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Practice Test 4

# Practice Test 4 - CEM 383 — Practice Test 4 Fall 2007...

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Unformatted text preview: CEM 383 — Practice Test 4 Fall 2007 1. (15 points) Consider the equilibrium 1 HgO(s) <:> Hg(g)+ 502(5)) If 1.0 g of solid HgO is placed in a ﬂask, evacuated , sealed and allowed to come to equilibrium, show that 2 W 31.5 total K: assume all gases are ideal ( MHg = 200 g/mole, M02 = 16 g/mole). 2. (20 points) Bicarbonate ion is an extremely poor acid with Ka = 4.8 x 10‘“. HCO;(aq) ﬁ H+(aq>+ CO§'(aq) Calculate the pH of a 0.01M solution of NaZCO3 in water. . (20 points) Using the dissociation equilibria for phosphoric acid given below, What volume of 12.0 M HCl would have to be added to 1.0 L of 0.1 M K3PO4 to buffer the solution at pH = 11.8? H2P0;(aq) (:9 H+(aq) + Hpoj‘mq) pK; = 7.21 HPO§‘(aq) ﬁ H+(aq) + Poi—(aq) pKa” = 12.32 Vol. HCl = . (25 points) a) The overall reaction in a lead storage battery involves a 2—electron oxidation of Pb(s) coupled to a 2—electron reduction of Pb02(s) in the presence of sulfuric acid to yield PbSO4(s). The overall reaction is Pb(s) + P1702 (s) + 2H + (aq) + 2HSO; (aq) (2) 2PI9SO4 (s) + 2H20(l) Calculate E, the cell voltage or emf, for this battery at 25 °C when [H] = [HSO4'] = 4.5 M. At 25 °C, E° = 2.04V for the lead storage battery. E: b) For the overall cell reaction, Afr" = —315kJ / mole and A3" = 263.5] / (K-mole). Calculate E for the cell at —20°C when the [H+] = [HSO4'] = 4.5 M. 5. (20 points) The decomposition of silver oxide is given by 2Ag20(s) <=> 4Ag(s) + 02(g) Given that Aﬁ" = 62.1kJ/mole and AS" = 132.81/(K-mol) at 298K, determine the temperature where the equilibrium pressure of 02(g) is 0.2 bar. Assume that the enthalpy and entropy of the reaction are independent of temperature. CEM 383 Eguation Sheet PV= nRT 17: V/n PV Ptotalz 1:;“13 3 Pi =)(iPtotal Z = E _ RT a _ P- _ :7, 2Xi=1 Z—1+BP+CP2+DP3+ _ 2 "1 z=ﬂ=1+3(;j+c(;) +14; ”total RT V V V Etransz 3nRT/2: Crms 2 3—K! V M — 8RT _ 2RT c= — cmp — —— 7rM M _ N Z N W247) 31(7) ,1:_Z_ i: 92 Z] r2 M1 w = _ Pex,dV w = -nRTln (Vz/VI) AU: q + w H = U+ PV 8U 8H Cv = _ C = _ [871]"), p (8T )Pm AUZCV(T2—T1) AH=Cp(T2—T1) Cp—Cv=nR AH=AU+A(PV) AU = qv = -Cbomb(T2 — T1) P1Vly = P2V2y Argo = 2vaf—Ho(pr0d)—ZvrAfEUO'eac) prod reac A,HT2 = A,HT1 + A,C,,(T2 — T1) A,H s ZBEUeact) _ 213m prod) As=qm AS=ann£ T VI T T AS=CPln—2 efﬁciency= 1——1-=|—1| T1 T2 ‘12 T1 ql C.O.P. = = — Ava = AS + ASSW 2 0 T2—.Tl w [Aﬂsli AmS=—RmAmxg+nﬂnXﬂ AMS= 7~ f A H T2 Amp 5 = ”P AS = —dCP T Tb T1 T __ _ _ T2 C A, S” = szo (prod) — ZVSOUeact) AS = —TKdT n T T AS=CVln—2 A,52 =A, SI+Ar CP 111—2— 1 1 AG = AH — TAS AA = AU — TAS dG = -SdT + VdP dA = -SdT — PdV 45" = vaAf 5" (prod) — Zv,AfE"(reac) P A? T P A}? T — T AG=nRTln —2 1’2—P1=—_1n —2 1n —2 =— 2 1 1 AV T1 P1 R T1T2 _ _ _ a _ V=n1V1+n2V2 Vi=[—V] Hi=Gi=[-a—G] ani T,P,nj ani T,P,nj Ame = nRT(X11nXl + X2 1n X2) P1 = X113: 0 P. P. ym. y.[i] ‘= , RT] ‘ .= I: X_ .=——1 ,:—l—l—=l , =, :1 nu“: nut + naz at I): Y! 1 at lbar a! ma [i]0 01(S) at(l) 13. = log. = K'mi A7} = Kfm-i ATb = Kbm-i 72: = RTM-i MsolvRszlb - Kf,b=—-———= z=1—a+av v=v++v_ (1000g / kg)A,rH um = (v4.1: + v_,uf)+ RT In a; ai = hm: mi = (m:+ m} )1’ " A,E=Ar5"+RT1nQ Aﬁ” =—RT1nK ln(—K—2] = Arﬁ[T2 — T1] K1 R T1T2 aA+bB ﬁ cC+ dD c d aCaD K: a 12 “A03 0 d =PCPD a b PAPB KP C d c d c+d—a—b XCXD .Pc+d—a—b —— nCnD. 13mm] xaxb total _ a b A B ”AnB n C d c d m'chPD a b a b YAYB PAPB total = 72Y\$.[C]”[D]" : [C]C[D]" 727\$ [A]"[B]” " [A]“[B]” K=KYKP= K = 1(ch KW :10” = KaKb = [H+][0H‘] pH = —log[H+] pOH = —log[OH'] pH+pOH = 14 '1) pK = —logK pH = pKa + logk—Oﬂﬂ [aczd] —A 5” RTan A §°(T—298) E0=Eo —E0 E0: ' = E0=E0 +_r___. cathode anode VF VF T 298 VF T c d . 2 c d E=E”—§—ln “Ca” =E" —Mlnacal’ vF ajaB VF alga; ...
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Practice Test 4 - CEM 383 — Practice Test 4 Fall 2007...

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