ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÖпÆÔº´óÁ¬»¯Ñ§ÎïÀíÑо¿ËùµÄÒ»Ïî×îгɹûʵÏÖÁ˼×Íé¸ßЧÉú²úÒÒÏ©£¬ÈçͼËùʾ£¬¼×ÍéÔÚ´ß»¯×÷ÓÃÏÂÍÑÇ⣬ÔÚ²»Í¬Î¶ÈÏ·ֱðÐγɵÈ×ÔÓÉ»ù£¬ÔÚÆøÏàÖо×ÔÓÉ»ù£ºCH2żÁª·´Ó¦Éú³ÉÒÒÏ©(¸Ã·´Ó¦¹ý³Ì¿ÉÄæ)
ÎïÖÊ | ȼÉÕÈÈ/(kJmol-1) |
ÇâÆø | 285.8 |
¼×Íé | 890.3 |
ÒÒÏ© | 1411.0 |
(1)ÒÑÖªÏà¹ØÎïÖʵÄȼÉÕÈÈÈçÉϱíËùʾ£¬Ð´³ö¼×ÍéÖƱ¸ÒÒÏ©µÄÈÈ»¯Ñ§·½³Ìʽ______________¡£
(2)ÏÖ´úʯÓÍ»¯¹¤²ÉÓÃAg×÷´ß»¯¼Á£¬¿ÉʵÏÖÒÒÏ©ÓëÑõÆøÖƱ¸X(·Ö×ÓʽΪC2H4O£¬²»º¬Ë«¼ü)¸Ã·´Ó¦·ûºÏ×îÀíÏëµÄÔ×Ó¾¼Ã£¬Ôò·´Ó¦²úÎïÊÇ____________(Ìî½á¹¹¼òʽ)
(3)ÔÚ400¡æʱ£¬Ïò³õʼÌå»ýΪ1LµÄºãѹÃܱշ´Ó¦£¬Æ÷ÖгäÈë1 mol CH4£¬·¢Éú(1)Öз´Ó¦£¬²âµÃƽºâ»ìºÏÆøÌåÖÐC2H4µÄÌå»ý·ÖÊýΪ25.0%¡£Ôò£º
¢ÙÔÚ¸ÃζÈÏ£¬Æäƽºâ³£ÊýKC£½____________¡£
¢ÚÈôÏò¸Ã·´Ó¦Æ÷ÖÐͨÈë¸ßÎÂË®ÕôÆø(²»²Î¼Ó·´Ó¦£¬¸ßÓÚ400¡æ)£¬ÔòC2H4µÄ²úÂÊ____________¡£(Ìî¡°Ôö´ó¡±¡°¼õС¡±¡°²»±ä¡±»ò¡°ÎÞ·¨È·¶¨¡±)£¬ÀíÓÉÊÇ_______________________________¡£
¢ÛÈô·´Ó¦Æ÷µÄÌå»ý¹Ì¶¨£¬²»Í¬Ñ¹Ç¿Ï¿ɵñ仯ÈçÓÒͼËùʾ£¬ÔòѹǿP1£¬P2µÄ´óС¹ØϵÊÇ____________¡£
(4)ʵ¼ÊÖƱ¸C2H4ʱ£¬Í¨³£´æÔÚ¸±·´Ó¦2CH4(g)C2H4(g)£«2H2(g)¡£·´Ó¦Æ÷ºÍCH4ÆðʼÁ¿²»±ä£¬²»Í¬Î¶ÈÏÂC2H6ºÍC2H4µÄÌå»ý·ÖÊýÓëζȵĹØϵÇúÏßÈçÓÒͼËùʾ¡£ÔÚζȸßÓÚ600¡æʱ£¬ÓпÉÄܵõ½Ò»Öֽ϶àµÄ˫̼Óлú¸±²úÎïµÄÃû³ÆÊÇ____________¡£
(5)C2H4¡¢C2H6³£³£×÷ΪȼÁϵç³ØµÄÔÁÏ£¬Çëд³öC2H4ÔÚNaOHÈÜÒºÖÐ×öȼÁϵç³ØµÄ¸º¼«µÄµç¼«·´Ó¦·½³Ìʽ________________________________________________¡£
¡¾´ð°¸¡¿2CH4(g)C2H4(g)+2H2(g) ¦¤H=+202.0kJ/mol) 1.0 Ôö´ó ¸Ã·´Ó¦ÎªÆøÌåÌå»ýÔö´óµÄÎüÈÈ·´Ó¦£¬Í¨Èë¸ßÎÂË®ÕôÆûÏ൱ÓÚ¼ÓÈÈ£¬Í¬Ê±Í¨ÈëË®ÕôÆø£¬·´Ó¦Æ÷µÄÌå»ýÔö´ó£¬Ï൱ÓÚ¼õСѹǿ£¬¾ùʹƽºâÓÒÒÆ£¬C2H4µÄ²úÂÊÔö´ó P1>P2 ÒÒȲ 16OH-+C2H4-12e-=2CO32-+10H2O
¡¾½âÎö¡¿
(1)¸ù¾Ý±í¸ñÖÐÊý¾ÝÓУº¢ÙH2(g)+O2(g)¨TH2O(l)¡÷H1=-285.8kJ/mol£»
¢ÚCH4(g)+2O2(g)=CO2(g)+2H2O(l)¡÷H2=-890.3kJ/mol£»
¢ÛC2H4(g)+3O2(g)=2CO2(g)+2H2O(l)¡÷H3=-1411.0kJ/mol£»
¼×ÍéÖƱ¸ÒÒÏ©µÄ»¯Ñ§·½³ÌʽΪ£º2CH4(g)C2H4(g)+2H2(g)£¬¸ù¾Ý¸Ç˹¶¨ÂÉ£¬½«¢Ú¡Á2-¢Û-¢Ù¡Á2µÃµ½£¬2CH4(g)C2H4(g)+2H2(g)¡÷H=2¡÷H2-¡÷H3-2¡÷H1=+202.0kJ/mol£»
(2)ÓÉÌâÒâÒÒÏ©ÓëÑõÆø´ß»¯ÖƱ¸X£¬XµÄ·Ö×ÓʽC2H4O£¬²»º¬Ë«¼ü£¬·´Ó¦·ûºÏ×îÀíÏëµÄÔ×Ó¾¼Ã¿ÉÖª£¬XµÄ½á¹¹¼òʽΪ£»
(3)¢Ù400¡æʱ£¬Ïò1LµÄºãÈÝ·´Ó¦Æ÷ÖгäÈë1molCH4£¬·¢ÉúÉÏÊö·´Ó¦£¬²âµÃƽºâ»ìºÏÆøÌåÖÐC2H4µÄÌå»ý·ÖÊýΪ25.0%£¬Éèת»¯µÄ¼×ÍéΪx£¬Óɴ˽¨Á¢ÈçÏÂÈý¶Îʽ£º
ÏàͬÌõ¼þÏÂÆøÌåµÄÌå»ý±ÈµÈÓÚÎïÖʵÄÁ¿Ö®±È£¬ËùÒÔÓÐ=25%£¬½âµÃx=mol£¬Æ½ºâʱÆøÌåµÄ×ÜÎïÖʵÄÁ¿Îª£¬ÈÝÆ÷ºãѹ£¬ÔòÆøÌåµÄÎïÖʵÄÁ¿Ö®±ÈµÈÓÚÌå»ýÖ®±È£¬ÔÌå»ýΪ1L£¬Ôò´ËʱÌå»ýӦΪL£¬ËùÒÔ»¯Ñ§Æ½ºâ³£ÊýΪƽºâ³£ÊýK==1.0£»
¢Ú¼×ÍéÖƱ¸ÒÒÏ©µÄ·´Ó¦ÎªÆøÌåÌå»ýÔö´óµÄÎüÈÈ·´Ó¦£¬Í¨Èë¸ßÎÂË®ÕôÆø£¬Ï൱ÓÚ¼ÓÈÈ£¬Æ½ºâÓÒÒÆ£¬²úÂÊÔö´ó£»Í¬Ê±Í¨ÈëË®ÕôÆø£¬ÈÝÆ÷µÄÌå»ýÔö´ó£¬Ï൱ÓÚ¼õСѹǿ£¬Æ½ºâÓÒÒÆ£¬²úÂÊÒ²Ôö´ó£¬Òò´ËC2H4µÄ²úÂʽ«Ôö´ó£»
¢ÛÈôÈÝÆ÷Ìå»ý¹Ì¶¨£¬¼×ÍéÖƱ¸ÒÒÏ©µÄ·´Ó¦ÎªÆøÌåÌå»ýÔö´óµÄ·´Ó¦£¬Î¶ÈÏàͬʱ£¬Ôö´óѹǿƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬CH4µÄƽºâת»¯ÂʽµµÍ£¬Òò´ËP1>P2£»
(4) ¼×ÍéÔÚ²»Í¬Î¶ÈÏ·ֱðÐγɵÈ×ÔÓÉ»ù£¬¾Ýͼ¿É֪ζȸßÓÚ600¡æʱ£¬Ó¦Óн϶àµÄ×ÔÓÉ»ùÉú³É£¬×ÔÓÉ»ù½áºÏÉú³É£¬Ôò˫̼Óлú¸±²úÎïΪÒÒȲ£»
(5)ȼÁϵç³ØµÄ¸º¼«»¯ºÏ¼ÛÉý¸ß£¬C2H4ÔÚNaOHÈÜÒºÖÐת»¯ÎªCO32-£¬µç¼«·´Ó¦Ê½Îª£ºC2H4-12e-+16OH-=2CO32-+10H2O¡£
¡¾ÌâÄ¿¡¿Ä³Ñо¿Ð¡×éͬѧÓû̽¾¿Ä³´ü³¨¿Ú·ÅÖÃÒ»¶Îʱ¼äµÄÃûΪ¡°ÁòËáÑÇÌú¼ÒÍ¥Ô°ÒÕ¾«Æ··ÊÁÏ¡±µÄ»¯·ÊµÄÖ÷Òª³É·Ö¼°Ïà¹ØÐÔÖÊ¡£Ê×ÏȶԸû¯·ÊµÄ³É·Ö½øÐÐÁËÈçϼÙÉ裺
a£®Ö»º¬ÓÐFeSO4
b£®º¬ÓÐFeSO4ºÍFe2(SO4)3
c£®Ö»º¬ÓÐFe2(SO4)3
½«»¯·Ê¹ÌÌå·ÛÄ©ÈÜÓÚË®Öеõ½ÈÜÒº(¼ÇΪX)£¬½øÐÐÈçÏÂʵÑ飺
ʵÑéÐòºÅ | ²Ù×÷ | ÏÖÏó |
¢¡ | È¡2 mLÈÜÒºX£¬¼ÓÈë1 mL 1 mol¡¤L£1 NaOHÈÜÒº | ²úÉúºìºÖÉ«³Áµí |
¢¢ | È¡2 mLÈÜÒºX£¬¼ÓÈë1µÎKSCN | ÈÜÒºÏÔºìÉ« |
£¨1£©ÇëÓÃÎÄ×Ö±íÊö×ö³ö¼ÙÉèbµÄÒÀ¾ÝÊÇ__________________________¡£
£¨2£©¶ÔʵÑ颡µÄÔ¤ÆÚÏÖÏóÊDzúÉú°×É«³Áµí¡¢±äΪ»ÒÂÌÉ«¡¢×îºó³öÏÖºìºÖÉ«³Áµí£¬Ô¤ÆÚ²úÉú¸ÃÏÖÏóµÄÒÀ¾ÝÊÇ(Óû¯Ñ§·½³Ìʽ»òÀë×Ó·½³Ìʽ±í´ï)_____¡¢_____¡£
£¨3£©ÓÉʵÑ颢µÃ³öµÄ½áÂÛÊÇ____________¡£½áºÏʵÑ颡¡¢¢¢£¬ÍƲâʵÑ颡ʵ¼ÊÏÖÏóÓëÔ¤ÆÚÏÖÏó²»·ûµÄÔÒò¿ÉÄÜÊÇ_____________________________¡£Îª½øÒ»²½ÑéÖ¤¼ÙÉ裬С×éͬѧ½øÐÐÁËÒÔÏÂʵÑ飺
ʵÑéÐòºÅ | ²Ù×÷ | ÏÖÏó |
¢£ | È¡2 mLÈÜÒºX£¬¼ÓÈë1µÎKSCN£¬ÔÙ¼ÓÈë1 mLË® | ÈÜÒºÏÔºìÉ« |
¢¤ | È¡2 mLÈÜÒºX£¬¼ÓÈë1µÎKSCN£¬ÔÙ¼ÓÈë1 mLÂÈË® | ÈÜÒºÏÔºìÉ«£¬ÑÕÉ«±È¢£Éî |
£¨4£©ÊµÑ颤ÖÐÂÈË®²Î¼Ó·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_______________________¡£
£¨5£©Í¨¹ýÒÔÉÏʵÑ飬¿ÉµÃµ½µÄ½áÂÛÊÇ_____________________________£¬ÇëÍêÕû±í´ï¸Ã½áÂÛÊÇÈçºÎµÃ³öµÄ_______________________________¡£