ÌâÄ¿ÄÚÈÝ

£¨12·Ö£©

£¨1£©±ê×¼×´¿öÏ£¬Ìå»ýԼΪ11.2 LµÄNH3ÖÐÔ¼º¬ÓР     ¸ö·Ö×Ó¡¢º¬ÓР    ¸öÖÊ×Ó¡£

£¨2£©Í¬ÎÂͬѹÏ£¬Í¬ÖÊÁ¿µÄ°±ÆøºÍÁò»¯ÇâÆøÌ壨H2S£©µÄÌå»ý±ÈΪ            ¡£Í¬ÎÂͬѹÏ£¬Èô¶þÕßÇâÔ­×ÓÊýÏàµÈ£¬ËüÃǵÄÌå»ý±ÈΪ             ¡£

£¨3£©ÔÚijζÈʱ£¬Ò»¶¨Á¿µÄÔªËØAµÄÇ⻯ÎïAH3ÔÚÒ»¶¨Ìå»ýÃܱÕÈÝÆ÷ÖпÉÍêÈ«·Ö½â³ÉÁ½ÖÖÆøÌ¬µ¥ÖÊ£¬´ËʱÃܱÕÈÝÆ÷ÖÐÆøÌå·Ö×Ó×ܵÄÎïÖʵÄÁ¿Ôö¼ÓÁË75%¡£ÔòAµ¥ÖʵÄÒ»¸ö·Ö×ÓÖÐÓÐ_____¸öAÔ­×Ó£¬AH3·Ö½â·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________¡£

 

¡¾´ð°¸¡¿

£¨1£©3.01¡Á1023£¨2·Ö£© 3.01¡Á1024£¨2·Ö£©

£¨2£©2©U1£¨2·Ö£© 2©U3£¨2·Ö£©

£¨3£©4£¨2·Ö£© 4AH3 = A4+6H2£¨2·Ö£©

¡¾½âÎö¡¿¿¼²éÎïÖʵÄÁ¿µÄÓйؼÆËã¡£

£¨1£©¸ù¾Ý±ê×¼×´¿öÏÂµÄÆøÌåĦ¶ûÌå»ý¿ÉÖª£¬11.2L°±ÆøµÄÎïÖʵÄÁ¿ÊÇ11.2L¡Â22.4L/mol£½0.5mol¡£Òò´Ëº¬ÓеķÖ×ÓÊýÊÇ10¡Á0.5mol¡Á6.02¡Á1023/mol£½3.01¡Á1023¡£°±Æø·Ö×ÓÖк¬ÓÐ10¸öÖÊ×Ó£¬ËùÒÔº¬ÓеÄÖÊ×ÓÊýÊÇ0.5mol¡Á6.02¡Á1023/mol£½3.01¡Á1024¡£

£¨2£©Í¬ÖÊÁ¿µÄ°±ÆøºÍÁò»¯ÇâÆøÌ壨H2S£©µÄÎïÖʵÄÁ¿Ö®±ÈÊÇ34©U17£½2©U1£¬ËùÒÔ¸ù¾Ý°¢·ü¼ÓµÂÂÞ¶¨ÂÉ¿ÉÖª£¬¶þÕßµÄÌå»ýÖ®±ÈÊÇ2©U1£»¸ù¾Ý¶þÕߵĻ¯Ñ§Ê½¿ÉÖª£¬Í¬ÎÂͬѹÏ£¬Èô¶þÕßÇâÔ­×ÓÊýÏàµÈ£¬ËüÃǵÄÌå»ý±ÈΪ1/3©U1/2£½2©U3¡£

£¨3£©¸ÃÇ⻯Îï·Ö½âµÄ·½³ÌʽÊÇ2xAH3 =2 Ax+3xH2£¬ÓÉÓÚÃܱÕÈÝÆ÷ÖÐÆøÌå·Ö×Ó×ܵÄÎïÖʵÄÁ¿Ôö¼ÓÁË75%¡££¬ÔòÓУ¬½âµÃx£½4£¬¼´Aµ¥ÖʵÄÒ»¸ö·Ö×ÓÖÐÓÐ4¸öÔ­×Ó£¬·´Ó¦µÄ·½³ÌʽÊÇ4AH3 = A4+6H2¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
COºÍH2ÓëÎÒÃǵÄÉú²úºÍÉú»îµÈ·½ÃæÃÜÇÐÏà¹Ø£®
£¨1£©½«Ãº×ª»¯ÎªË®ÃºÆøÊÇͨ¹ý»¯Ñ§·½·¨½«Ãº×ª»¯Îª½à¾»È¼Áϵķ½·¨Ö®Ò»£®
ÒÑÖª£ºC£¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5kJ£®mol-1
H2£¨g£©+
1
2
O2£¨g£©=H2O£¨g£©¡÷H=-242.0kJ£®mol-1
CO£¨g£©+
1
2
O2£¨g£©=CO2£¨g£©¡÷H=-283.0kJ£®mol-1
ÔòC£¨s£©ÓëË®ÕôÆø·´Ó¦ÖÆÈ¡COºÍH2µÄÈÈ»¯Ñ§·½³ÌʽΪ
C£¨s£©+H2O£¨g£©=CO£¨g£©+H2£¨g£©¡÷H=+131.5kJ£®mol-1
C£¨s£©+H2O£¨g£©=CO£¨g£©+H2£¨g£©¡÷H=+131.5kJ£®mol-1
£®±ê×¼×´¿öÏ£¬V£¨ CO£©£ºV£¨H2£©=1£ºlµÄË®ÃºÆø22.4L£¬ÍêȫȼÉÕÉú³ÉCO2ºÍË®ÕôÆø£¬·Å³öµÄÈÈÁ¿Îª
262.5kJ
262.5kJ
£®
£¨2£©Ò»¶¨Ìõ¼þÏ£¬ÔÚÈÝ»ýΪ3LµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦£ºCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©£¬´ïƽºâ״̬£®¸ù¾Ýͼʾ»Ø´ð£º
¢Ù500¡æÊ±£¬´Ó·´Ó¦¿ªÊ¼ÖÁ´ïµ½Æ½ºâ״̬V£¨H2£©=
2nB
3tB
mol£®L-1£®min-1
2nB
3tB
mol£®L-1£®min-1
£¨ÓÃnB¡¢tB±íʾ£©
¢ÚKA ºÍKBµÄ¹ØÏµÊÇ£ºKA
£¾
£¾
KB£¬¸Ã·´Ó¦µÄ¡÷H
£¼
£¼
0£¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
¢Û300¡æ´ïƽºâʱ£¬½«ÈÝÆ÷ÈÝ»ýѹËõµ½Ô­À´µÄ
1
2
£¬ÆäËûÌõ¼þ²»±ä£¬Ôòv£¨Õý£©
£¾
£¾
v£¨Ä棩£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©£®
£¨3£©ÊÒÄÚÃºÆøÐ¹Â©Ôì³ÉÈËÌåÖж¾£¬ÊÇÒòΪCOÎüÈë·ÎÖÐÓëÊäÑõѪºìµ°°×£¨HbO2£©·¢Éú·´Ó¦£ºCO+HbO2?O2+HbCO£¬37¡æÊ±£¬K=220£®µ±[HbCO]£º[HbO2]¡Ý0.02ʱ£¬¼´ÎüÈëCOÓëO2ÎïÖʵÄÁ¿Å¨¶ÈÖ®±È¡Ý
1£º11000
1£º11000
ʱ£¬È˵ÄÖÇÁ¦»áÊÜË𣻰ÑCOÖж¾µÄ²¡ÈË·ÅÈë¸ßѹÑõ²ÕÖнⶾµÄÔ­ÀíÊÇ
ÑõÆøÅ¨¶ÈÔö´ó£¬ÉÏÊö»¯Ñ§Æ½ºâÄæÏòÒÆ¶¯£¬Ê¹CO´ÓѪºìµ°°×ÖÐÍÑÀë³öÀ´
ÑõÆøÅ¨¶ÈÔö´ó£¬ÉÏÊö»¯Ñ§Æ½ºâÄæÏòÒÆ¶¯£¬Ê¹CO´ÓѪºìµ°°×ÖÐÍÑÀë³öÀ´
£®

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

¾«Ó¢¼Ò½ÌÍø