ÌâÄ¿ÄÚÈÝ

·´Ó¦SO2(g)+ NO2(g)SO3(g)+NO(g) ,ÈôÔÚÒ»¶¨Î¶ÈÏ£¬½«ÎïÖʵÄÁ¿Å¨¶È¾ùΪ2mol/LµÄSO2(g)ºÍNO2(g)×¢ÈëÒ»ÃܱÕÈÝÆ÷ÖУ¬µ±´ïµ½Æ½ºâ״̬ʱ£¬²âµÃÈÝÆ÷ÖÐSO2(g)µÄת»¯ÂÊΪ50%£¬ÊÔÇó£ºÔÚ¸ÃζÈÏ¡£

£¨1£©´Ë·´Ó¦µÄƽºâ³£Êý¡£

£¨2£©ÈôSO2(g)µÄ³õʼŨ¶ÈÔö´óµ½3mol/L, NO2(g)µÄ³õʼŨ¶ÈÈÔΪ2mol/L£¬ÔòSO2(g)¡¢NO2(g)µÄת»¯ÂʱäΪ¶àÉÙ£¿

 

K=1£¨4·Ö£©×ª»¯ÂÊ40%£¬60%£¨4·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾»¯Ñ§¡ª¡ª»¯Ñ§Óë¼¼Êõ¡¿£¨15·Ö£©ÁòËṤҵÉú²úÓ¦¿¼ÂÇ×ۺϾ­¼ÃÐ§ÒæÎÊÌâ¡£

£¨1£©Èô´ÓÏÂÁÐËĸö³ÇÊÐÖÐÑ¡ÔñÒ»´¦Ð½¨Ò»×ùÁòËá³§£¬ÄãÈÏΪ³§Ö·ÒËÑ¡ÔÚ

          µÄ½¼Çø£¨ÌîÑ¡ÏîµÄ±êºÅ£©

A£®Óзḻ»ÆÌú¿ó×ÊÔ´µÄ³ÇÊР   B£®·ç¹âÐãÀöµÄÂÃÓγÇÊÐ

C£®ÏûºÄÁòËá½Ï¶àµÄ¹¤Òµ³ÇÊР   D£®È˿ڳíÃܵÄÎÄ»¯¡¢ÉÌÒµÖÐÐijÇÊÐ

£¨2£©¾Ý²âË㣬½Ó´¥·¨ÖÆÁòËá¹ý³ÌÖУ¬Èô·´Ó¦Èȶ¼Î´±»ÀûÓã¬ÔòÿÉú²ú1t 98%ÁòËáÐèÏûºÄ3.6¡Á105kJÄÜÁ¿¡£Èô·´Ó¦:SO2(g)+1/2O2(g)=SO3(g)  ¡÷H=£­98£®3kJ¡¤mol£­1£»·Å³öµÄÈÈÁ¿ÄÜÔÚÉú²ú¹ý³ÌÖеõ½³ä·ÖÀûÓÃ, Çëͨ¹ý¼ÆËãÅжÏÿÉú²ú1t98%ÁòËáÖ»ÐèÍâ½çÌṩ£¨»ò¿ÉÏòÍâ½çÊä³ö£©          Ç§½¹ÄÜÁ¿£»

£¨3£©CuFeS2ÊÇ»ÆÌú¿óµÄÁíÒ»³É·Ö£¬ìÑÉÕʱ£¬CuFeS2ת»¯ÎªCuO¡¢Fe2O3ºÍSO2£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                              ¡£

       £¨4£©ÓÉÁòËá³§·ÐÌÚ¯ÅųöµÄ¿óÔüÖк¬ÓÐFe2O3¡¢CuO¡¢CuSO4£¨ÓÉCuOÓëSO3ÔÚ·ÐÌÚ¯Öл¯ºÏ¶ø³É£©£¬ÆäÖÐÁòËáÍ­µÄÖÊÁ¿·ÖÊýËæ·ÐÌگζȲ»Í¬¶ø±ä»¯£¨¼ûÏÂ±í£©

·ÐÌگζÈ/¡æ

600

620

640

660

¿óÔüÖÐCuSO4µÄÖÊÁ¿·ÖÊý/%

9.3

9.2

9.0

8.4

ÒÑÖªCuSO4ÔÚµÍÓÚ660¡æÊ±²»»á·Ö½â£¬Çë¼òÒª·ÖÎöÉϱíÖÐCuSO4µÄÖÊÁ¿·ÖÊýËæÎ¶ÈÉý¸ß¶ø½µµÍµÄÔ­Òò                                                      ¡£

 

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

¾«Ó¢¼Ò½ÌÍø