[OH-] so answer choice 3 is incorrect So only true answer is II

  22. [H+]=Kw/[OH-] 1.7x10^-7= 1x10^-14/X Solve for X X = 6.7X10^-8 Answer choice 4

  23. BaF2 [Ba2+] 2[F-] Ksp = X * (2X)2 Ksp = X * 4X2 Ksp= 4X3 2x10^-6 = 4X3 Solve for X X= .0079 mol Answer choice 4

  24. Just rank them based on Ksp Largest Ksp will be most soluble Thus the correct order is #3

  25. Weakest base here will be the strongest acid…. And the strongest acid will have the largest Ka D- will therefore be the weakest Base

  26. Ca=[H+] for strong acids Cb=[OH-] for strong bases

  27. Weak acids and bases [H+]= (CaKa)^(1/2) [OH-]= (CbKb)^(1/2) [H+]= (CaKa)^(1/2) [H+]= [(0.10)(1X10^-6)]^(1/2) [H+]= 3X10^-4 -log(3X10^-4) = 3.5 Answer #5

  28. Weak acids and bases [H+]= (CaKa)^(1/2) [OH-]= (CbKb)^(1/2) We have NaF which gives us F- in solution which is the conjugate base of HF So we need to get the Kb from HF’s Ka Kw/Ka=Kb 1X10^-14/7.2X10^-4)= kb Kb= 1.3x10^-11 [OH-]= (CbKb)^(1/2) [H+]= [(0.30)(1.3X10^-11)]^(1/2) [OH-]= 2 X 10^-6 -log(2X10^-6) =5.6 which is our pOH So pH = 14- 5.6= 8.3 Answer #1

  29. This is a buffer calc because we have B and BH+ Kb=1.8X10^-5 pKb= 4.74 4.74 + log(.3/.1)= pOH 4.74 +.38 =pOH 5.12= pOH 10^-5.12= [OH-] 7.5X10^-6 = [OH-]

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