ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿£¨1£©ÊÀ½çË®²úÑøֳлá½éÉÜÁËÒ»ÖÖÀûÓõ绯ѧÔÀí¾»»¯Óã³ØÖÐË®Öʵķ½·¨£¬Æä×°ÖÃÈçͼËùʾ¡£Çëд³öÒõ¼«µÄµç¼«·´Ó¦Ê½ _________________________¡£
£¨2£©¼×´¼¿ÉÀûÓÃˮúÆøºÏ³É£ºCO(g)+2H2(g) CH3OH(g) ¡÷H<0. Ò»¶¨Ìõ¼þÏ£¬½«1molCOºÍ2molH2ͨÈëÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£¬µ±¸Ä±äζȻòѹǿʱ£¬Æ½ºâºóCH3OHµÄÌå»ý·ÖÊý ¦ÕCH3OH£©±ä»¯Ç÷ÊÆÈçͼËùʾ¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ___________¡£
A.״̬M£¬Æ½ºâʱµÄCOת»¯ÂÊΪ10%
B.ͼÖÐѹǿµÄ´óС¹ØϵÊÇa<b<c<d
C.ºãκãѹʱ£¬ÔÚÔƽºâÌåϵÖÐÔÙ³äÈëÊÊÁ¿¼×´¼£¬ÖØÐÂƽºâºóÌåϵÖм״¼µÄÌå»ý·ÖÊý²»±ä
D.µ±ÌåϵÖÐ n(CO)/n(H2)µÄÖµ²»Ëæʱ¼ä±ä»¯Ê±£¬Ìåϵ´ïµ½Æ½ºâ
£¨3£©¶þÑõ»¯ÁòºÍµâË®»á·¢ÉúÈç϶þ²½·´Ó¦:
·´Ó¦ | »î»¯ÄÜ | |
µÚÒ»²½ | SO2+I2+2H2O4H++SO42¡ª+2I¡ª | 9.2kJ¡¤mo1-1 |
µÚ¶þ²½ | I2+ I¡ªI3 ¡ª | 23.5kJ¡¤mo1-1 |
Ò»¶¨Ìõ¼þÏ£¬1mol SO2·Ö±ð¼ÓÈëµ½Ìå»ýÏàͬ¡¢Å¨¶È²»Í¬µÄµâË®ÖУ¬Ìåϵ´ïµ½Æ½ºâºó£¬H+¡¢I3¡ª¡¢SO42¡ªµÄÎïÖʵÄÁ¿Ëæn(I2)/n(SO2)µÄ±ä»¯ÇúÏßÈçͼ (ºöÂÔ·´Ó¦Ç°ºóµÄÌå»ý±ä»¯)¡£
¢ÙÓÐÈËÈÏΪXµãµÄI¡ªÅ¨¶ÈСÓÚYµã£¬ÄãÈÏΪ¸Ã¹ÛµãÊÇ·ñÕýÈ·________£¬ÔÒòÊÇ_________________¡£
¢Úµ±n(I2)/n(SO2)=4ʱ,ÇëÔÚÏÂͼ»³öÌåϵÖÐn (I¡ª)·´Ó¦Ê±¼äµÄ±ä»¯ÇúÏß¡£________
¢Û»¯Ñ§ÐËȤС×éÄâ²ÉÓÃÏÂÊö·½·¨À´²â¶¨I2+I¡ªI3¡ªµÄƽºâ³£Êý£¨ÊÒÎÂÌõ¼þϽøÐУ¬ÊµÑéÖÐÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ)£º
ÒÑÖª£ºI¡ªºÍI3 ¡ª²»ÈÜÓÚCCl4£»:Ò»¶¨Î¶ÈÏ£¬µâµ¥ÖÊÔÚËÄÂÈ»¯Ì¼ºÍË®»ìºÏÒºÌåÖУ¬µâµ¥ÖʵÄŨ¶È±ÈÖµ ¼´ÊÇÒ»¸ö³£Êý(ÓÃKd±íʾ£¬³ÆΪ·ÖÅäϵÊý)£¬ÊÒÎÂÌõ¼þÏ Kd=85¡£ÊµÑé²âµÃÉϲãÈÜÒºÖÐc(I3 ¡ª)=0.049mol/L£¬Ï²ãÒºÌåÖÐc(I2)=0.085mol¡¤L-1¡£½áºÏÉÏÊöÊý¾Ý£¬¼ÆËãÊÒÎÂÌõ¼þÏÂI2+ I¡ªI3¡ªµÄƽºâ³£ÊýK=_______(±£ÁôÈýλÓÐЧÊý×Ö)¡£
¡¾´ð°¸¡¿2NO3££«12H£«£«10e£=N2£«6H2OBC²»ÕýȷͼÖÐbÏß±íʾI3££¬Ëæ×Ån(I2)/n(SO2)µÄÖµÔö´ó£¬ÓÐÀûÓÚI2£«I£I3£Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬Æ½ºâºóI£Å¨¶È½µµÍ961
¡¾½âÎö¡¿
£¨1£©¸ù¾Ýµç½âµÄÔÀí£¬Òõ¼«Éϵõ½µç×Ó£¬·¢Éú»¹Ô·´Ó¦£¬¸ù¾Ý×°ÖÃͼ£¬NO3£ÔÚÒõ¼«ÉϷŵ磬µç½âÖÊ»·¾³ÎªËáÐÔ£¬Òò´ËÒõ¼«·´Ó¦Ê½Îª2NO3££«12H£«£«10e£=N2£«6H2O£»£¨2£©A¡¢×´Ì¬Mʱ£¬¼×´¼µÄÌå»ý·ÖÊýΪ10%£¬Áî¸ÃÌõ¼þÏ£¬ÏûºÄCOµÄÎïÖʵÄÁ¿Îªxmol£¬ÔòƽºâʱCOµÄÎïÖʵÄÁ¿Îª(1£x)mol£¬»ìºÏÆøÌå×ÜÎïÖʵÄÁ¿Îª(3£2x)mol£¬Òò´ËÓСÁ100%=10%£¬½âµÃx=7/8£¬ÔòCOµÄת»¯ÂÊΪ7/8£¬¹ÊA´íÎó£»B¡¢¸ù¾ÝͼÏñ£¬´Ó×óÏòÓÒ¼×´¼µÄÌå»ý·ÖÊýÔö´ó£¬ËµÃ÷ƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬¸ù¾ÝÀÕÏÄÌØÁÐÔÀí£¬Ñ¹Ç¿´óС¹ØϵÊÇa<b<c<d£¬¹ÊBÕýÈ·£»C¡¢ÈÝÆ÷ΪºãκãѹÏ£¬ÔÙ³äÈëÊÊÁ¿¼×´¼£¬ÓëƽºâΪµÈЧƽºâ£¬ÔòÖØдﵽƽºâºó¼×´¼µÄÌå»ý·ÖÊý²»±ä£¬¹ÊCÕýÈ·£»D¡¢ÒòΪÊÇ°´ÕÕ»¯Ñ§¼ÆÁ¿ÊýͶÈ룬¸Ã±ÈֵʼÖÕ±£³Ö²»±ä£¬¹ÊD´íÎ󣻣¨3£©¸ù¾Ý·´Ó¦²½Ö裬ÍƳöaΪH£«£¬BΪI3££¬CΪSO42££¬¢ÙËæ×Ån(I2)/n(SO2)µÄÔö¼Ó£¬ËµÃ÷ÔÚÔÀ´µÄ»ù´¡ÉÏÔö¼ÓI2µÄÎïÖʵÄÁ¿£¬µÚ¶þ²½µÄƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬I£µÄת»¯ÂʽµµÍ£¬Òò´Ë¸Ã¹Ûµã²»ÕýÈ·£¬ÔÒòÊÇͼÖÐbÏß±íʾI3££¬Ëæ×Ån(I2)/n(SO2)µÄÖµÔö´ó£¬ÓÐÀûÓÚI2£«I£I3£Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬Æ½ºâºóI£Å¨¶È½µµÍ£»¢ÚÒòΪ¿ªÊ¼Í¨ÈëSO2Ϊ1mol£¬µ±n(I2)/n(SO2)=4ʱ£¬n(I2)=4mol£¬Éú³ÉI£ÎïÖʵÄÁ¿×î´óֵΪ2mol£¬µ«²»ÄÜ´ïµ½2mol£¬µÚÒ»²½µÄ»î»¯ÄÜС£¬·´Ó¦ËÙÂʿ죬µÚ¶þ²½»î»¯Äܴ󣬷´Ó¦ËÙÂʽÏÂý£¬Òò´ËͼÏñÊÇ£»¢Ûƽºâ³£ÊýÖ»ÊÜζȵÄÓ°Ï죬ËÄÂÈ»¯Ì¼µÄÃܶȴóÓÚË®£¬Ï²ãΪI2(CCl4)£¬ÉϲãΪI2(H2O)£¬¸ù¾ÝKdµÄ¶¨Ò壬ÍƳö£º=Kd£¬´úÈëÊýÖµ£¬¼ÆËã³öc[I2(H2O)]=0.001mol¡¤L£1£¬10mLÈÜÒºÖÐc(I3£)=0.049mol¡¤L£1£¬ÎïÖʵÄÁ¿Îª0.00049mol£¬·´Ó¦µÄI£ÎïÖʵÄÁ¿Îª0.00049mol£¬Æ½ºâʱI£ÎïÖʵÄÁ¿Îª(0.1¡Á0.01£0.00049)mol=0.00051mol¡¤L£1£¬Ôòc(I£)=0.051mol¡¤L£1£¬¸ù¾Ýƽºâ³£ÊýµÄ±í´ïʽ£¬K==961¡£