ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿¼×´¼¿É²ÉÓöàÖÖ·½·¨ÖƱ¸£¬ÆäÓÃ;¹ã·º£¬ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ¡£

¢ñ£®ÀûÓÃºÏ³ÉÆø£¨CO¡¢CO2¡¢H2£©ÔÚ´ß»¯¼Á×÷ÓÃϺϳɼ״¼£¬·¢Éú·´Ó¦ÈçÏ£º

¢ÙCO(g)+2H2(g)CH3OH(g)

¢ÚCO2(g)+3H2(g)CH3OH(g)+H2O(g)

£¨1£©·´Ó¦¢Ù¹ý³ÌÖÐÄÜÁ¿±ä»¯ÈçÏÂͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ______________(ÌîÑ¡Ïî±êºÅ)¡£

A£®¢Ù·´Ó¦µÄ¡÷H=Äæ·´Ó¦»î»¯ÄÜ-Õý·´Ó¦»î»¯ÄÜ

B£®°Ñ1molCO(g)ºÍ2molH2(g)³äÈëÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦£¬´ïƽºâʱ·Å³öµÄÈÈÁ¿Îª91kJ

C£®´ÓͼÖÐÐÅÏ¢¿ÉÖª¼ÓÈë´ß»¯¼ÁÄܸı䷴ӦÀú³ÌºÍÈÈЧӦ

D£®ÏàͬÌõ¼þÏ£¬CO(g)ºÍH2(g)·´Ó¦Éú³É1molҺ̬CH3OH·Å³öµÄÈÈÁ¿´óÓÚ91kJ

£¨2£©ÏÖÀûÓâٺ͢ÚÁ½¸ö·´Ó¦ºÏ³ÉCH3OH£¬ÒÑÖªCO¿Éʹ·´Ó¦µÄ´ß»¯¼ÁÊÙÃüϽµ¡£

ÈôÇâ̼±È±íʾΪf= [(n(H2)-n(CO2)]/[(n(CO)-n(CO2)],ÔòÀíÂÛÉÏf=______________ʱ£¬Ô­ÁÏÆøµÄÀûÓÃÂÊ×î¸ß¡£µ«Éú²úÖÐÍùÍù²ÉÓÃÂÔ¸ßÓÚ¸ÃÖµµÄÇâ̼±È£¬ÀíÓÉÊÇ£º______________________________¡£

¢ò£®¼×´¼¿ÉÓÃÓÚÖÆÈ¡¼×Ëá¼×õ¥£¬Æä·´Ó¦·½³ÌʽΪ£ºCH3OH(g)+CO(g)HCOOCH3(g)¡÷H<0¡£¿ÆÑÐÈËÔ±µÄ²¿·ÖÑо¿½á¹ûÈçÏ£º

£¨3£©´Ó·´Ó¦Ñ¹Ç¿¶Ô¼×´¼×ª»¯ÂʵÄÓ°Ï조ЧÂÊ¡±Í¼ºÍÉú²ú³É±¾½Ç¶È·ÖÎö£¬¹¤ÒµÖÆÈ¡¼×Ëá¼×õ¥Ó¦Ñ¡ÔñµÄ×î¼ÑѹǿÊÇ______________________(Ìî¡°3.5¡Á106Pa¡±¡°4.0¡Á106Pa¡±»ò¡°5.0¡Á106Pa¡±)¡£

£¨4£©Êµ¼Ê¹¤ÒµÉú²úÖвÉÓõÄζÈÊÇ80¡æ,ÆäÀíÓÉÊÇ_______________________________________¡£

¢ó£®¼×´¼»¹¿ÉÒÔÓÃÓںϳɶþ¼×ÃÑ£¬·¢ÉúµÄ·´Ó¦Îª2CH3OH(g)CH3OCH3(g)+H2O(g)

¼ºÖª¸Ã·´Ó¦ÔÚijζÈÏÂµÄÆ½ºâ³£ÊýΪ900,´ËζÈÏ£¬ÔÚÃܱÕÈÝÆ÷ÖмÓÈëCH3OH,·´Ó¦µ½Ä³Ê±¿Ì²âµÃ¸÷×é·ÖŨ¶ÈÈçÏ£º

ÎïÖÊ

CH3OH

CH3OCH3

H2O

Ũ¶È(mol/L)

1.25

0.9

0.9

£¨5£©±È½Ï´ËʱÕý¡¢Äæ·´Ó¦ËÙÂʵĴóС£ºvÕý___________vÄæ(Ìî¡°>¡°<¡±»ò¡°=¡±)¡£

£¨6£©Èô¼ÓÈëCH3OHºó£¬¾­6min·´Ó¦´ïµ½Æ½ºâ£¬Ôò¸Ãʱ¼äÄÚÆ½¾ù·´Ó¦ËÙÂÊv(CH3OH)=___________mol/(L¡¤min)¡£

¡¾´ð°¸¡¿ D 2 ʹCO³ä·Ö·´Ó¦£¬±ÜÃâ·´Ó¦´ß»¯¼ÁÊÙÃüϽµ 4.0¡Á106Pa µÍÓÚ80¡æ¡¢·´Ó¦ËÙʽÏС£¬¸ßÓÚ80¡æ,ζȶԷ´Ó¦ËÙÂÊÓ°Ïì½ÏС,ÇÒ·´Ó¦·ÅÈÈ£¬ÉýÎÂÆ½ºâÄæÏòÒÆ¶¯£¬×ª»¯ÂʽµµÍ > 0.5

¡¾½âÎö¡¿£¨1£©A¡¢ÓÉͼ¿ÉÖª£¬¢Ù·´Ó¦Îª·ÅÈÈ·´Ó¦£¬¡÷H=Õý·´Ó¦»î»¯ÄÜ-Äæ·´Ó¦»î»¯ÄÜ£¬Ñ¡ÏîA´íÎó£»B£®·´Ó¦Îª¿ÉÄæ·´Ó¦£¬²»¿ÉÄÜÍêȫת»¯£¬¹Ê°Ñ1molCO(g)ºÍ2molH2(g)³äÈëÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦£¬´ïƽºâʱ·Å³öµÄÈÈÁ¿Ð¡ÓÚ91kJ£¬Ñ¡ÏîB´íÎó£»C£®´ÓͼÖÐÐÅÏ¢¿ÉÖª¼ÓÈë´ß»¯¼ÁÄܸı䷴ӦÀú³Ìµ«²»ÄܸıäÈÈЧӦ£¬Ñ¡ÏîC´íÎó£»D£®ÆøÌ¬CH3OHת»¯ÎªÒºÌ¬Ê±·ÅÈÈ£¬¹ÊÏàͬÌõ¼þÏ£¬CO(g)ºÍH2(g)·´Ó¦Éú³É1molҺ̬CH3OH·Å³öµÄÈÈÁ¿´óÓÚ91kJ£¬Ñ¡ÏîDÕýÈ·£»´ð°¸Ñ¡D£»£¨2£©¸ù¾Ý¢ÙºÍ¢ÚÁ½¸ö·´Ó¦ºÏ³ÉCH3OH·½³Ìʽ£¬ÔòÀíÂÛÉÏf= [(n(H2)-n(CO2)]/[(n(CO)-n(CO2)]=ʱ£¬Ô­ÁÏÆøµÄÀûÓÃÂÊ×î¸ß£»µ«Éú²úÖÐÍùÍù²ÉÓÃÂÔ¸ßÓÚ¸ÃÖµµÄÇâ̼±È£¬ÀíÓÉÊÇ£ºÊ¹CO³ä·Ö·´Ó¦£¬±ÜÃâ·´Ó¦´ß»¯¼ÁÊÙÃüϽµ£»£¨3£©´Ó·´Ó¦Ñ¹Ç¿¶Ô¼×´¼×ª»¯ÂʵÄÓ°Ï조ЧÂÊ¡±Í¼ºÍÉú²ú³É±¾½Ç¶È·ÖÎö£¬¹¤ÒµÖÆÈ¡¼×Ëá¼×õ¥Ó¦Ñ¡ÔñµÄ×î¼ÑѹǿÊÇ4.0¡Á106Pa£»£¨4£©Êµ¼Ê¹¤ÒµÉú²úÖвÉÓõÄζÈÊÇ80¡æ,ÆäÀíÓÉÊǵÍÓÚ80¡æ¡¢·´Ó¦ËÙʽÏС£¬¸ßÓÚ80¡æ,ζȶԷ´Ó¦ËÙÂÊÓ°Ïì½ÏС,ÇÒ·´Ó¦·ÅÈÈ£¬ÉýÎÂÆ½ºâÄæÏòÒÆ¶¯£¬×ª»¯ÂʽµµÍ£»

£¨5£©Q=£¬Æ½ºâÕýÏòÒÆ¶¯£¬vÕý>vÄæ£»£¨6£©ÉèÆ½ºâʱÉú³ÉÎïµÄŨ¶ÈΪ0.9+x£¬Ôò¼×´¼µÄŨ¶ÈΪ£¨1.25-2x£©ÓУº900=£¬½âµÃx=0.6 mol/L£¬¹Ê1.25-2x =0.05 mol/L£¬

Óɱí¿ÉÖª£¬¼×´¼µÄÆðʼŨ¶È¶ÈΪ(1.25+1.8) mol/L=3.05 mol/L£¬ÆäƽºâŨ¶ÈΪ0.05 mol/L£¬

6min±ä»¯µÄŨ¶ÈΪ3.0 mol/L£¬¹Ê¸Ãʱ¼äÄÚÆ½¾ù·´Ó¦ËÙÂÊv(CH3OH)==0.5mol/(L¡¤min)¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ó÷½Ç¦¿ó¾«¿ó(Ö÷ҪΪPbS)ºÍÈíÃÌ¿ó(Ö÷ҪΪMnO2,»¹ÓÐÉÙÁ¿Fe2O3£¬Al2O3µÈÔÓÖÊ)ÖÆ±¸PbSO4ºÍMn3O4µÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£º¢ÙPbS+MnO2+4H+=Mn2++Pb2++S+2H2O

¢Ú25¡æÊ±£¬Ksp(PbCl2)=1.6¡Á10-5£¬Ksp(PbSO4)=1.8¡Á10-8

¢ÛPbCl2(s)+2Cl-(aq)PbCl42-(aq)¡÷H>0

£¨1£©80¡æÓÃÑÎËá´¦ÀíÁ½ÖÖ¿óʯ£¬Îª¼Ó¿ìËá½þËÙÂÊ£¬»¹¿É²ÉÓõķ½·¨ÊÇ_______________(ÈÎдһÖÖ)¡£

£¨2£©ÏòËá½þÒºÖмÓÈë±¥ºÍʳÑÎË®µÄÄ¿µÄÊÇ_________________£»¼ÓÈëÎïÖÊX¿ÉÓÃÓÚµ÷½ÚËá½þÒºµÄpHÖµ£¬ÎïÖÊX¿ÉÒÔÊÇ___________(Ìî×Öĸ)¡£

A£®MnCO3 B£®NaOH C£®ZnO D£®PbO

£¨3£©ÂËÔüÖк¬ÓÐÁ½ÖÖ½ðÊôÔÓÖÊÐγɵϝºÏÎÆä³É·ÖΪ_____________(д»¯Ñ§Ê½);

ÇëÉè¼Æ·ÖÀëÁ½ÖÖ½ðÊô»¯ºÏÎïµÄ·Ïßͼ(Óû¯Ñ§Ê½±íʾÎïÖÊ£¬ÓüýÍ·±íʾת»¯¹ØÏµ£¬¼ýÍ·ÉÏ×¢Ã÷ÊÔ¼ÁºÍ·ÖÀë·½·¨)¡£______________________________________

£¨4£©ÏòÂËÒº2ÖÐͨÈëNH3ºÍO2·¢Éú·´Ó¦£¬Ð´³ö×Ü·´Ó¦µÄÀë×Ó·½³Ìʽ¡£___________________________________

£¨5£©ÓÃMn3O4ΪԭÁÏ¿ÉÒÔ»ñµÃ½ðÊôÃÌ£¬Ñ¡ÔñºÏÊʵÄÒ±Á¶·½·¨Îª___________(Ìî×Öĸ)¡£

A£®ÈÈ»¹Ô­·¨ B£®µç½â·¨ C£®Èȷֽⷨ

£¨6£©Çó25¡æÂÈ»¯Ç¦³Áµíת»¯ÎªÁòËáǦ³Áµí·´Ó¦µÄƽºâ³£ÊýK=___________(±£Áôµ½ÕûÊýλ)¡£

¡¾ÌâÄ¿¡¿ÑÇÏõËáÄÆ(NaNO2)ÊÇÒ»ÖÖ¹¤ÒµÑΣ¬Íâ¹ÛÓëʳÑÎÏàËÆ¡£ÏÂÃæÊÇijѧϰС×éÉè¼ÆµÄNaNO2ÖÆÈ¡ÊµÑéºÍ´¿¶È¼ìÑéʵÑé¡£¸ÃС×éÊÕ¼¯ÁËÏà¹Ø×ÊÁÏ£º

¢ÙSO2ºÍHNO3ÈÜÒº·´Ó¦Éú³ÉNOxºÍH2SO4

¢Ú3NO2-+2H+=2NO¡ü+NO3-+H2O

¢ÛNO2-+Ag+=AgNO2¡ý(AgNO2Ϊµ­»ÆÉ«½Ó½ü°×É«¹ÌÌ壬ÔÚË®ÖÐÐγɳÁµí)

¢ñ£®ÑÇÏõËáÄÆµÄÖÆÈ¡ÊµÑé

£¨1£©ÒÇÆ÷aµÄÃû³ÆÎª________________________£¬A×°ÖÃÖз¢ÉúµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ_____________________________________________¡£

£¨2£©B×°ÖÃÖжà¿×ÇòÅݵÄ×÷ÓÃÊÇ_________________________________________________¡£

£¨3£©Èô×°ÖÃBÖÐÒݳöµÄNOÓëNO2ÆøÌåÎïÖʵÄÁ¿Ö®±ÈΪ2¡Ã1£¬Ôò×°ÖÃBÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_________________________________________________________¡£

£¨4£©ÊµÑé¹ý³ÌÖÐÐè¿ØÖÆC×°ÖÃÖÐÈÜÒºpH>7£¬·ñÔòCÖÐÉú³ÉµÄNaNO2µÄ²úÁ¿»áϽµ£¬ÀíÓÉÊÇ_____________________________________________________¡£

£¨5£©Çë¸ù¾ÝÌâ¸ÉËù¸øÐÅÏ¢Éè¼ÆÊµÑéÖ¤Ã÷C×°ÖÃÖÐÓÐNO2²úÉú£º_________________________________¡£(ÏÞÑ¡ÓõÄÊÔ¼Á£ºÏ¡ÏõËá¡¢ÏõËáÒøÈÜÒº¡¢NaOHÈÜÒº)

¢ò£®ÑÇÏõËáÄÆµÄ´¿¶È¼ìÑé

ÒÑÖª£ºNO2-+MnO4-+H+¡úNO3-+Mn2++H2O

£¨6£©·´Ó¦½áÊøºóCÖÐÈÜҺͨ¹ý½á¾§»ñµÃNaNO2´Ö²úÆ·mg,ÈܽâºóÏ¡ÊÍÖÁ250mL£¬·Ö±ðÈ¡25.00mLÓÃcmol/LµÄËáÐÔKMnO4ÈÜҺƽÐеζ¨Èý´Î£¬Æ½¾ùÿ´ÎÏûºÄËáÐÔKMnO4ÈÜÒºµÄÌå»ýΪVmL¡£Ôò´Ö²úÆ·ÖÐNaNO2µÄÖÊÁ¿·ÖÊýΪ_____________(Óú¬c¡¢V¡¢mµÄʽ×Ó±íʾ)¡£

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

¾«Ó¢¼Ò½ÌÍø