ijÊжԴóÆø½øÐмà²â£¬·¢ÏÖ¸ÃÊÐÊ×ÒªÎÛȾÎïΪ¿ÉÎüÈë¿ÅÁ£ÎïPM2.5£¨Ö±¾¶Ð¡ÓÚµÈÓÚ2.5µÄÐü¸¡¿ÅÁ£Î£¬ÆäÖ÷ÒªÀ´Ô´ÎªÈ¼Ãº¡¢»ú¶¯³µÎ²ÆøµÈ¡£Òò´Ë£¬¶ÔPM2.5¡¢SO2¡¢NOxµÈ½øÐÐÑо¿¾ßÓÐÖØÒªÒâÒå¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½«PM2.5Ñù±¾ÓÃÕôÁóË®´¦ÀíÖƳɴý²âÊÔÑù¡£
Èô²âµÃ¸ÃÊÔÑùËùº¬Àë×ӵĻ¯Ñ§×é·Ö¼°ÆäŨ¶ÈÈçÏÂ±í£º

Àë×Ó
H+
K+
Na+
NH4+
SO42£­
NO3£­
Cl£­
Ũ¶È/mol¡¤L£­1
δ²â¶¨
4¡Á10£­6
6¡Á10£­6
2¡Á10£­5
4¡Á10£­5
3¡Á10£­5
2¡Á10£­5
 
¸ù¾Ý±íÖÐÊý¾ÝÅжÏÊÔÑùµÄpH=         ¡£
£¨2£©Îª¼õÉÙSO2µÄÅÅ·Å£¬³£²ÉÈ¡µÄ´ëÊ©ÓУº
¢Ù½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ¡£
ÒÑÖª£ºH2£¨g£©+1/2O2£¨g£©=H2O£¨g£© ¡÷H=£­241.8kJ¡¤mol£­1
C£¨s£©+1/2O2£¨g£©="CO" £¨g£©       ¡÷H=£­110.5kJ¡¤mol£­1
д³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º             ¡¡¡¡                 ¡£
¢ÚÏ´µÓº¬SO2µÄÑÌÆø¡£ÒÔÏÂÎïÖÊ¿É×÷Ï´µÓ¼ÁµÄÊÇ           ¡£
A£®Ca(OH) 2   B£®Na2CO3  C£®CaCl2D£®NaHSO3
£¨3£©Æû³µÎ²ÆøÖÐÓÐNOxºÍCOµÄÉú³É¼°×ª»¯
¢Ù Èô1mol¿ÕÆøº¬0.8molN2ºÍ0.2molO2£¬Æû¸×ÖеĻ¯Ñ§·´Ó¦Ê½ÎªN2 (g)+O2(g)2NO(g) ¡÷H0
1300¡æʱ½«1mol¿ÕÆø·ÅÔÚÃܱÕÈÝÆ÷ÄÚ·´Ó¦´ïµ½Æ½ºâ£¬²âµÃNOΪ8¡Á10£­4mol¡£¼ÆËã¸ÃζÈϵÄƽºâ³£ÊýK=               ¡£
Æû³µÆô¶¯ºó£¬Æû¸×ζÈÔ½¸ß£¬µ¥Î»Ê±¼äÄÚNOÅÅ·ÅÁ¿Ô½´ó£¬ÆäÔ­ÒòÊÇ           ¡£
¢ÚÄ¿Ç°£¬ÔÚÆû³µÎ²ÆøϵͳÖÐ×°Öô߻¯×ª»¯Æ÷¿É¼õÉÙCOºÍNOxµÄÎÛȾ£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ        ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡                    ¡£
¢Û Æû³µÈ¼ÓͲ»ÍêȫȼÉÕʱ²úÉúCO£¬ÓÐÈËÉèÏë°´ÏÂÁз´Ó¦³ýÈ¥CO£¬2CO£¨g£©=2C£¨s£©+O2£¨g£©
ÒÑÖª¸Ã·´Ó¦µÄ¡÷H0£¬ÅжϸÃÉèÏëÄÜ·ñʵÏÖ²¢¼òÊöÆäÒÀ¾Ý£º                   ¡£

£¨1£©PH=4
£¨2£©¢ÙC(s)+H2O(g)=H2(g)+CO(g) ¡÷H=+131.3kJ/mol ¢Ú  A  B
£¨3£©¢Ù4¡Á10-6£¬Î¶ÈÔ½¸ß£¬·´Ó¦ËÙÂÊÔ½´ó  ¢Ú2XCO+2NOX2XCO2+N2
¢Û²»ÄÜʵÏÖ£¬ÒòΪ¸Ã·´Ó¦µÄ¡÷H>0,¡÷S<0,ËùÒÔ¡÷G>0

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¡¾»¯Ñ§¡ª¡ªÑ¡ÐÞ2£º»¯Ñ§Óë¼¼Êõ¡¿£¨15·Ö£©
µÚÎåÖ÷×åµÄÁ×µ¥Öʼ°Æ仯ºÏÎïÔÚ¹¤ÒµÉÏÓй㷺ӦÓá£
£¨1£©Í¬Á×»ÒʯÔÚ¸ßÎÂÏÂÖƱ¸»ÆÁ×µÄÈÈ»¯Ñ§·½³ÌʽΪ£º
4Ca5(PO4)3F(s)+21SiO2(s)+30C(s)=3P4(g)+20CaSiO3(s)+30CO(g)+SiF4(g) H
ÒÑÖªÏàͬÌõ¼þÏ£º
4Ca3(PO4)2F(s)+3SiO2(s)=6Ca3(PO4)2(s)+2CaSiO3(s)+SiF4(g) ¡÷H1
2Ca3(PO4)2(s)+10C(s)=P4(g)+6CaO(s)+10CO(g) ¡÷H2
SiO2(s)+CaO(s)=CaSiO3(s) ¡÷H3
Óá÷H1¡¢¡÷H2ºÍ¡÷H3±íʾH£¬ÔòH=                   £»
£¨2£©Èý¾ÛÁ×Ëá¿ÉÊÓΪÈý¸öÁ×Ëá·Ö×Ó£¨Á×Ëá½á¹¹Ê½Èçͼ£©Ö®¼äÍÑÈ¥Á½
¸öË®·Ö×Ó²úÎÆä½á¹¹Ê½Îª                               £¬Èý¾Û
Á×ËáÄÆ£¨Ë׳ơ°ÎåÄÆ¡±£©Êdz£ÓõÄË®´¦Àí¼Á£¬Æ仯ѧʽΪ           £»

£¨3£©´ÎÁ×ËáÄÆ£¨NaH2PO2£©¿ÉÓÃÓÚ¹¤ÒµÉϵĻ¯Ñ§¶ÆÄø¡£
¢Ù»¯Ñ§¶ÆÄøµÄÈÜÒºÖк¬ÓÐNi2+ºÍH2PO2£­£¬ÔÚËáÐÔµÈÌõ¼þÏ·¢ÉúÏÂÊö·´Ó¦£º
£¨a£©      Ni2+ +      H2PO2£­+       ¡ú    Ni++      H2PO3£­+     
£¨b£©6H2PO-2 +2H+ =2P+4H2PO3+3H2¡ü     ÇëÔÚ´ðÌ⿨ÉÏд³ö²¢Åäƽ·´Ó¦Ê½£¨a£©£»
¢ÚÀûÓâÙÖз´Ó¦¿ÉÔÚËÜÁ϶Ƽþ±íÃæ³Á»ýÄø¡ªÁ׺Ͻ𣬴Ӷø´ïµ½»¯Ñ§¶ÆÄøµÄÄ¿µÄ£¬ÕâÊÇÒ»ÖÖ³£¼ûµÄ»¯Ñ§¶Æ¡£Çë´ÓÒÔÏ·½Ãæ±È½Ï»¯Ñ§¶ÆÓëµç¶Æ¡£
·½·¨ÉϵIJ»Í¬µã£º                                                     £»
Ô­ÀíÉϵIJ»Í¬µã£º                                                     £»
»¯Ñ§¶ÆµÄÓŵ㣺                                                       ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

(1)ij·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯ÈçͼËùʾ£º

д³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                                               ¡£
(2)0.3 molÆø̬¸ßÄÜȼÁÏÒÒÅðÍé(B2H6)ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5 kJµÄÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ                                            ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ºìÁ×P(s)ºÍCl2(g)·¢Éú·´Ó¦Éú³ÉPCl3(g)ºÍPCl5(g)¡£·´Ó¦¹ý³ÌºÍÄÜÁ¿¹ØϵÈçͼËùʾ(ͼÖеĦ¤H±íʾÉú³É1 mol²úÎïµÄÊý¾Ý)¡£¸ù¾ÝÏÂͼ»Ø´ðÏÂÁÐÎÊÌ⣺

(1)PºÍCl2·´Ó¦Éú³ÉPCl3µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                            £»
(2)PºÍCl2·ÖÁ½²½·´Ó¦Éú³É1 mol PCl5µÄ¦¤H3£½                   £¬PºÍCl2Ò»²½·´Ó¦Éú³É1 mol PCl5µÄ¦¤H4           ¦¤H3(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±¡¢»ò¡°µÈÓÚ¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨16·Ö£©¹¤ÒµºÏ³É°±ÓëÖƱ¸ÏõËáÒ»°ã¿ÉÁ¬ÐøÉú²ú£¬Á÷³ÌÈçÏ£º

£¨1£©¹¤ÒµÉú²úʱ£¬ÖÆÈ¡ÇâÆøµÄÒ»¸ö·´Ó¦Îª£ºCO(g)+H2O(g)CO2(g)+H2(g)¡£t¡æʱ£¬Íù10LÃܱÕÈÝÆ÷ÖгäÈë2mol COºÍ3molË®ÕôÆø¡£·´Ó¦½¨Á¢Æ½ºâºó£¬ÌåϵÖÐc(H2)=0.12mol¡¤L-1¡£Ôò¸ÃζÈÏ´˷´Ó¦µÄƽºâ³£ÊýK=    £¨Ìî¼ÆËã½á¹û£©¡£
£¨2£©ºÏ³ÉËþÖз¢Éú·´Ó¦N2(g)+3H2(g)2NH3(g) ¡÷H<0¡£Ï±íΪ²»Í¬Î¶Èϸ÷´Ó¦µÄƽºâ³£Êý¡£ÓÉ´Ë¿ÉÍÆÖª£¬±íÖÐT1   300¡æ£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£

T/¡æ
T1
300
T2
K
1.00¡Á107
2.45¡Á105
1.88¡Á103
 
£¨3£©°±ÆøÔÚ´¿ÑõÖÐȼÉÕÉú³ÉÒ»ÖÖµ¥ÖʺÍË®£¬¿Æѧ¼ÒÀûÓôËÔ­Àí£¬Éè¼Æ³É¡°°±Æø-ÑõÆø¡±È¼Áϵç³Ø£¬ÔòͨÈë°±ÆøµÄµç¼«ÊÇ         £¨Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±£©£»¼îÐÔÌõ¼þÏ£¬¸Ãµç¼«·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½Îª                      ¡£
£¨4£©Óð±ÆøÑõ»¯¿ÉÒÔÉú²úÏõËᣬµ«Î²ÆøÖеÄNOx»áÎÛȾ¿ÕÆø¡£Ä¿Ç°¿Æѧ¼Ò̽Ë÷ÀûÓÃȼÁÏÆøÌåÖеļ×ÍéµÈ½«µªµÄÑõ»¯ÎﻹԭΪµªÆøºÍË®£¬·´Ó¦»úÀíΪ£º
CH4(g)+4NO2(g)£½4NO(g)+CO2(g)+2H2O(g)  ¡÷H= £­574kJ¡¤mol£­1
CH4(g)+4NO(g)£½2N2(g)+CO2(g)+2H2O(g)   ¡÷H= £­1160kJ¡¤mol£­1
Ôò¼×ÍéÖ±½Ó½«NO2»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ                                       ¡£
£¨5£©Ä³Ñо¿Ð¡×éÔÚʵÑéÊÒÒÔ¡°Ag-ZSM-5¡±Îª´ß»¯¼Á£¬²âµÃ½«NOת»¯ÎªN2µÄת»¯ÂÊËæζȱ仯Çé¿öÈçͼ¡£¾Ýͼ·ÖÎö£¬Èô²»Ê¹ÓÃCO£¬Î¶ȳ¬¹ý775K£¬·¢ÏÖNOµÄת»¯ÂʽµµÍ£¬Æä¿ÉÄܵÄÔ­ÒòΪ                                   £»ÔÚn(NO)/n(CO)=1µÄÌõ¼þÏ£¬Ó¦¿ØÖƵÄ×î¼ÑζÈÔÚ      ×óÓÒ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

2013ÄêÎíö²ÌìÆø¶à´ÎËÁÅ°ÎÒ¹úÖж«²¿µØÇø¡£ÆäÖÐÆû³µÎ²ÆøºÍȼúβÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£
£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO(g)+2CO2CO2(g)+N2(g)
¢Ù¶ÔÓÚÆøÏà·´Ó¦£¬ÓÃij×é·Ö£¨B£©µÄƽºâѹǿ£¨PB£©´úÌæÎïÖʵÄÁ¿Å¨¶È£¨CB£©Ò²¿ÉÒÔ±íʾƽºâ³£Êý£¨¼Ç×÷KP£©£¬Ôò¸Ã·´Ó¦µÄKP=-                ¡£
¢Ú¸Ã·´Ó¦ÔÚµÍÎÂÏÂÄÜ×Ô·¢½øÐУ¬¸Ã·´Ó¦µÄ¦¤H            0¡££¨Ñ¡Ìî¡°>¡±¡¢¡°<¡±£©
¢ÛÔÚijһ¾øÈÈ¡¢ºãÈݵÄÃܱÕÈÝÆ÷ÖгäÈëÒ»¶¨Á¿µÄNO¡¢CO·¢ÉúÉÏÊö·´Ó¦£¬²âµÃÕý·´Ó¦µÄËÙÂÊËæʱ¼ä±ä»¯µÄÇúÏßÈçͼËùʾ£¨ÒÑÖª£ºt2 --tl=t3£­t2£©¡£

ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ         ¡££¨Ìî±àºÅ£©
A£®·´Ó¦ÔÚcµãδ´ïµ½Æ½ºâ״̬
B£®·´Ó¦ËÙÂÊaµãСÓÚbµã
C£®·´Ó¦ÎïŨ¶Èaµã´óÓÚbµã
D£®NOµÄת»¯ÂÊ£ºtl¡«t2>t2¡«t3
£¨2£©ÃºµÄ×ÛºÏÀûÓá¢Ê¹ÓÃÇå½àÄÜÔ´µÈÓÐÀûÓÚ¼õÉÙ»·¾³ÎÛȾ¡£ºÏ³É°±¹¤ÒµÔ­ÁÏÆøµÄÀ´Ô´Ö®Ò»Ë®ÃºÆø·¨£¬ÔÚ´ß»¯¼Á´æÔÚÌõ¼þÏÂÓÐÏÂÁз´Ó¦£º
¢Ù
¢Ú
¢Û
¢Ù¡÷H3ºÍ¡÷H1¡¢¡÷H2µÄ¹ØϵΪ¡÷H3=            ¡£
¢ÚÔÚºãÎÂÌõ¼þÏ£¬½«l mol COºÍ1 mol H2O£¨g£©³äÈëij¹Ì¶¨ÈÝ»ýµÄ·´Ó¦ÈÝÆ÷£¬´ïµ½Æ½ºâʱÓÐ50%µÄCOת»¯ÎªCO2¡£ÔÚtlʱ±£³ÖζȲ»±ä£¬ÔÙ³äÈë1 mol H2O£¨g£©£¬ÇëÔÚͼÖл­³ötlʱ¿ÌºóH2µÄÌå»ý·ÖÊý±ä»¯Ç÷ÊÆÇúÏß¡£

¢Û¼×´¼ÆûÓÍ¿É¡¯ÒÔ¼õÉÙÆû³µÎ²Æø¶Ô»·¾³µÄÎÛȾ¡£
ij»¯¹¤³§ÓÃˮúÆøΪԭÁϺϳɼ״¼£¬ºãÎÂÌõ¼þÏ£¬ÔÚÌå»ý¿É±äµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£ºCO(g)+2H2(g) CH3OH(g)µ½´ïƽºâʱ£¬²âµÃCO¡¢H2¡¢CH3OH·Ö±ðΪ1 mol¡¢1 mol¡¢1 mol£¬ÈÝÆ÷µÄÌå»ýΪ3L£¬ÏÖÍùÈÝÆ÷ÖмÌÐøͨÈË3 mol CO£¬´Ëʱv£¨Õý£©         v£¨Ä棩£¨Ñ¡Ìî¡®¡®>¡±¡¢¡°<¡¯¡¯»ò¡°=¡±£©£¬ÅжϵÄÀíÓÉ        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

̼ÊÇÐγɻ¯ºÏÎïÖÖÀà×î¶àµÄÔªËØ£¬Æäµ¥Öʼ°»¯ºÏÎïÊÇÈËÀàÉú²úÉú»îµÄÖ÷ÒªÄÜÔ´ÎïÖÊ¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÓлúÎïM¾­¹ýÌ«Ñô¹â¹âÕÕ¿Éת»¯³ÉN£¬×ª»¯¹ý³ÌÈçÏ£º
¦¤H£½£«88.6 kJ¡¤mol£­1
ÔòM¡¢NÏà±È£¬½ÏÎȶ¨µÄÊÇ               ¡£
(2)ÒÑÖªCH3OH(l)µÄȼÉÕÈÈΪ238.6 kJ¡¤mol£­1£¬CH3OH(l)£«O2(g)=CO2(g)£«2H2(g)  ¦¤H£½£­a kJ¡¤mol£­1£¬Ôòa       238.6(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)¡£
(3)ʹCl2ºÍH2O(g)ͨ¹ý×ÆÈȵÄÌ¿²ã£¬Éú³ÉHClºÍCO2£¬µ±ÓÐ1 mol Cl2²ÎÓ뷴ӦʱÊͷųö145 kJÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£º                       ¡£
(4)»ð¼ýºÍµ¼µ¯±íÃæµÄ±¡²ãÊÇÄ͸ßÎÂÎïÖÊ¡£½«Ê¯Ä«¡¢ÂÁ·ÛºÍ¶þÑõ»¯îÑ°´Ò»¶¨±ÈÀý»ìºÏÔÚ¸ßÎÂÏÂìÑÉÕ£¬ËùµÃÎïÖÊ¿É×÷Ä͸ßβÄÁÏ£¬4Al(s)£«3TiO2(s)£«3C(s)=2Al2O3(s)£«3TiC(s)¡¡¦¤H£½£­1 176 kJ¡¤mol£­1£¬Ôò·´Ó¦¹ý³ÌÖУ¬Ã¿×ªÒÆ1 molµç×ӷųöµÄÈÈÁ¿Îª         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖªH2(g)¡¢CO(g)ºÍCH3OH(l)µÄȼÉÕÈÈ·Ö±ðΪ285.8 kJ¡¤mol£­1¡¢283.0 kJ¡¤mol£­1ºÍ726.5 kJ¡¤mol£­1¡£Çë»Ø´ðÏÂÁÐÎÊÌâ¡£
(1)ÓÃÌ«ÑôÄÜ·Ö½â5 molҺ̬ˮÏûºÄµÄÄÜÁ¿ÊÇ     kJ¡£
(2)Һ̬¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ÆøÌåºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ            ¡£
(3)ÔÚÒÔ¼×´¼ÎªÈ¼ÁϵÄȼÁϵç³ØÖУ¬µç½âÖÊÈÜҺΪËáÐÔ£¬ÔòÕý¼«µÄµç¼«·´Ó¦Ê½Îª                                                                       ¡£
ÀíÏë״̬Ï£¬¸ÃȼÁϵç³ØÏûºÄ2 mol¼×´¼ËùÄܲúÉúµÄ×î´óµçÄÜΪ1 404.2 kJ£¬Ôò¸ÃȼÁϵç³ØµÄÀíÂÛЧÂÊΪ     ¡£(ȼÁϵç³ØµÄÀíÂÛЧÂÊÊÇÖ¸µç³ØËù²úÉúµÄ×î´óµçÄÜÓëȼÁϵç³Ø·´Ó¦ËùÄÜÊͷŵÄÈ«²¿ÄÜÁ¿Ö®±È)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¶þÑõ»¯Ì¼ÊÇÒ»ÖÖ±¦¹óµÄ̼Ñõ×ÊÔ´¡£ÒÔCO2ºÍNH3ΪԭÁϺϳÉÄòËØÊǹ̶¨ºÍÀûÓÃCO2µÄ³É¹¦·¶Àý¡£ÔÚÄòËغϳÉËþÖеÄÖ÷Òª·´Ó¦¿É±íʾÈçÏ£º
·´Ó¦¢ñ£º2NH3(g)£«CO2(g)NH2CO2NH4(s)   ¡÷H1="a" kJ¡¤mol£­1
·´Ó¦¢ò£ºNH2CO2NH4(s)CO(NH2)2(s)£«H2O(g)   ¡÷H2=£«72.49kJ¡¤mol£­1
×Ü·´Ó¦¢ó£º2NH3(g)£«CO2(g)CO(NH2)2(s)£«H2O(g)   ¡÷H3=£­86.98kJ¡¤mol£­1
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©·´Ó¦¢ñµÄ¡÷H1=__________kJ¡¤mol£­1(ÓþßÌåÊý¾Ý±íʾ)¡£
£¨2£©·´Ó¦¢òµÄ¡÷S______(Ì»ò£¼)0£¬Ò»°ãÔÚ__________Çé¿öÏÂÓÐÀûÓڸ÷´Ó¦µÄ½øÐС£
£¨3£©·´Ó¦¢óÖÐÓ°ÏìCO2ƽºâת»¯ÂʵÄÒòËغܶ࣬ÏÂͼ1ΪijÌض¨Ìõ¼þÏ£¬²»Í¬Ë®Ì¼±Èn(H2O)/n(CO2)ºÍζÈÓ°ÏìCO2ƽºâת»¯Âʱ仯µÄÇ÷ÊÆÇúÏß¡£
¢ÙÆäËûÌõ¼þÏàͬʱ£¬ÎªÌá¸ßCO2µÄƽºâת»¯ÂÊ£¬Éú²úÖпÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ________(ÌîÌá¸ß»ò½µµÍ)ˮ̼±È¡£
¢Úµ±Î¶ȸßÓÚ190¡æºó£¬CO2ƽºâת»¯ÂʳöÏÖÈçͼ1ËùʾµÄ±ä»¯Ç÷ÊÆ£¬ÆäÔ­ÒòÊÇ__________¡£

£¨4£©·´Ó¦¢ñµÄƽºâ³£Êý±í´ïʽK1=____________________£»Èç¹ûÆðʼζÈÏàͬ£¬·´Ó¦¢ñÓÉÔÚºãÎÂÈÝÆ÷½øÐиÄΪÔÚ¾øÈÈ(ÓëÍâ½çûÓÐÈÈÁ¿½»»»)ÈÝÆ÷ÖнøÐУ¬Æ½ºâ³£ÊýK1½«__________(ÌîÔö´ó¡¢¼õÉÙ¡¢²»±ä)¡£
£¨5£©Ä³Ñо¿Ð¡×éΪ̽¾¿·´Ó¦¢ñÖÐÓ°Ïìc(CO2)µÄÒòËØ£¬ÔÚºãÎÂϽ«0.4molNH3ºÍ0.2molCO2·ÅÈëÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬t1ʱ´ïµ½Æ½ºâ¹ý³ÌÖÐc(CO2)Ëæʱ¼ät±ä»¯Ç÷ÊÆÇúÏßÈçÉÏͼ2Ëùʾ¡£ÈôÆäËûÌõ¼þ²»±ä£¬t1ʱ½«ÈÝÆ÷Ìå»ýѹËõµ½1L£¬Çë»­³öt1ºóc(CO2)Ëæʱ¼ät±ä»¯Ç÷ÊÆÇúÏß(t2´ïµ½ÐµÄƽºâ)¡£
£¨6£©ÄòËØÔÚÍÁÈÀÖлᷢÉú·´Ó¦CO(NH2)2£«2H2O(NH4)2CO3¡£ÏÂÁÐÎïÖÊÖÐÓëÄòËØÓÐÀàËÆÐÔÖʵÄÊÇ______¡£

A£®NH2COONH4B£®H2NOCCH2CH2CONH2
C£®HOCH2CH2OHD£®HOCH2CH2NH2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸