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chm_122_mt_2s

# chm_122_mt_2s - CHM 122 2“d Midterm KEY<61 Last Name...

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Unformatted text preview: CHM 122 2“d Midterm KEY [<61 Last Name First Name Student ID (Please Print) — — — — — — — IF, NO WORK 0R REASONING IS SHOWN, THEN, N0 CREDIT WILL BE GIVEN All answers must be M. You may use the backside of a page or request additional pages from the instructor for additional room, but all answers must be on the front of each page and circled. Signature CHM 122- 2nd Midterm Problem 1. (15 Points) I would like you to make a buffer of pH = 10.00. The reagents you have available are NH4C1 and NH3. How would you make this buffer? Kb(NH3) = 1.8x10'5 -17 Pkg + 4009 £64m) , K : 1%, : M35553: ;- susx/ "‘__.'—" Q lrg'X/(fJ CACMA) ( [4b ® ._ [56147) [0:00 ' 9.626 2’: 46:9 MW!) 60 ____ "‘- 515 049 3 ‘10; 4542335 “’7 cm ' ' Problem 2. (15 Points) You have a buffer that has twice as much acid, HA, as conjugate base, A'. If Ka for this acid is 3.5x10'6, what is the pH of this buffer solution? If I titrate one-third (1/3) of your conjugate base ~ with a strong acid> what is the new pH of your buffer solution? [all-#3 1‘ 0'2 ~lo‘ 0 CHM 122- 2“d Midterm ' .- Problem 3. (20 Points) How much energy is needed to change 10.00 gr. of water from solid ice at m1 00°C to gaseous water at 200.0°C? (Hints: MW (H20) m 18.0 gI./mole & watch your units) AHqu(H20) = 6.01 kJ/mole, AHvaP(H20) = 40.7 kJ/Inole. S(H20(s)) = 2.10 J/(gT_°c), 3(H20 (1)) : 4_18 J/(gnoc), whom) = 201 mger +9 .th Q N Q2. by d 03 m] 0%; ")5 “#9 MM” ‘7 Mr“ #3 7 CE" WWW 2 f? a: . He 1‘ (91741.2 a Hear ‘7 F6 gm [00 c n q t’ I00 c “’0 C 0%. 0 a . ﬁe; ' _ T a n— .r .? #em" éiolx/035/[10mO ) r.— BISVX/OJ 16.x? \ C231 3( \0 . __. ' EJ Q Q3 :(Ll,/3ﬁ](I0IO€;/j(loop/ ___.. LII/3X10 610006 ‘77 -- - 3 [0.00 ,— _—; 52.26 x20 Qt] “'- (90:4X/0 ._— I” Q‘Olyloff Problem 4. (15 Points) Find AHo for F60 (S) + CO (g) 9 Fe (3) + CO; (g) given the followin. (I) F6203(s) + 3c0 (g) —> 2Fe(s) + 3c02 (g) A °=-2s.o kJ @ 313.2103 (3) + co (g) —-> 2F6304(s) + co2 (g) AH°=—59.0 H (33 13.3304 (5) + (:0 (g) a 3FcO(s) + c02 (g) AH°=+38.0 k1 3‘ “Take 3%) “3&203 + ?Co “7 éFQ + (7604. AH I “576.01c7‘ To kill F2103 x x #0.. +3909; ‘ ' 0. 3F 0. 4C0 O ’ ' ' r0 I K a“. v 4- ; rake -Q,(® a * 6F€0+3COJ +‘fco 76FQ+§CO+WOJL 4": ’0’“- “O 6 520(5) + Q cow” M7 6 Fats? + écow/ 0H: “lo/iv Dam/18 ___7 F4200) 1" CO €77 'W'"? FQIS/ + Cog’ﬁ/ We CHM 122. 2nd Midterm Problem 5. (15 Points) Calculate the standard entropy change at 25“C for the following reaction: CH3CH20H (I) + 02 (g) 9 CH3COOH (l) + H20 (1) Given that the standard entropies of the substances in J/(K mol) at 25°C are: CH3CH20H (l) = 161.0, 02 (g) = 205 .0, CH3COOH (l) 2 160.0 and H20 (1) = 69.9 A\$ .3 2: n (15 (fax/j -- 4, in ésti’vecﬂ @ (/wio + (\$7.7) -« (Hp/,0 1* 5205/0) Problem 6. (20 Points) a) Calculate the value of the thermodynamic equilibrium constant at 25°C for the reaction N204 (g) ‘3’ 2 N02 (:9 Given that 256% of the substances in kJ/mol at 25°C are N0; = 51.30, N204 = 97.82 and that R = 8.314 J/(K mol). A60 3 ﬂ QGMPMN "‘ (TF4?ch --— a(§’r30/I :— 91?,31 £360 :qi?§]€7 5:62” : "KTan :7 9.?3x/037 '1" “(913/7/(298N07k) [-17 ‘3 “ if ? b) At what temperature does this reaction become spontaneous? :: AGO + KTAHK "*7 5"“. 777” W {6‘0 rim” 7‘0 me ﬁe ems» we)”. L/r'qlsxms + (3,3;9/(7) [120.2537 ":3 O ...
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chm_122_mt_2s - CHM 122 2“d Midterm KEY<61 Last Name...

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