ÌâÄ¿ÄÚÈÝ

£¨14·Ö£©»¯Ñ§ÔÚÄÜÔ´¿ª·¢ÓëÀûÓÃÖÐÆð×ÅÊ®·ÖÖØÒªµÄ×÷Óá£
£¨1£©Ô̲ØÔÚº£µ×µÄ¿Éȼ±ùÊǸßѹÏÂÐγɵÄÍâ¹ÛÏñ±ùµÄ¼×ÍéË®ºÏÎï¹ÌÌ壮±»³Æ֮Ϊ¡°Î´À´ÄÜÔ´¡±¡£ÔÚ25¡æ¡¢101 kPaÏ£¬1g¼×ÍéÍêȫȼÉÕÉú³ÉºÍҺ̬ˮʱ·ÅÈÈ55.6 kJ¡£¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ ______£ºÏàͬÌõ¼þÏ£¬356 g¿Éȼ±ù£¨·Ö×ÓʽΪCH4¡¤9H2O£¬Mr£½178£©Êͷŵļ×ÍéÆøÌåÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³öµÄÈÈÁ¿Îª_______kJ¡£
£¨2£©¶þ¼×ÃÑ(CH3OCH3)ÊÇÎÞÉ«ÆøÌ壬¿É×÷ΪһÖÖÐÂÐÍÄÜÔ´£¬¾ßÓÐÇå½à¡¢¸ßЧµÄÓÅÁ¼ÐÔÄÜ¡£ÓɺϳÉÆø(×é³ÉΪH2¡¢COºÍÉÙÁ¿µÄCO2)Ö±½ÓÖƸ÷¶þ¼×ÃÑ£¬ÆäÖеÄÖ÷Òª¹ý³Ì°üÀ¨ÒÔÏÂËĸö·´Ó¦£º
¼×´¼ºÏ³É·´Ó¦£º
£¨¢ñ£©CO£¨g£©+2H2£¨g£©¨TCH3OH£¨g£©  ¡÷H1£½£­90.1kJ?mol-1
£¨¢ò£©CO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£© ¡÷H2£½£­49.0kJ?mol-1
ˮúÆø±ä»»·´Ó¦£º£¨¢ó£©CO£¨g£©+H2O£¨g£©¨TCO2£¨g£©+H2 £¨g£© ¡÷H3£½£­41.1kJ?mol-1
¶þ¼×ÃѺϳɷ´Ó¦£º£¨¢ô£©2CH3OH£¨g£©¨TCH3OCH3£¨g£©+H2O£¨g£©¡÷H4£½£­24.5kJ?mol-1
¢Ù·ÖÎö¶þ¼×ÃѺϳɷ´Ó¦(iv)¶ÔÓÚCOת»¯ÂʵÄÓ°Ïì___________________________________¡£
¢ÚÓÉH2ºÍCOÖ±½ÓÖƱ¸¶þ¼×ÃÑ£¨ÁíÒ»²úÎïΪˮÕôÆø£©µÄÈÈ»¯Ñ§·½³ÌʽΪ£º__________________¡£¸ù¾Ý»¯Ñ§·´Ó¦Ô­Àí£¬·ÖÎöÔö¼Óѹǿ¶ÔÖ±½ÓÖƱ¸¶þ¼×ÃÑ·´Ó¦µÄÓ°Ïì_________________________________¡£
£¨3£©¶þ¼×ÃÑÖ±½ÓȼÁϵç³Ø¾ßÓÐÆô¶¯¿ì¡¢Ð§ÂʸߵÈÓŵ㡣Èôµç½âÖÊΪ¼îÐÔ£¬¶þ¼×ÃÑÖ±½ÓȼÁϵç³ØµÄ¸º¼«·´Ó¦Îª______________________£¬Ò»¸ö¶þ¼×ÃÑ·Ö×Ó¾­¹ýµç»¯Ñ§Ñõ»¯£¬¿ÉÒÔ²úÉú________µç×ӵĵçÁ¿¡£
£¨1£©2O2£¨g£©+CH4£¨g£©¨TCO2£¨g£©+H2O£¨l£© ¡÷H£½£­889.6kJ/mol   1779.2
£¨2£©¢ÙÏûºÄ¼×´¼£¬´Ù½ø¼×´¼ºÏ³É·´Ó¦£¨¢ñ£©Æ½ºâÓÒÒÆ£¬COת»¯ÂÊÔö´ó£»Éú³ÉµÄH2O£¬Í¨¹ýˮúÆø±ä»»·´Ó¦£¨¢ó£©ÏûºÄ²¿·ÖCO
¢Ú2CO£¨g£©+4H2£¨g£©£½CH3OCH3£¨g£©+H2O£¨g£© ¡÷H£½£­204.7kJ?mol-1
¸Ã·´Ó¦·Ö×ÓÊý¼õÉÙ£¬Ñ¹Ç¿Éý¸ßʹƽºâÓÒÒÆ£¬COºÍH2ת»¯ÂÊÔö´ó£¬CH3OCH3²úÂÊÔö¼Ó£®Ñ¹Ç¿Éý¸ßʹCOºÍH2Ũ¶ÈÔö¼Ó£¬·´Ó¦ËÙÂÊÔö´ó
£¨3£©CH3OCH3£«16OH£­£­12e£­£½2CO32£­£«11H2O£»12

ÊÔÌâ·ÖÎö£º£¨1£©1g¼×ÍéÍêȫȼÉÕÉú³ÉºÍҺ̬ˮʱ·ÅÈÈ55.6 kJ£¬Ôò16g¼×Íé¼´1mol¼×ÍéÍêȫȼÉÕÉú³ÉºÍҺ̬ˮʱ·ÅÈÈ55.6 kJ¡Á16£½889.6kJ£¬Òò´Ë¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ2O2£¨g£©+CH4£¨g£©¨TCO2£¨g£©+H2O£¨l£© ¡÷H£½£­889.6kJ/mol¡£356 g¿Éȼ±ù£¨·Ö×ÓʽΪCH4¡¤9H2O£¬Mr£½178£©Öм×ÍéµÄÖÊÁ¿ÊÇ356g¡Á£½32g£¬¼´º¬ÓÐ2mol¼×Í飬ËùÒÔÊͷŵļ×ÍéÆøÌåÍêȫȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³öµÄÈÈÁ¿Îª889.6kJ/mol¡Á2mol£½1779.2kJ¡£
£¨2£©¢Ù¶þ¼×ÃѺϳɷ´Ó¦£¨¢ô£©¶ÔÓÚCOת»¯ÂʵÄÓ°Ï죬ÏûºÄ¼×´¼£¬´Ù½ø¼×´¼ºÏ³É·´Ó¦£¨¢ñ£©CO£¨g£©+2H2£¨g£©¨TCH3OH£¨g£©Æ½ºâÓÒÒÆ£¬COת»¯ÂÊÔö´ó£»Éú³ÉµÄH2O£¬Í¨¹ýˮúÆø±ä»»·´Ó¦£¨¢ó£©CO£¨g£©+H2O£¨g£©¨TCO2£¨g£©+H2 £¨g£©ÏûºÄ²¿·ÖCO£¬¹Ê´ð°¸Îª£ºÏûºÄ¼×´¼£¬´Ù½ø¼×´¼ºÏ³É·´Ó¦£¨¢ñ£©Æ½ºâÓÒÒÆ£¬COת»¯ÂÊÔö´ó£»Éú³ÉµÄH2O£¬Í¨¹ýˮúÆø±ä»»·´Ó¦£¨¢ó£©ÏûºÄ²¿·ÖCO£»
¢Ú¢ñ¡¢CO£¨g£©+2H2£¨g£©¨TCH3OH£¨g£©¡÷H1=-90.1kJ?mol-1¡¢¢ô¡¢2CH3OH£¨g£©¨TCH3OCH3£¨g£©+H2O£¨g£©¡÷H4=-24.5kJ?mol-1£¬ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢ñ¡Á2+¢ôµÃµ½£º2CO£¨g£©+4H2£¨g£©=CH3OCH3+H2O£¨g£© ¡÷H£½£­204.7kJ?mol-1¡£¸Ã·´Ó¦ÊÇÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬Ôö¼ÓѹǿƽºâÕýÏò½øÐУ¬·´Ó¦ËÙÂÊÔö´ó£¬COºÍH2ת»¯ÂÊÔö´ó£¬CH3OCH3²úÂÊÔö¼Ó£»¹Ê´ð°¸Îª£º2CO£¨g£©+4H2£¨g£©£½CH3OCH3£¨g£©+H2O£¨g£©¡÷H£½£­204.7kJ?mol-1£»¸Ã·´Ó¦·Ö×ÓÊý¼õÉÙ£¬Ñ¹Ç¿Éý¸ßʹƽºâÓÒÒÆ£¬COºÍH2ת»¯ÂÊÔö´ó£¬CH3OCH3²úÂÊÔö¼Ó£®Ñ¹Ç¿Éý¸ßʹCOºÍH2Ũ¶ÈÔö¼Ó£¬·´Ó¦ËÙÂÊÔö´ó£»
£¨3£©Ô­µç³ØÖиº¼«Ê§È¥µç×Ó£¬Ôò¶þ¼×ÃÑÔÚ¸º¼«·¢ÉúÑõ»¯·´Ó¦¡£Èôµç½âÖÊΪ¼îÐÔ£¬¶þ¼×ÃÑÖ±½ÓȼÁϵç³ØµÄ¸º¼«·´Ó¦Îª¶þ¼×ÃÑʧµç×ÓÉú³É̼Ëá¸ù£¬½áºÏÔ­×ÓÊغãºÍµçºÉÊغãд³öµç¼«·´Ó¦Îª£ºCH3OCH3£«16OH£­£­12e£­£½2CO32£­£«11H2O£» ¸ù¾Ý¸º¼«µç¼«·´Ó¦Ê½¿ÉÖªÒ»¸ö¶þ¼×ÃÑ·Ö×Ó¾­¹ýµç»¯Ñ§Ñõ»¯£¬¿ÉÒÔ²úÉú12¸öµç×ӵĵçÁ¿¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
(15·Ö)£¨1£©ÒÑÖªÔÚ³£Î³£Ñ¹Ï£º
¢Ù 2CH3OH(l)+3O2(g)  2CO2(g)+4H2O(g)  ¡÷H=" -1275.6" kJ¡¤mol£­1
¢Ú H2O(l)     H2O(g)                   ¡÷H=" +" 44.0 kJ.mol£­1
д³ö±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ                               ¡£
£¨2£©¼×´¼ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÓÖÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓпª·¢ºÍÓ¦ÓõĹãÀ«Ç°¾°¡£
·½·¨Ò»
CO(g) +2H2(g) CH3OH(g)
·½·¨¶þ
CO2(g) +3H2(g) CH3OH(g) +H2O(g)
¹¤ÒµÉÏ¿ÉÓÃÈçÏ·½·¨ºÏ³É¼×´¼£º
     
·½·¨Ò»                                      ·½·¨¶þ
¢Ù·½·¨Ò»£º¸Ã·´Ó¦µÄ¡÷S      0£¨Ìî¡°£¾¡±»ò¡°£¼¡±£©¡£Í¼ÖÐÇúÏßaµ½ÇúÏßbµÄ´ëÊ©ÊÇ           
               £¬ºãκãÈÝʱ£¬ÏÂÁÐ˵·¨ÄÜ˵Ã÷·´Ó¦µ½´ïƽºâ״̬µÄÊÇ            ¡£
A.ÌåϵµÄƽ¾ùĦ¶ûÖÊÁ¿²»Ôٸı䠠              B. V(CO)= V(CH3OH)
C. H2µÄת»¯ÂÊ´ïµ½ÁË×î´óÏ޶Ƞ                D. ¡÷H²»Ôٸıä
¢Ú·½·¨¶þ£º½«CO2ºÍH2°´ÎïÖʵÄÁ¿Ö®±È1:3³äÈëÌå»ýΪ2.0LµÄºãÈÝÃܱÕÈÝÆ÷Öз´Ó¦£¬ÈçͼÁ½ÌõÇúÏß·Ö±ð±íʾѹǿΪ0.1 MPaºÍ5.0 MPaÏÂCO2ת»¯ÂÊËæζȵı仯¹Øϵ£¬ÆäÖÐaµãµÄƽºâ³£Êý±í´ïʽΪ£º             £»a£¬bÁ½µã»¯Ñ§·´Ó¦ËÙÂʱðÓÃVa¡¢Vb±íʾ£¬ÔòVa       Vb£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©¡£ ÒÑÖªÔ­×ÓÀûÓÃÂÊ=ÆÚÍû²úÎïµÄ×ÜÖÊÁ¿ÓëÉú³ÉÎïµÄ×ÜÖÊÁ¿Ö®±È£¬Ôò·½·¨Ò»µÄÔ­×ÓÀûÓÃÂÊÊÇ·½·¨¶þµÄÔ­×ÓÀûÓÃÂʵĠ           ±¶(±£ÁôÁ½Î»Ð¡Êý).
£¨3£©¼×´¼¶ÔË®ÖÊ»áÔì³ÉÒ»¶¨µÄÎÛȾ£¬ÓÐÒ»Öֵ绯ѧ·¨¿ÉÏû³ýÕâÖÖÎÛȾ£¬ÆäÔ­ÀíÊÇͨµçºó½«Co2£«Ñõ»¯³ÉCo3£«£¬È»ºóÒÔCo3£«×öÑõ»¯¼Á°ÑË®Öеļ״¼Ñõ»¯³ÉCO2¶ø¾»»¯¡£Ð´³ö³ýÈ¥¼×´¼µÄÀë×Ó·½³Ìʽ              ¡£ 
£¨16·Ö£©ÄòËØ¿É×÷ΪH2O2µÄÎȶ¨ÔØÌ壬Éú²úÒ»ÖÖ¹Ì̬¡¢ÄÍ´¢´æ¡¢Ò×ÔËÊäµÄÐÂÐÍÑõ»¯¼ÁºÍÏû¶¾¼Á¡ª¹ýÑõ»¯ÄòËØ[CO(NH2)2¡¤H2O2]£¬ÆäºÏ³É¹¤ÒÕÁ÷³ÌÈçÏ£º

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷I¡¢IIµÄÃû³Æ·Ö±ðÊÇ______¡¢_______¡£CO(NH2)2¡¤H2O2·Ö×ÓÖÐÄòËغ͹ýÑõ»¯ÇâÖ®¼äÒÔ________½áºÏ£¬Æä½áºÏÁ¦½ÏÈõ£¬¾ßÓÐÄòËغ͹ýÑõ»¯ÇâË«ÖØÐÔÖÊ¡£
£¨2£©¹¤ÒµÉÏÉú²úÄòËØËùÐèµÄÔ­ÁÏÆø¿ÉÓÉÌìÈ»ÆøÓëË®·´Ó¦ÖƱ¸£¬ÒÑÖª£º
¢Ù¼×Íé¡¢ÇâÆøµÄȼÉÕÈÈ·Ö±ðΪ890£®3KJ/mol¡¢285£®8kJ/mol
¢Ú
д³öCH4ÓëË®ÕôÆø×÷ÓÃÉú²úCO2ºÍH2µÄÈÈ»¯Ñ§·½³Ìʽ£º_______________¡£
£¨3£©ºÏ³É¹ýÑõ»¯ÇâÄòËؼÓÈëµÄÎȶ¨¼Á¿ÉÒÔÊÇË®ÑîËá¡¢¾ÆʯËáµÈ¡£¾ÆʯËá·Ö×ÓʽΪC4H6O6£¬ÆäºË´Å¹²ÕñÇâÆ×Ö»ÓÐ3Öַ棬ֻº¬ôÈ»ùºÍôÇ»ùÁ½ÖÖ¹ÙÄÜÍÅ£¬Ôò¾ÆʯËáµÄ½á¹¹¼òʽΪ_____¡£
£¨4£©Îª²â¶¨²úÆ·ÖÐH2O2µÄº¬Á¿£¬³ÆÈ¡¸ÉÔïÑùÆ·12£®0gÅä³É250mLÈÜÒº£¬È¡25£®00mL
ÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÊÊÁ¿ÁòËáËữ£¬ÓÃ0£®20mol/LKMnO4±ê×¼ÈÜÒºµÎ¶¨£¬Èý´ÎµÎ¶¨Æ½¾ùÏûºÄKMnO4ÈÜÒº20£®00mL£®£¨KMO4ÈÜÒºÓëÄòËز»·´Ó¦£©£®
¢ÙÍê³É²¢Åäƽ·½³Ìʽ£ºMnO4-+    H2O2+   _________=Mn2++ O2¡ü+___£»
¢Ú¼ÆËã³ö²úÆ·ÖÐH2O2µÄÖÊÁ¿·ÖÊýΪ_______¡£
£¨5£©µç½âÄòËصķÏË®¼È¿ÉÒÔ´¦Àí·ÏË®£¬ÓÖ¿ÉÖƵô¿Ç⣬µç½âÔ­ÀíÈçͼËùʾ¡£µç½â³ØÖиôĤ½ö×èÖ¹ÆøÌåͨ¹ý£¬BÁ½¼«¾ùΪ¶èÐԵ缫¡£B¼«Á¬½ÓµçÔ´µÄ____¼«£¨Ìî
¡°Õý¡±»ò¡°¸º¡±£©£¬Ñô¼«µç¼«·´Ó¦·½³ÌΪ_____________¡£
£¨2014½ìºÓÄÏÊ¡ÖÐÔ­ÃûУ¸ßÈýÏÂѧÆÚµÚÒ»´ÎÁª¿¼»¯Ñ§ÊÔ¾í£©
ÑÇÏõËáÄÆÊÇÒ»ÖÖ¹¤ÒµÑΣ¬ÔÚÉú²ú¡¢Éú»îÖÐÓ¦Óù㷺¡£ÏÖÓÃÏÂͼËùʾÒÇÆ÷£¨¼Ð³Ö×°ÖÃÒÑÊ¡ÂÔ£©¼°Ò©Æ·£¬Ì½¾¿ÑÇÏõËáÄÆÓëÁòËá·´Ó¦¼°Éú³ÉÆøÌå²úÎïµÄ³É·Ö¡£ÒÑÖª£º
¢ÙNO£«NO2£«2OH-£½2NO2-£«2H2O
¢ÚÆøÌåÒº»¯µÄζȣºNO2  21¡æ¡¢NO £­152¡æ

£¨1£©ÎªÁ˼ìÑé×°ÖÃAÖÐÉú³ÉµÄÆøÌå²úÎÒÇÆ÷µÄÁ¬½Ó˳ÐòΪ£¨´Ó×óÏòÓÒÁ¬½Ó£©£ºA¡úC¡ú_______¡ú_______¡ú_______£»×é×°ºÃÒÇÆ÷ºó£¬½ÓÏÂÀ´½øÐеIJÙ×÷ÊÇ________________¡£
£¨2£©¹Ø±Õµ¯»É¼Ð£¬´ò¿ª·ÖҺ©¶·»îÈû£¬µÎÈë70£¥µÄÁòËáºó, AÖвúÉúºì×ØÉ«ÆøÌå.
¢ÙÈ·ÈÏAÖвúÉúÆøÌ庬ÓÐNO£¬ÒÀ¾ÝµÄÏÖÏóÊÇ_____________________________.
¢Ú×°ÖÃEµÄ×÷ÓÃÊÇ_______________________________________________________
£¨3£©Èç¹ûÏòDÖÐͨÈë¹ýÁ¿O2,Ôò×°ÖÃBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________________.
Èç¹ûûÓÐ×°ÖÃC£¬¶ÔʵÑé½áÂÛÔì³ÉµÄÓ°ÏìÊÇ______________________________¡£
£¨4£©Í¨¹ýÉÏÊöʵÑé̽¾¿¹ý³Ì£¬¿ÉµÃ³ö×°ÖÃAÖз´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ_______________¡£
£¨5£©¹¤ÒµÉú²úÖеªÑõ»¯ÎïµÄÅÅ·Å»áÔì³É»·¾³ÎÛȾ£¬¿É²ÉÓÃÈçÏ·½·¨´¦ÀíµªÑõ»¯Î
CH4£¨g£©£«2NO2£¨g£©£½N2£¨g£©£«CO2£¨g£©£«2H2O£¨g£©   ¡÷H£½£­867kJ¡¤mol-1
CH4£¨g£©£«4NO£¨g£©£½2N2£¨g£©£«CO2£¨g£©£«2H2O£¨g£©   ¡÷H£½£­1160kJ¡¤mol-1
ÔòCH4½«NO2»¹Ô­ÎªNOµÄÈÈ»¯Ñ§·½³ÌʽΪ£º___________________________.

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø