DOC PREVIEW
U-M CHEM 260 - Exam 2 Key

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Exam 2 Key1. a) Linear Combination of Atomic Orbitals to form Molecular Orbitalsb) Energy is conservedc) The entropy of the universe (or a isolated system) is constantly increasing.d) A function that is independent of pathe) State functions: E,H,S Not state functions: w,q2. a) N2 = He2(1σ)2(2σ*)2(1π)4(3σ)2NO = He2(1σ)2(2σ*)2(1π)4(3σ)1 O2 = He2(1σ)2(2σ*)2(1π)4(3σ)2(2π*)2C2 = He2(1σ)2(2σ*)2(1π)4F2 = He2(1σ)2(2σ*)2(1π)4(3σ)2(2π*)4CN = He2(1σ)2(2σ*)2(1π)4(3σ)1b) Electric dipoles : O→N ; N→Cc) Stabilized by addition of an electron: C2, CN Stabilized by loss of an electron: F2, O2, NO3. Entropy∆Svap = ∆Hvap/ Tb∆Hvap = ∆Svap Tb = (85J/molK)(332K)(1mol) = 28220 JEnthalpy∆H=∆E+p∆V (constant pressure)∆E==∆H-p∆Vfrom here there is more than one possibility:#1: ∆E = ∆H-∆nRT (constant temp and # of moles for vaporization), so∆E = ∆H-nRT = 28220 – (1mole)(8.31J/molK)(332K) = 28220J -2759J = 25.46kJ#2 ∆E==∆H-p∆V∆E==∆H-p(Vf-VI) ==∆H-p(nRTf/p – 0) since compared to the gas, the volume ofthe solid is negligible. So, ∆E =(28220J)-(1mole)(8.31J/molK)(332K) = 25.46kJ#3 ∆E=q+wSince q = ∆H at constant pressure, and w = -pv, ∆E = ∆H –(p∆v) p∆v= 2759J, so ∆E = 28220J-2759J = 25.46kJWorkw = -pv = -nRT =(1mole)(8.31J/molK)(332K) = +2759J (work done on thebromine is positive, by the bromine is negative)Heat addedq = ∆H at constant pressure, so q = 28220JOr, this same result can be obtained by ∆E=q+w = 25.46kJ + 2.759kJ = 28.22kJ4. a) To sublimate diamond the reaction is: C(s) (diamond) Å C(g) 1 mole, at 298 K∆Hf(reaction) = ∆Hf(products) – ∆Hf(reactants)∆Hf(reaction) = (716.68 kJ/mol *1 mol) – (1.895 kJ/mol * 1mol)= 714.785 kJ b) Set up reaction pathway that we can use to calculate the reaction of interestWhere ∆HD = ∆HA + ∆HB + ∆HC∆HA = nCp(C(graphite)) ∆T = 1 mol * 8.527 J/K-mol*(298 K- 473 K) = -1499 J∆HB = n(∆Hsub) = n(∆Hf(products) – ∆Hf(reactants)) = 1 mol (∆Hf (C(diamond)) – ∆Hf(C(graphite)) = 1 mol*(1.895 kJ/mol - 0)= 1.895 kJ∆HC = nCp(C(diamond)) ∆T = 1 mol*6.133 J/K-mol*(473 K - 298 K) = 1073 J∆HD = -1499 J + 1.895*103 J + 1073 J = 1469 J = 1.47 kJTherefore the enthalpy of this reaction is 1.47 kJc) Need to determine at what temperature the enthalpy of this reaction is equal to zero.Use the same set-up as part b, but replace ∆HD = 0 and 473 K with x and solve:Where ∆HD = ∆HA + ∆HB + ∆HC∆HD= nCp(C(graphite))∆T + 1 mol (∆Hf (C(diamond)) – ∆Hf(C(graphite)) + nCp(C(diamond))∆T0 = 1 mol [8.527 J/K-mol*(298 K- x ) + 1.895 kJ/mol - 0) + 6.133 J/K-mol*(x - 298 K)]0 = 2541 – 8.527x+1895+6.133x – 1827 x = 1089 KThe final temperature is 1089 Kd) Again you need to set up a pathway DC(s) (graphite) at 473 K Å C(s) (diamond) at 473 KACC(s)(graphite) at 298 K Å C(s) (diamond) at 298 K B DC(s) (graphite) at 273 K Å C(g) at 1273 KACC(s)(graphite) at 298 K Å C(g) at 298 K BWhere ∆SD = ∆SA + ∆SB + ∆SC∆SA = nCp(C(graphite)) ln( Tf/Ti) = 1 mol * 8.527 J/K-mol ln (298K/273K) = .747 J/K∆SB = n(∆Hsub)/T = 1 mol*(716.68 kJ/mol - 0)/298 K = 2404 kJ/K∆SC = nCp(C(g)) ln(Tf/Ti) = 1 mol*20.833 J/K-mol*ln (1298 K/298 K) = 30.256 J/K∆SD = .747 J/K + 2404 J/K + 30.256 J/K = 2440 J/K = 2.44 kJ/KTherefore the entropy for this process is 2.44 kJ/molExtra credit: The easy way to do this problem is to recognize that ∆S = q/T at constanttemperature, and at constant pressure ∆S = ∆H/T. For entropy to be zero, we could haveT=0 K, or ∆H = 0, the temperature where this occurs was calculated in part C = 1089 K.Problem 5The key to this problem is to recognize that Cp is not constant with changingtemperature. This means that ∆H = Cp∆T and ∆S = Cpln(Tf/Ti) are both invalid sincethey assume that the Cp is constant. The correct way is as


View Full Document

U-M CHEM 260 - Exam 2 Key

Download Exam 2 Key
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Exam 2 Key and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Exam 2 Key 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?