ÌâÄ¿ÄÚÈÝ

t ¡æʱ£¬½«3 mol AºÍ1 mol BÆøÌåͨÈëÌå»ýΪ2 LµÄÃܱÕÈÝÆ÷ÖÐ(ÈÝ»ý²»±ä)£¬·¢Éú·´Ó¦£º3A(g)£«B(g)xC(g)¡£2 minʱ·´Ó¦´ïµ½Æ½ºâ״̬(ζȲ»±ä)£¬Ê£ÓàÁË0.8 mol B£¬²¢²âµÃCµÄŨ¶ÈΪ0.4 mol¡¤L£­1£¬ÇëÌîдÏÂÁпհףº
(1)´Ó¿ªÊ¼·´Ó¦ÖÁ´ïµ½Æ½ºâ״̬£¬Éú³ÉCµÄƽ¾ù·´Ó¦ËÙÂÊΪ________¡£
(2)x£½________£»Æ½ºâ³£ÊýK£½________¡£
(3)Èô¼ÌÐøÏòԭƽºâ»ìºÏÎïµÄÈÝÆ÷ÖÐͨÈëÉÙÁ¿º¤Æø(¼ÙÉ躤ÆøºÍA¡¢B¡¢C¶¼²»·´Ó¦)ºó£¬»¯Ñ§Æ½ºâ________(Ìîд×ÖĸÐòºÅ)¡£
A£®ÏòÕý·´Ó¦·½ÏòÒƶ¯
B£®ÏòÄæ·´Ó¦·½ÏòÒƶ¯
C£®²»Òƶ¯
(4)ÈôÏòԭƽºâ»ìºÏÎïµÄÈÝÆ÷ÖÐÔÙ³äÈëa mol C£¬ÔÚt ¡æʱ´ïµ½ÐµÄƽºâ£¬´ËʱBµÄÎïÖʵÄÁ¿Îªn(B)£½________mol¡£

(1)0.2 mol¡¤L£­1¡¤min£­1¡¡(2)4¡¡0.037
(3)C¡¡(4)(0.8£«0.2a)

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨17·Ö£©½µµÍÌ«ÆøÖÐCO2º¬Á¿¼°ÓÐЧ¿ª·¢ÀûÓÃCO2£¬ÊÇ¿Æѧ¼ÒÑо¿µÄÖØÒª¿ÎÌâ¡£
£¨1£©½«È¼Ãº·ÏÆøÖеÄCO2ת»¯Îª¶þ¼×Ãѵķ´Ó¦Ô­ÀíΪ£º
2CO2£¨g£©+6H2£¨g£©CH3OCH3£¨g£©+3H2O£¨g£© ¡÷H=-122£®4kJ¡¤mol£­1
¢ÙijζÈÏ£¬½«2£®0molCO2£¨g£©ºÍ6£®0molH2£¨g£©³äÈëÌå»ý¿É±äµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦µ½´ïƽºâʱ£¬¸Ä±äѹǿºÍζȣ¬Æ½ºâÌåϵÖÐCH3OCH3£¨g£©µÄÎïÖʵÄÁ¿·ÖÊý±ä»¯ÈçϱíËùʾ¡£

ÔòPl      P3£¨Ìî¡°>¡±¡°<¡±»ò¡°=¡±£¬ÏÂͬ£©¡£ÈôT1¡¢Pl£¬T3¡¢P3ʱƽºâ³£Êý·Ö±ðΪK1¡¢K3£¬
ÔòK1     K3¡£T1¡¢PlʱH2µÄƽºâת»¯ÂÊΪ         ¡£
¢ÚÒ»¶¨Ìõ¼þÏ£¬tÉÏÊö·´Ó¦ÔÚÃܱÕÈÝÆ÷Öдïƽºâ¡£µ±½ö¸Ä±äÓ°Ïì·´Ó¦µÄÒ»¸öÌõ¼þ£¬ÒýÆðµÄÏÂÁб仯ÄÜ˵Ã÷ƽºâÒ»¶¨ÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÊÇ____      ¡£
A£®·´Ó¦ÎïµÄŨ¶È½µµÍ        B£®ÈÝÆ÷ÄÚѹǿÔö´ó
C£®Õý·´Ó¦ËÙÂÊ´óÓÚÄæ·´Ó¦ËÙÂÊ    D£®»¯Ñ§Æ½ºâ³£ÊýKÔö´ó   
£¨2£©Ì¼ËáÇâ¼ØÈÜÒº¼ÓˮϡÊÍ£¬       £¨Ìî¡°Ôö´ó¡±¡°²»±ä¡±»ò¡°¼õС¡±£©¡£ÓÃ̼Ëá¼ØÈÜÒºÎüÊÕ¿ÕÆøÖÐCO2£¬µ±ÈܶɳÊÖÐÐÔʱ£¬ÏÂÁйØϵ»ò˵·¨ÕýÈ·µÄÊÇ        ¡£
A£®c£¨K+£©=2c£¨CO£©+c£¨HCO£©+c£¨H2CO3£©
b£®c£¨HCO£©c£¨CO£©
c£®½µµÍζȣ¬c£¨H+£©¡¤c£¨OH£­£©²»±ä
£¨3£©ÏòÊ¢ÓÐFeCl3ÈÜÒºµÄÊÔ¹ÜÖеμÓÉÙÁ¿Ì¼Ëá¼ØÈÜÒº£¬Á¢¼´²úÉúÆøÌ壬ÈÜÒºÑÕÉ«¼ÓÉÓü¤¹â±ÊÕÕÉäÄܲúÉú¶¡´ï¶ûЧӦ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ                      ¡£

ÁòËáÓÃ;ʮ·Ö¹ã·º£¬¹¤ÒµÉϺϳÉÁòËáʱ£¬½«SO2ת»¯Îª´ß»¯Ñõ»¯ÊÇÒ»¸ö¹Ø¼ü²½Öè¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã·´Ó¦ÔÚºãκãÈÝÃܱÕÈÝÆ÷ÖнøÐУ¬ÅжÏÆä´ïµ½Æ½ºâ״̬µÄ±êÖ¾ÊÇ¡¡¡¡¡££¨Ìî×Öĸ£©
a£®SO2ºÍSO3Ũ¶ÈÏàµÈ
b£®SO2°Ù·Öº¬Á¿±£³Ö²»±ä
c£®ÈÝÆ÷ÖÐÆøÌåµÄѹǿ²»±ä
d£®SO3µÄÉú³ÉËÙÂÊÓëSO2µÄÏûºÄËÙÂÊÏàµÈ
e£®ÈÝÆ÷ÖлìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
£¨2£©Ä³Î¶ÈÏ£¬2SO2(g)£«O2(g£©2SO3(g£© ¡÷H£½£­196 kJ?mol£­1¡£ÔÚÒ»¸ö¹Ì¶¨ÈÝ»ýΪ5 LµÄÃܱÕÈÝÆ÷ÖгäÈë0.20 mol SO2ºÍ0.10 mol O2£¬°ë·ÖÖÓºó´ïµ½Æ½ºâ£¬²âµÃÈÝÆ÷Öк¬SO3Ϊ0.18 mol£¬Ôòv(O2)£½¡¡  ¡¡mol?L£­1?min£­1£¬·Å³öµÄÈÈÁ¿Îª¡¡         ¡¡ kJ¡£
£¨3£©Ò»¶¨Î¶Èʱ£¬SO2µÄƽºâת»¯ÂÊ(¦Á)ÓëÌåϵ×Üѹǿ(p)µÄ¹ØϵÈçͼËùʾ¡£ÊÔ·ÖÎö¹¤ÒµÉú²úÖвÉÓó£Ñ¹µÄÔ­ÒòÊÇ¡¡               ¡¡¡£

£¨4£©½«Ò»¶¨Á¿µÄSO2ºÍ0.7 molO2·ÅÈëÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚ550¡æºÍ´ß»¯¼Á×÷ÓÃÏ·¢Éú·´Ó¦¡£·´Ó¦´ïµ½Æ½ºâºó£¬½«ÈÝÆ÷ÖеĻìºÏÆøÌåͨ¹ý¹ýÁ¿NaOHÈÜÒº£¬ÆøÌåÌå»ý¼õÉÙÁË21.28 L£»ÔÙ½«Ê£ÓàÆøÌåͨ¹ý½¹ÐÔûʳ×ÓËáµÄ¼îÐÔÈÜÒºÎüÊÕO2£¬ÆøÌåµÄÌå»ýÓÖ¼õÉÙÁË5.6 L(ÒÔÉÏÆøÌåÌå»ý¾ùΪ±ê×¼×´¿öϵÄÌå»ý)¡£Ôò¸Ã·´Ó¦´ïµ½Æ½ºâʱSO2µÄת»¯ÂÊÊǶàÉÙ£¿£¨ÒªÐ´³ö¼ÆËã¹ý³Ì£¬¼ÆËã½á¹û±£ÁôһλСÊý¡££©

£¨1£©ÒÑÖª£ºO2 (g)= O2£« (g)+e£­ ¡÷H1=" +1175.7" kJ¡¤mol£­1
PtF6(g)+ e£­= PtF6£­(g)    ¡÷H2=" -" 771.1 kJ¡¤mol£­1
O2+PtF6£­(s)=O2+(g)+PtF6£­ (g)  ¡÷H3="+482.2" kJ¡¤mol£­1
Ôò·´Ó¦£ºO2£¨g£©+ PtF6 (g) = O2+PtF6(s)µÄ¡÷H="_____" kJ¡¤mol-1¡£
ÈçͼΪºÏ³É°±·´Ó¦ÔÚʹÓÃÏàͬµÄ´ß»¯¼Á£¬²»Í¬Î¶ȺÍѹǿÌõ¼þϽøÐз´ Ó¦£¬³õʼʱN2ºÍH2µÄÌå»ý±ÈΪ1:3ʱµÄƽºâ»ìºÏÎïÖа±µÄÌå»ý·ÖÊý£º

¢Ù ÔÚÒ»¶¨µÄζÈÏ£¬ÏòÌå»ý²»±äµÄÃܱÕÈÝÆ÷ÖгäÈ뵪ÆøºÍÇâÆø·¢ÉúÉÏÊö·´Ó¦£¬ÏÂÁÐÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ       ¡£
a£®ÌåϵµÄѹǿ±£³Ö²»±ä        b£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
c£®N2ºÍH2µÄÌå»ý±ÈΪ1:3      d£®»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿²»±ä
¢Ú·Ö±ðÓÃvA£¨NH3£©ºÍvB£¨NH3£©±íʾ´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ״̬A¡¢BʱµÄ·´Ó¦ËÙÂÊ£¬ÔòvA£¨NH3£©    vB£¨NH3£©£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£¬¸Ã·´Ó¦µÄµÄƽºâ³£ÊýkA     kB£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£¬ÔÚ250 ¡æ¡¢1.0¡Á104kPaÏ´ﵽƽºâ£¬H2µÄת»¯ÂÊΪ     %£¨¼ÆËã½á¹û±£ÁôСÊýµãºóһ룩£»
£¨3£©25¡æʱ£¬½«a mol NH4NO3ÈÜÓÚË®£¬ÈÜÒº³ÊËáÐÔ£¬Ô­Òò                                   £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£Ïò¸ÃÈÜÒºÖмÓÈëbL°±Ë®ºóÈÜÒº³ÊÖÐÐÔ£¬ÔòËù¼Ó°±Ë®µÄŨ¶ÈΪ          mol/L£¨Óú¬a¡¢bµÄ´úÊýʽ±íʾ£¬NH3¡¤H2OµÄµçÀëƽºâ³£ÊýΪKb=2¡Á10-5£©
£¨4£©ÈçͼËùʾ£¬×°ÖâñΪ¼×ÍéȼÁϵç³Ø£¨µç½âÖÊÈÜҺΪKOHÈÜÒº£©£¬Í¨¹ý×°ÖâòʵÏÖÌú°ôÉ϶ÆÍ­¡£µç¶ÆÒ»¶Îʱ¼äºó£¬×°ÖâñÖÐÈÜÒºµÄpH      £¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±£©£¬a¼«µç¼«·´Ó¦·½³ÌʽΪ                 £»Èôµç¶Æ½áÊøºó£¬·¢ÏÖ×°ÖâòÖÐÒõ¼«ÖÊÁ¿±ä»¯ÁË25.6g£¨ÈÜÒºÖÐÁòËáÍ­ÓÐÊ£Óࣩ£¬Ôò×°ÖâñÖÐÀíÂÛÉÏÏûºÄ¼×Íé     L£¨±ê×¼×´¿öÏ£©¡£

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

¾«Ó¢¼Ò½ÌÍø