Question

¡¾ÌâÄ¿¡¿Îíö²ÓɶàÖÖÎÛȾÎïÐγɣ¬ÆäÖаüº¬¿ÅÁ£Î°üÀ¨PM2.5ÔÚÄÚ£©¡¢µªÑõ»¯ÎNOx£©¡¢CO¡¢SO2µÈ¡£»¯Ñ§ÔÚ½â¾öÎíö²ÎÛȾÖÐÓÐ×ÅÖØÒªµÄ×÷Óá£

£¨1£©ÒÑÖª£º2SO2£¨g£©+O2£¨g£©2SO3£¨g£© ¦¤H=£­196.6 kJ¡¤mol -1

2NO£¨g£©+O2£¨g£©2NO2£¨g£© ¦¤H=£­113.0 kJ¡¤mol-1

Ôò·´Ó¦NO2£¨g£©+SO2£¨g£©SO3£¨g£©+NO£¨g£© ¦¤H= kJ¡¤mol-1¡£

Ò»¶¨Ìõ¼þÏ£¬½«NO2ÓëSO2ÒÔÌå»ý±È1:2ÖÃÓÚºãκãÈݵÄÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬²âµÃÉÏÊö·´Ó¦Æ½ºâʱNO2ÓëSO2Ìå»ý±ÈΪ1:5£¬Ôòƽºâ³£ÊýK£½ ¡£

£¨2£©ÏÂͼÊÇÒ»ÖÖÓÃNH3ÍѳýÑÌÆøÖÐNOµÄÔ­Àí¡£

¢Ù¸ÃÍÑÏõÔ­ÀíÖУ¬NO×îÖÕת»¯ÎªH2OºÍ_____________£¨Ìѧʽ£©¡£

¢Úµ±ÏûºÄ1 mol NH3ºÍ0.5 molO2ʱ£¬³ýÈ¥µÄNOÔÚ±ê×¼×´¿öϵÄÌå»ýΪ__________L¡£

£¨3£©NOÖ±½Ó´ß»¯·Ö½â£¨Éú³ÉN2ºÍO2£©Ò²ÊÇÒ»ÖÖÍÑÏõ;¾¶¡£ÔÚ²»Í¬Ìõ¼þÏ£¬NOµÄ·Ö½â²úÎﲻͬ¡£ÔÚ¸ßѹÏ£¬NOÔÚ40¡æÏ·ֽâÉú³ÉÁ½ÖÖ»¯ºÏÎÌåϵÖи÷×é·ÖÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÇúÏßÈçÓÒͼËùʾ£¬Ð´³öNO·Ö½âµÄ»¯Ñ§·½³Ìʽ£º_______________¡£

Answer 1

¡¾´ð°¸¡¿£¨1£©-41.8 £» 1.8 »ò9/5 £¨2£©¢ÙN2 £» ¢Ú11.2 £» £¨3£©3NO£½N2O + NO2

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©ÒÑÖª£º¢Ù2SO2£¨g£©+O2£¨g£©2SO3£¨g£© ¦¤H=£­196.6 kJ¡¤mol £­1

¢Ú2NO£¨g£©+O2£¨g£©2NO2£¨g£© ¦¤H=£­113.0 kJ¡¤mol£­1

Ôò¸ù¾Ý¸Ç˹¶¨ÂÉ¿ÉÖª£¨¢Ù¡ª¢Ú£©¡Â2¼´µÃµ½·´Ó¦

NO2£¨g£©+SO2£¨g£©SO3£¨g£©+NO£¨g£©µÄ¦¤H=£­41.8kJ¡¤mol£­1

NO2£¨g£©+SO2£¨g£©SO3£¨g£©+NO£¨g£©

ÆðʼŨ¶È£¨mol/L£© 1 2 0 0

ת»¯Å¨¶È£¨mol/L£© x x x x

ƽºâŨ¶È£¨mol/L£©1¡ªx 2¡ªx x x

Ôò£¨1¡ªx£©£º£¨2¡ªx£©£½1£º5

½âµÃx£½0.75

ËùÒÔƽºâ³£ÊýK£½¡£

£¨2£©¢Ù¸ù¾ÝʾÒâͼ¿ÉÖª¸ÃÍÑÏõÔ­ÀíÖУ¬NO×îÖÕת»¯ÎªH2OºÍN2¡£

¢Úµ±ÏûºÄ1 mol NH3ºÍ0.5 molO2ʱ£¬¸ù¾Ýµç×ÓµÃʧÊغã¿ÉÖªÏûºÄNOµÄÎïÖʵÄÁ¿£½£¬ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ0.5mol¡Á22.4L/mol£½11.2L¡£

£¨3£©¸ù¾ÝͼÏñ¿ÉÖªYºÍZÊÇÉú³ÉÎÇÒ¶þÕßµÄÎïÖʵÄÁ¿ÏàµÈ£¬3molNO·Ö½â¾ùÉú³É1molYºÍZ£¬Ôò¸ù¾ÝµªÔ­×ÓÊغã¿ÉÖªÁ½ÖÖ»¯ºÏÎïÓ¦¸ÃÊÇN2OºÍNO2£¬ËùÒÔNO·Ö½âµÄ»¯Ñ§·½³ÌʽΪ3NO N2O+NO2¡£