ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿»·¾³ÎÊÌâÒѳÉΪȫÇòÆÕ±éµÄÈÈÃÅ»°Ìâ¡£

£¨1£©µªÑõ»¯ÎïÊÇ¿ÕÆøµÄÖ÷ÒªÎÛȾԴ֮һ¡£¿ÉÀûÓü×Íé´ß»¯»¹Ô­µªÑõ»¯Îï¡£ÒÑÖª£º

CH4(g)+2NO2(g)= N2(g)+CO2(g)+2H2O(g) ¦¤H1= -867 kJ¡¤mol-1

CH4(g)+4NO2(g)= 4NO(g)+CO2(g)+2H2O(g) ¦¤H2= -564 kJ¡¤mol-1

ÔòCH4½«NO»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ ¡£

£¨2£©´ß»¯·´Ïõ»¯·¨ºÍµç»¯Ñ§½µ½â·¨¿ÉÓÃÓÚÖÎÀíË®ÖÐÏõËáÑεÄÎÛȾ¡£´ß»¯·´Ïõ»¯·¨ÖУ¬ÓÃH2ÔÚ´ß»¯¼Á±íÃ潫NO3£­»¹Ô­ÎªN2£¬Ò»¶Îʱ¼äºó£¬ÈÜÒºµÄ¼îÐÔÃ÷ÏÔÔöÇ¿¡£Ôò·´Ó¦µÄÀë×Ó·½³ÌʽΪ ¡£

£¨3£©¾ÓÊÒ×°ÐÞ²ÄÁÏ»á»Ó·¢³ö¼×È©£¬ÎÛȾ¿ÕÆø¡£Í¨¹ý´«¸ÐÆ÷¿ÉÒÔ¼à²â¿ÕÆøÖм×È©µÄº¬Á¿¡£Ò»ÖÖȼÁϵç³ØÐͼ×È©ÆøÌå´«¸ÐÆ÷µÄ¹¤×÷Ô­ÀíÈçͼËùʾ£¬Æ为¼«µç¼«·´Ó¦Ê½Îª ¡£

£¨4£©Åŷŵ½´óÆøÖеÄSO2»áÐγÉÁòËáÐÍËáÓ꣬SO2ÐγÉËáÓêµÄ»¯Ñ§·½³ÌʽΪ ¡£È¼Ãº»ðµç·¢µç³§µÄβÆøÖк¬SO2£¬³£ÓÃNa2SO3ÈÜÒº»ò°±Ë®ÎüÊÕβÆø£¬Éú³ÉNaHSO3 »òNH4HSO3£¬ÒÑÖªNaHSO3ÈÜÒºÏÔËáÐÔ£¬ÔòÆäÈÜÒºÖÐÀë×ÓŨ¶È´óС¹ØϵΪ ¡£

£¨5£©ÒÑÖª25¡æʱ£¬Ksp(BaSO4) =1.0 ¡Á10£­10mol2¡¤L-2 ¡£ÏÖ½«4.0¡Á10£­4mo1/L µÄNa2SO4 ÈÜÒºÓëÒ»¶¨Å¨¶ÈµÄBaCl2ÈÜÒºµÈÌå»ý»ìºÏÉú³É³Áµí£¬¼ÆËãÓ¦¼ÓÈëBaCl2ÈÜÒºµÄ×îСŨ¶ÈΪ_________¡£

¡¾´ð°¸¡¿£¨1£©CH4(g)+4NO(g)=2N2(g)+CO2(g)+2H2O(g)¦¤H=-1170kJ¡¤mol-1

£¨2£©¢Ù2NO3-+5H2N2+4H2O+2OH-

£¨3£©HCHO+H2O-4e-=CO2¡ü+4H+

£¨4£©SO2+H2OH2SO3£»2H2SO3+O2=2H2SO4

£¨»ò2SO2+O22SO3SO3+H2O=H2SO4£©

c(Na+)£¾c(HSO3£­)£¾c(H+)£¾c(SO32£­)£¾c(OH£­)

£¨5£©1.0¡Á10-6mo1/L

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©¢ÙCH4(g)+2NO2(g)= N2(g)+CO2(g)+2H2O(g) ¦¤H1= -867 kJ¡¤mol-1

¢ÚCH4(g)+4NO2(g)= 4NO(g)+CO2(g)+2H2O(g) ¦¤H2= -564 kJ¡¤mol-1£»

¸ù¾Ý¸Ç˹¶¨ÂÉ¢Ù¡Á2£­¢ÚµÃCH4½«NO»¹Ô­ÎªN2µÄÈÈ»¯Ñ§·½³ÌʽΪ

CH4(g)+4NO(g)= 2N2(g)+CO2(g)+2H2O(g)¦¤H= -1170 kJ¡¤mol-1£»

£¨2£©¢ÙH2ÔÚ´ß»¯¼Á±íÃ潫NO3£­»¹Ô­ÎªN2£¬ÈÜÒºµÄ¼îÐÔÃ÷ÏÔÔöǿ˵Ã÷Éú³ÉOH-£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ2NO3-+5H2 N2+4H2O+2OH-£»£¨3£©¸ù¾Ýͼʾ£¬¼×È©ÔÚ¸º¼«±»Ñõ»¯Îª¶þÑõ»¯Ì¼£¬¸º¼«µç¼«·´Ó¦Ê½ÎªHCHO + H2O- 4e-=CO2¡ü+4H+£»£¨4£©SO2ÐγÉËáÓêµÄ»¯Ñ§·½³ÌʽΪSO2 + H2O H2SO3£»2H2SO3 +O2 = 2H2SO4£¨»ò2SO2 + O2 2SO3 SO3 +H2O = H2SO4£©£»NaHSO3ÈÜÒºÏÔËáÐÔ£¬ËµÃ÷HSO3£­µÄµçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬ÔòÆäÈÜÒºÖÐÀë×ÓŨ¶È´óС¹ØϵΪc(Na+) £¾ c(HSO3£­) £¾ c(H+) £¾ c(SO32£­) £¾ c(OH£­)£»£¨5£©ÉèBaCl2ÈÜÒºµÄ×îСŨ¶ÈΪc mo1/L£¬Ôò=£¬c>1.0¡Á10-6mo1/L¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø