MOOREHW8 - Chapter 6: Energy and Chemical Reactions 19

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Unformatted text preview: Chapter 6: Energy and Chemical Reactions 19 AIm:'ctt-:Etpln.llnn‘all.' Describe and explain your choice of the system and the surrotmdnigs. and describe 21 transfer of energy and materials into and out of the system. The system is identified as pi'eCtsely what we are snutying. The stu'rountitngs are everything: else The important aspects of the sturoriusdmgs are usually those things Ln contact with the system or in close proxinuty tat The System: The plant tstem. leaves. mots. etc] The Sm'ronndmgs: Anytlnng not the plant [ant soil. water. sun. etc.) ihl To study the plant gromng. we must isolate it and see [1011' it interacts with its stm'c-Lmdings. tct Light energy and eartnn die-tilde are absm‘bed by the leases and are converted to other molecules storing the energy as chenncal energy and using :1 to increase the size of the plant. Knit-tents are absofoed through the soil (innierals and water] to assist in the chemical processes. The plant expels oxygen and other waste materials into the sorrounclnigs I Reasonable .itmrer Check: This defnntton of re systern dlffft‘Enllal'fi between the live ot'gatnsm and the materials and energy required for It to stay aliye. Strate’gt and Elpt'anrtttnn: Describe and explain your choice of the system and the surroundings. describe transfer of energy and materials into and an of the system. and determine if the process 15 exothennic or endothennic. The system is identified as precisely what we are sutdying. The stnrotuidnigs are everything else The important aspects of the sturotuidnigs are usually those things Lfl contact with the system or in close proximity Ike prncess is exothelnnc ifthe system loses energy The process 1s endothermic tf the system grants energy. tat The System: N'HJC'l This choice was made following the discussion in Section IS 2' {page 220i when describing a reaction. the system ts usually defined as all the atoms that make up the reactants." Here the equation for the reaction being smdiecl is NHJCII. H —I- NHfI aq) * C'l‘taq‘l The Sturotmdntps: Anything! not NHJE']. tncludhig: the water The chance to consider water as part of the summoning-s instead of part of the system is based upon the fact that it is neither a reactant 13013 product in giyen the dissolying ecptatton Although water undeniably has a strong interaction With the products of the reaction. there are still no H30 molecules in this equation. Defniing the system to inchicle the water can also be done. Data collected from dassolt'uig experunents Will not actually be able to fitnctimally separate the water from the material dissohinst. ibt To study the release of energy during the phase change cfthis ionic compound. we must isolate it and see how it interacts with the surrcrundtings. Ic :- The system's interactton With the stn'rcundings causes heat energy to be transferred into the surroundings and out of the system. There is no maternal transfer in this process. but there is a change 111 the speed}: interaction between the water and system. tdt Since changes in the system cause energy to be gained by the system. the process is endothermic v’ Reawtmbfe AIL-me: C‘fteck' This definition of the system is restrlctrye enough to allow us to learn more about the relationship between the energy required to break the ionic bonds in the solid and the nmy' inyolyed with the pruhtcts' ntcrea sen interaction With the stare-trifling solvent mnlecules 4.1 Attnrer.‘ 5.00 x 105 J 53. .S‘tt'areg'. and Explananon: then the mass and tuttial temperatute of a liquid substance. lite bothttg point of the liquid. the final temperature of the gaseous SLIlJSlflIlE‘E. the specific heat capamty and the enthalpy of t'apot'tzattou. determine the energy (tn Joules: requued to complete the tratttittan. Rean'angtng Equation 6.2. determtne the heat energy requn'ed to mite the tenperauue to the boiling pomt Then use the etuhalpy of yapznrizatimt 13111313] to detenhine the energy required to boil the hqtuct at the benhng point. Define the tystem as the liquid benzene. To change the temperature ofthe system. ute q = c x tn 1 .5T qT-me = tbenzene I “haulene 3" AIMEE IGUD a =tl."rt-Jg_1:C*11xt'lOllkg'tx 11 ~ “301 "CeEO.U=C‘I=lEI5:-c1IJ-‘TJ 'e To change a phate in the tystetn. use q = tn a AH“? 1000 a “lull = alternate“; 3" AH.-¢_Mzm{ = l 1.00 kg}: 1 kg“ x [5951g)=3.95. x inf-J The sum gu‘efi the total heat energy tequh-ecl gm z qhm -qbcd=tl.05 HlCISJIJ*l3.951<lOL-'Ji' 5.00 x105] V Reasonable Armrest Check: To make the final product. the ltquid' 5 tetnperattn'e needed to be caused. and then the ltqtud needed to be yapouzed Both of these etmtges require energy to he added to: the ay'atem A?.LS‘lt'€?'.' (a) Bits} {h} the enthalpy of fusiun let positive Stt'rttegt and £1}: t'man'ott. {at The line on the graph with the steepest vertical slope has the largeet temperature ehange 1when a fixed amount of heat energy is added The substantial temperatute tnrtease means the l.pet'ltlte heat capaelty of the substance 11] that phase is smaller than u]. the other [313363 Smee the slope of the line teptesentmg the heating of the Wltd ts steeper than the slopes of the hues. rept'etenttng the heatrug of the Liquid or the gas It“ has the largest specific heat capacity [b5 The hunzuntai lines on the graph represent phaae ti'attaittuns The longest horizontal lute on the graph inchcate'. that more energy 1% tequned for that phase namitmn than the other clues Since the line representing the melting: of the éD-lttl tcn ltqtnd it shmtet than the [me renew-min? the yapotizatim efthe liquid to gas. the enthalpy nffuttot] is smaller than the enthalpy of vaporization. Tlm is not sutptishtg. runce the t'apmizaticut of a liquid tequit'e‘a enough energy to eyet‘eotue the all the attt'aetmns between the molecules in the ltqtud state: u'het'eaa the melting procees only requh'ea mine of the attractn'e fcnreea be overcome We always find that the entlmlpy effusion 15 smaller than the enthalpy of t'atnrizatiotl. I'm The algebratc sign of the enthalpy of yap-etnzatton 1‘: always posittye. Suite heat energy it tequit'ed to overcome all of the attractions between the molecules in the ltqutti state. 81. .trtr‘u'ct'.‘ (I) 1.4 x 10‘ J transferred (other answers are possible depending on the choice of system) [b] — 42 In! Srrnrcgt and Exp-human: Given the mass of a soluble tonic solid and the yohnne and temperanrre of a sample of water detennine the heat energy t1'a11sfe1 hunt the syste111t'o the sun'oundings and the enthalpy change {AI-{'31 by calculating the enerfi change per mol of tonic solid. We can define the system as the Natle. as described in Question 11 and 111 Section 6.1 {page 220). with AHmsohm: = —qmm at constant pressure Use the temperature of the sun'otuidmgs [the water solution: and the specific heat capacity of water solution to calculate qwmh. First. look up the spectftc heat capac1ty ot'water. Rearrangmg Eqnahon 6.3. ‘ini’r heat energy gamed by the u'ater. Heat energy from the reaction 15 used to raise the temperature of the water. so relate the heat energy gamed by the water to that lost in the reaction. hHmwhug. anlly. dn'u‘te the 311mm” by the mass of the ionic compound and convert grams to moles. Iai- The specific heat capacity [c1 cf water is 4.184 I g“ 3C“ according to Table 6 1. For at temperature change. rearrange Equation 6.1 q = L x m x AT. 1flle water represents the pan of the sutrotmdrngs affected by it change in the system 131' c 13' | In (line: = I [limite1 x T I . IFOJEC—IIfi'C-BI‘C {warm _ 1.'.I1 Her “'1 [E1 1311' I ‘1'ka qnw-{tteu 9-1 :C—Itx 140-3 0mm x 11'“; a- 131 I“) z 1.4 s: 10‘ J transfeited t‘mm the system to the stn'roundmgs Ibl Heat energy 1s lost by the reaction. so (minivan. is negatlye Heat energy is gained by the water. so qmmm! 1s posrtiye. The quantity of heat energy gained by the water is rec-(timed by the leacnon 314111.111 qdrssolt'lng _ qwat-et The reaction DCCIII‘S at constant pressure and there 1': no work done 80 AH I Chi-“mm. 1.11:1 1111.11 of NaOH _ —14.c10J J 39997111 NaCIH 111-3 kJ hH- : 1-: ~_ 1 -_4: _ 13.0 g Nam-I 1 met REVOH 1000 J r1101 3125011- U‘ Rmsormbie Answer Check: The ionization of an icenc cumlpound iin'olt'es segttratimg the canons and amcns. then hydrating them. The enthalpy of this change cauld conceivably be posmt‘e or negatn'e. but II is expect that it Will be small compared to reactions tt'here name significant bond rearrangement is lmppening. —42 kJ mol is smaller {closer to zero] than the rest of the AH“ yalnes for other kinds of reacnons recently studied. Note that the cmlntg of the solute 111 the process has been neglected. This may or111aynot be flppt'npliflrt‘. It would be better if we had the specific heat capacity of the solution. then he could use the entire kins-mt mass of the solution. rather than just the water Notice. A yet-y cotlnnon. name—empirical approach to sntymg tl11s question 1s to idennfy the water and the salt as the system With that detnnnon for the system dEgfim = cumming v (tum. If that 1s the case. the answer to the questicn 111 {21" “'1” be different. tat If The system is insolated. 11E 1 0. so there is no transfer oi energy Etom the system to the surmtmdnrgs. (1:13 qmmhmg = 11mm. Measuring the system's temperature clmnge indicates a gain of thermal energy by the water and qwm: 1s poutis'e. Since the reactmn occurs at the same time and no heat energy escapes the ulsolated system. it preyes that the dissolving reaction produces energy. is ‘Leactlm negattye. and the reaction. 1s exothermic The numerical result is the smne as pl’D't‘tdBd aboye. ...
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This note was uploaded on 06/03/2008 for the course CHEM 1A taught by Professor Nitsche during the Fall '08 term at University of California, Berkeley.

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MOOREHW8 - Chapter 6: Energy and Chemical Reactions 19

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