Question

13£®¼õÉÙCO₂µÄÅÅ·ÅÒÔ¼°CO₂µÄ×ÊÔ´»¯ÀûÓþßÓÐÖØÒªÒâÒ壮
£¨1£©Óð±Ë®²¶¼¯ÑÌÆøÖеÄCO₂Éú³Éï§ÑΣ¬ÊǼõÉÙCO₂ÅŷŵĿÉÐдëÊ©Ö®Ò»£®
¢Ùд³ö°±Ë®²¶¼¯ÑÌÆøÖеÄCO₂Éú³Éï§ÑεÄÖ÷Òª»¯Ñ§·½³ÌʽCO₂+NH₃•H₂O¨TNH₄HCO₃£®
¢Ú·Ö±ðÓò»Í¬pHµÄÎüÊÕ¼ÁÎüÊÕÑÌÆøÖеÄCO₂£¬CO₂ÍѳýЧÂÊÓëÎüÊÕ¼ÁµÄpH¹ØϵÈçͼ¢ñËùʾ£¬CO₂ÍѳýЧÂÊÓëÎüÊÕ¼ÁµÄpH¹ØϵÊÇCO₂ÍѳýЧÂÊËæÎüÊÕ¼ÁµÄpHÔö´ó¶øÔö´ó£®ÑÌÆøÖÐCO₂µÄº¬Á¿Îª12%£¬ÑÌÆøͨÈË°±Ë®µÄÁ÷Á¿Îª0.052m³/h£¨±ê×¼×´¿ö£©£¬ÓÃpHΪ12.81µÄ°±Ë®ÎüÊÕÑÌÆø30min£¬ÍѳýµÄCO₂µÄÎïÖʵÄÁ¿Îª0.13mol£¨¾«È·µ½0.01£©£®
¢ÛCO₂ÍѳýЧÂÊÓëζȵĹØϵÈçͼ¢òËùʾ£®´Ó40¡æµ½45¡æÍѳýCO₂ЧÂʽµµÍµÄÖ÷ÒªÔ­ÒòÊÇ̼ËáÇâï§ÊÜÈÈÒ×·Ö½âÉú³É¶þÑõ»¯Ì¼£®
£¨2£©½«CO₂ºÍ¼×ÍéÖØÕûÖƺϳÉÆø£¨COºÍH₂£©ÊÇCO₂×ÊÔ´»¯ÀûÓõÄÓÐЧ;¾¶£®ºÏ³ÉÆøÓÃÓÚÖƱ¸¼×´¼µÄ·´Ó¦Îª
2H₂£¨g£©+CO£¨g£©?CH₃OH£¨g£©¡÷H=-90.1kJ/mol£®
¢ÙÔÚT¡æʱ£¬ÈÝ»ýÏàͬµÄ¼×¡¢ÒÒ¡¢±ûÈý¸öºãÈÝÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶ÈË·´Ó¦Î²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ£®

ÈÝÆ÷		¼×	ÒÒ	±û
Æðʼ·´Ó¦ÎïͶÈëÁ¿		2mol H2¡¢l mol CO	1mol CH3OH	2mol CH3OH
ƽ	c£¨CH£¬0H£©/mol/L	C1	C2	c3
ºâ	·´Ó¦µÄÄÜÁ¿±ä»¯ZkJ	x	y	z
Êý	Ìåϵѹǿ/Pa	P1	P2	P3
¾Ý	·´Ó¦Îïת»¯ÂÊ	a1	a2	a3

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇD£¨Ìî×Öĸ¾ŽºÅ£©£®
A.2c₁£¾c₃           B£®|x|+|y|£¼90.1        C.2p₂£¼p₃        D£®a₁+a₃£¼1
¢ÚÔÚT¡æ¡¢ºãÈÝÌõ¼þÏ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýKp=6.0¡Á10^-3£¨kPa£©^-2£®Èô¼×ÈÝÆ÷Öз´Ó¦´ïµ½Æ½ºâʱp£¨CH₃OH£©=24.0kPa£¬Ôòƽºâ»ìºÏÆøÌåÖÐCH₃OHµÄÎïÖʵÄÁ¿·ÖÊýΪ44.4%£¨KpÊÇÓÃƽºâ·Öѹ´úÌæƽºâŨ¶È¼ÆËãËùµÃƽºâ³£Êý£¬·Öѹ=×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý£©£®

Answer 1

·ÖÎö £¨1£©¢Ù°±Ë®ÓëCO₂·´Ó¦Éú³É̼ËáÇâ泥¬¾Ý´ËÊéд£»
¢Ú¸ù¾Ýͼ1·ÖÎö£¬CO₂ÍѳýЧÂÊËæÎüÊÕ¼ÁµÄpHÔö´ó¶øÔö´ó£»¸ù¾ÝÑÌÆøͨÈË°±Ë®µÄÁ÷Á¿Îª0.052m³/h£¨±ê×¼×´¿ö£©£¬ÓÃpHΪ12.81µÄ°±Ë®ÎüÊÕÑÌÆø30min£¬½áºÏÑÌÆøÖÐCO₂µÄº¬Á¿Îª12%¼ÆËã³öCO₂µÄÁ¿£»
¢Û¸ù¾ÝζȸßÉú³ÉµÄ̼ËáÇâï§Ò×·Ö½âÉú³É¶þÑõ»¯Ì¼·ÖÎö£»
£¨2£©¢ÙA¡¢¼×¡¢±ûÏà±È½Ï£¬°Ñ¼×µÈЧΪ¿ªÊ¼¼ÓÈë1molCH₃OH£¬±ûÖм״¼µÄÎïÖʵÄÁ¿Îª¼×µÄ2±¶£¬Ñ¹Ç¿Ôö´ó£¬¶ÔÓÚ·´Ó¦CH₃OH£¨g£©?CO£¨g£©+2H₂£¨g£©£¬Æ½ºâÏòÉú³É¼×´¼µÄ·½ÏòÒƶ¯£»
B¡¢ÓÉÌâÄ¿¿ÉÖªÉú³É1molCH₃OHµÄÄÜÁ¿±ä»¯Îª90.8kJ£¬¼×¡¢ÒÒƽºâ״̬Ïàͬ£¬Áîƽºâʱ¼×´¼Îªnmol£¬¼ÆËã·´Ó¦ÈÈÊýÖµ£¬¾Ý´ËÅжϣ»
C¡¢±È½ÏÒÒ¡¢±û¿ÉÖª£¬±ûÖм״¼µÄÎïÖʵÄÁ¿ÎªÒÒµÄ2±¶£¬Ñ¹Ç¿Ôö´ó£¬¶ÔÓÚ·´Ó¦CH₃OH£¨g£©?CO£¨g£©+2H₂£¨g£©£¬Æ½ºâÏòÉú³É¼×´¼µÄ·½ÏòÒƶ¯£»
D¡¢¼×¡¢ÒÒ´¦ÓÚÏàͬµÄƽºâ״̬£¬Ôò¦Á₁+¦Á₂=1£¬ÓÉCµÄ·ÖÎö¿ÉÖª¦Á₂£¾¦Á₃£¬¾Ý´ËÅжϣ»
¢ÚÓÉ2H₂£¨g£©+CO£¨g£©?CH₃OH£¨g£©Æðʼ·´Ó¦ÎïͶÈëÁ¿2mol H₂¡¢l mol CO£¬É迪ʼʱÇâÆøѹǿΪ2P£¬COѹǿΪP£¬ÁÐÈý¶Îʽ£¬¸ù¾Ýƽºâ³£ÊýKp=6.0¡Á10^-3£¨kPa£©^-2£¬¼ÆËã³öƽºâʱµÄ×Üѹ£¬×îºó¸ù¾ÝCH₃OHµÄÎïÖʵÄÁ¿·ÖÊýΪ$\frac{·Öѹ}{×Üѹ}$½øÐмÆË㣮

½â´ð ½â£º£¨1£©¢Ù°±Ë®ÓëCO₂·´Ó¦Éú³É̼ËáÇâ泥¬·´Ó¦µÄ·½³ÌʽΪCO₂+NH₃•H₂O¨TNH₄HCO₃£»¹Ê´ð°¸Îª£ºCO₂+NH₃•H₂O¨TNH₄HCO₃£»
¢Ú¸ù¾Ýͼ1·ÖÎö£¬CO₂ÍѳýЧÂÊËæÎüÊÕ¼ÁµÄpHÔö´ó¶øÔö´ó£»ÒòΪÑÌÆøͨÈË°±Ë®µÄÁ÷Á¿Îª0.052m³/h£¨±ê×¼×´¿ö£©£¬ËùÒÔ30minͨÈëµÄÑÌÆøÖꬶþÑõ»¯Ì¼Ìå»ýΪ0.052m³/h¡Á0.5h¡Á12%=0.00312m³£¬ÔòÎïÖʵÄÁ¿Îª$\frac{0.00312¡Á10{\;}^{3}L}{22.4L/mol}$¡Á91.6%=0.13mol£¬¹Ê´ð°¸Îª£ºCO₂ÍѳýЧÂÊËæÎüÊÕ¼ÁµÄpHÔö´ó¶øÔö´ó£»0.13mol£»
¢ÛÒòΪζȸßÉú³ÉµÄ̼ËáÇâï§Ò×·Ö½âÉú³É¶þÑõ»¯Ì¼£¬ËùÒÔ´Ó40¡æµ½45¡æÍѳýCO₂ЧÂʽµµÍ£¬¹Ê´ð°¸Îª£ºÌ¼ËáÇâï§ÊÜÈÈÒ×·Ö½âÉú³É¶þÑõ»¯Ì¼£»
£¨2£©¢ÙA¡¢¼×¡¢±ûÏà±È½Ï£¬°Ñ¼×µÈЧΪ¿ªÊ¼¼ÓÈë1molCH₃OH£¬±ûÖм״¼µÄÎïÖʵÄÁ¿Îª¼×µÄ2±¶£¬Ñ¹Ç¿Ôö´ó£¬¶ÔÓÚ·´Ó¦CH₃OH£¨g£©?CO£¨g£©+2H₂£¨g£©£¬Æ½ºâÏòÉú³É¼×´¼µÄ·½ÏòÒƶ¯£¬¹Ê2c₁£¼c₃£¬¹ÊA´íÎó£»
B¡¢ÓÉÌâÄ¿¿ÉÖªÉú³É1molCH₃OHµÄÄÜÁ¿±ä»¯Îª90.8kJ£¬¼×¡¢ÒÒƽºâ״̬Ïàͬ£¬Áîƽºâʱ¼×´¼Îªnmol£¬¶ÔÓÚ¼×ÈÝÆ÷£¬a=90.8n£¬µÈÓÚÒÒÈÝÆ÷b=90.8£¨1-n£©£¬¹Êa+b=90.8£¬¹ÊB´íÎó£»
C¡¢±È½ÏÒÒ¡¢±û¿ÉÖª£¬±ûÖм״¼µÄÎïÖʵÄÁ¿ÎªÒÒµÄ2±¶£¬Ñ¹Ç¿Ôö´ó£¬¶ÔÓÚ·´Ó¦CH₃OH£¨g£©?CO£¨g£©+2H₂£¨g£©£¬Æ½ºâÏòÉú³É¼×´¼µÄ·½ÏòÒƶ¯£¬¹Ê2p₂£¾p₃£¬¹ÊC´íÎó£»
D¡¢¼×¡¢ÒÒ´¦ÓÚÏàͬµÄƽºâ״̬£¬Ôò¦Á₁+¦Á₂=1£¬ÓÉCµÄ·ÖÎö¿ÉÖª¦Á₂£¾¦Á₃£¬ËùÒÔa₁+a₃£¼1£¬¹ÊDÕýÈ·£¬
¹ÊÑ¡D£»
¢ÚÓÉ      2H₂£¨g£©+CO£¨g£©?CH₃OH£¨g£©Æðʼ·´Ó¦ÎïͶÈëÁ¿2mol H₂¡¢l mol CO£¬É迪ʼʱÇâÆøѹǿΪ2P£¬COѹǿΪP£¬
¿ªÊ¼£¨kPa£© 2P       P        0
ת»¯£¨kPa£©48.0     24.0    24.0
ƽºâ£¨kPa£©2P-48    P-24    24
Kp=$\frac{24}{£¨2P-48£©{\;}^{2}¡Á£¨P-24£©}$=6.0¡Á10^-3£¨kPa£©^-2£¬½âµÃP=34.0kPa£¬Ôòƽºâ»ìºÏÆøÌåÖÐCH₃OHµÄÎïÖʵÄÁ¿·ÖÊýΪ$\frac{·Öѹ}{×Üѹ}$=$\frac{24}{30+24}$¡Á100%=44.4%£¬¹Ê´ð°¸Îª£º44.4%£»

µãÆÀ ±¾Ì⿼²é¶þÑõ»¯Ì¼µÄÎüÊÕЧÂʵÄʵÑé̽¾¿¡¢µÈЧƽºâÎÊÌâÒÔ¼°Æ½ºâ³£ÊýµÄÏà¹Ø¼ÆËãµÈ£¬ÄѶȽϴó£¬×¢Òâƽºâ״̬ÔÚÀí½âµÈЧƽºâÎÊÌâµÄÓ¦Óã®