Practice Problems 7 KEY

Practice Problems 7 KEY - CHEMISTRY 1414 Spring 2011...

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Unformatted text preview: CHEMISTRY 1414 Spring 2011 Practice Problem Set 7 Question 1 How much energy (in kJ) must be transferred to raise the temperature of a 250 mL cup of coffee from 20.5 °C to 95.6 °C? Assume that water and coffee have the same density (1.00 g/mL) and heat capacity (4.184 J/g-°C). - C m AT AT:01§.(,°L~L0.§°L‘~7S.I°L 44‘ lo W” M70”! : 7‘ a at: Lid“ j/7~°c >‘ L507 “ 75vl°L ‘ 731m” 3’: 7%:«1’ Question 2 Sucrose (table sugar) can be burned to give carbon dioxide and water according to the following equation: C12H22011(s) + 12 02 (g) 9 12 C02 (g) + 11 H20 (1) Aern = - 5645 kJ/mol. M w c 3‘1L.‘so 7mm What is the enthalpy change if 7.50 g of sugar is burned? ‘7'“)? (ILHLLDh/(L ): Orozvlol Mol 541.30? A”: _§U1§kj)LD.DZ|a|/v\0lt — ll” “3— Mol Question3 V MW; ’2’?" “Wu. The combustion of 1.500 g of benzoic acid (C6H5C02H) in a bomb calorimeter results in a change in temperature from 22.50 °C to 31.69 °C. Calculate the change in energy in k] per mole of benzoic acid for this process if the heat capacity of the calorimeter is 893 J/°C. 5(a): (MOM/«AW *AT 1., ‘ M? 7/‘1 V “W °c ’ 82m to? 3 AE= “XV : WU um”? ‘— - 0.1!“) LSOOj ((H’SLOL“( M“ x: 0,0)11%Mn| (Lilla) AT : m0; °c~ usow : TWC Ag: 351W; : ‘éhbflm/W) ‘_—.——_ D. D NAB/«d Question 4 Mixing sulfuric acid H2804 and water is highly exothermic. To measure the enthalpy change, 5.20 g of concentrated H2804 was added to 135 g of water in a coffee-cup calorimeter. The initial temperature was 20.2 °C and the final temperature was 28.8 °C. Calculate the enthalpy change in k] per mole of sulfuric acid for this process if the heat capacity of the final solution is 4.22 J/g°°C. 4F:Crmm.~AT AT=1%,%°L-Lo-z°c = +3.5°L exam-Mm! mm”: W‘HLTOKJr MHLO 21140.17 «(.1411 34.6 1 No.1? #7561 I 30%}: = 5.0% U AH: JW 1 "5,001 a] £107 Him )‘~ (LOUD/m qhvt‘fic) t ‘ 9001‘“? h : ~@ .0 m . logic/M (7 A“ “‘50” Questions In a coffee-cup calorimeter, 0.0552 g of magnesium metal is treated with 125 mL of 1.10 M hydrochloric acid. The reaction that occurs is: Mgo) +1HC1(aq) 9 Hug) + MgC12(aq) ‘/ BWLLI‘ __ The temperature changes from 22.21 °C to 24.46 °C. Calculate the enthalpy change for this process in kJ per mole of magnesium, given that the heat capacity of the solution is 4.20 J/g-°C and the density of the final solution is 1.15 g/mL. - D { M—VS‘Q \(Ov‘ M Gaiuc’ MINA kWh/1’ A MM] 7%} I I N°T 4 ’I'Ml‘Tiflfi Mam- «u = L.» Mm» AT AT= Him ~ u“ L ‘ “\rold f" ,‘ » P’LUGIM. .IUML le‘cthL: “1'17 01?: Hui/7.x ,wm *2.1S°C = '240 T r 136” 4H: -0”). ~l-3l,k) Mel ‘\ O‘O§§11(‘L“l-3‘) )-. 0.0011§7MO\4H=EI.3L1J 0.00115Mu + Z-L§°( Mlle/inc : ~ 43310/ mo! My Question 6 A 125 mL aliquot of a 0.250 M CsOH solution is mixed with 50.0 mL of a 0.625 M HF solution in a coffee-cup calorimeter. The temperature of both solutions rises to 24.40 °C from 21.50 °C before mixing. What is the enthalpy of the reaction per mole of CsOH if the density of the final solution is 1.12 g/mL and the heat capacity of the solution is 4.21 J/g-°C. The reaction that occurs is: CSOH(aq) + HF(aq> 9 CSF(aq) + H200) \/ BA Mme! 4T: lW’iO‘I ~ Ll-SD °L= +2.30 " C "ism; ‘~ LlZSAnL + 50.0»1L)7~l")'?/ ML ‘16: ‘1.Ll7/7-.§Cx< 151';9 >t Ilium :10”) j = 7397.3 -. 1&0. a: [ll—l: “tr: ~ 1.301 m mow CrOHe O-llgLsLo .Uom = 0.03:3M.‘ AHHLZMB : -7175 W/Mu CsOH 003122“ Question 7 The enthalpy changes for the three following reactions have been determined: 1) Nug) + 3H2 (g) 9 2NH3(g) AHrxn =—91.8kJ I) 4NH3 (g) + 5 02 (g) 9 4N0(g) + 6 H20 (9 Aern = -906.2 k] 3) Hug) + 1/2 02(3) —> H20 (g) AH,xn =-241.8 kJ Use these data to calculate the enthalpy change for the reaction NW) + 02 (g) -> 2N0cg) Add I) + U Nu“ + 3 HLM) ‘0 L/Vl‘lam ‘l' Ll’UHMq) +8~07~wn ‘3 L1 NOW) l- 6HL0L1) MM [: Ll “Ham; l gotta!) 1‘ N104) + “hm ‘3 ZMHal7,th0t1)+éHL9m AH=‘<1%.0‘47 ) ZNHW) LSOzm Ham NHL”, 9 HMO”, +4149”) 41],: 23321390- Hd I): ’Ulm + 3NM) ’3) lA/Ilw) m [LA/'74)”) + sou“ ‘+ 2JUL“) 1‘ “711”? Ll'WV’I) *6 Hme lulu; m AH : ‘ '03” 7 0 w 49w 6% S Lm) +LNufi) 4— (NM?) ‘9 [hi/om) #6 “you” AH 5 _ '0“.ka Mm» AH yoL‘q)+LNLL‘H +éHim +6HLOm —0‘4N0‘5]+6I~ILDW)+-élhtq) +301qu ,1.g~1o.gb\j Mum Jr DLUH 41 NO”) 0' H: 41.703 “a f Question 8 Given the following data, calculate AH for the following reaction: P4010 (s) +6PC15 (g) '9 10 C13PO (g) NUo‘ T0 8 4 Met a). P46) + 6c12® —> 41301”) AH=-1225.6kJ b). P46) + 502(3) 9 P4010“) AH=-2967.3 kJ c).PC13(g) + chm) —> PC15(g) AH=-84.2 kJ d).PC13(g) + V20”) 9 Cl3PO (g) AH=-285.7 kJ -AMIO‘A) + AtVWQ B lbx(P(’3L01)+'/LDLU{\~)Cl3f0(‘i)) )OX~L9)S.7KJ + PHD“: ‘A fit“) + SOLO“ + 1‘167’3 5L3 \1\ UPC [31 ‘3”) + YOL“) 1k P‘fio‘oltl “‘7 [0 0300b?) Jr F‘lm +§01W 4H: “0'”: 3‘ ‘3 (OC'sPOm) +PVm Album?“ Io um;th Palm r Um) ID x Mun [ox/Hum M - All-WEB W] 16) c P C) F [Wu ‘L Woman + P£1§¢Cfl ‘7 lochpom +Plus) {‘ [DPL'3(9)+{OUU1] \( (D\\ 40H 5 gm 1 ’0 (broil) P‘Hr) 4’ '0 ([104) A : {1'3 '4 P‘1\)I + ()CIZlq) \) A} M P o 3 f‘tol H PC fi _ [NJ on) 0 \YM) +“\ [O Cl3P0H)4 b1 chm + 4,1 Paw” AH — 2.73.310 MM L \ (3*y‘ W 0 Pg \0(:\ 7" '0 “Huh + ‘1 FUN“ + Wimp) [OCl3PD 5 {)w DINA + A rug”) ‘3 ID (lg/)0“) 4H Afll‘éIOJKI :‘éIDJL‘j V \ ...
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Practice Problems 7 KEY - CHEMISTRY 1414 Spring 2011...

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