ÌâÄ¿ÄÚÈÝ

ÓÐÈËÉè¼Æ³öÀûÓÃCH4ºÍO2µÄ·´Ó¦£¬Óò¬µç¼«ÔÚKOHÈÜÒºÖй¹³ÉÔ­µç³Ø¡£µç³ØµÄ×Ü·´Ó¦ÀàËÆÓÚCH4ÔÚO2ÖÐȼÉÕ£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨    £©
¢ÙÿÏûºÄ1molCH4¿ÉÒÔÏòÍâµç·Ìṩ8mole-
¢Ú¸º¼«ÉÏCH4ʧȥµç×Ó£¬µç¼«·´Ó¦Ê½CH4+10OH-£­8e-=CO32-+7H2O
¢Û¸º¼«ÉÏÊÇO2»ñµÃµç×Ó£¬µç¼«·´Ó¦Ê½Îª O2+2H2O+4e-=4OH-
¢Üµç³Ø·Åµçºó£¬ÈÜÒºpH²»¶ÏÉý¸ß
A£®¢Ù¢ÚB£®¢Ù¢ÛC£®¢Ù¢ÜD£®¢Û¢Ü
A

ÊÔÌâ·ÖÎö£º¼×ÍéµÄÑõ»¯²úÎïÊÇCO2£¬·´Ó¦ÖÐ×ªÒÆ8¸öµç×Ó£¬¢ÙÕýÈ·£»ÓÉÓÚÈÜÒºÊÇÏÔ¼îÐԵģ¬ÔòCO2±»ÇâÑõ»¯¼ØÎüÊÕÉú³É̼Ëá¼Ø£¬¢ÚÕýÈ·£»ÑõÆøÔÚÕý¼«µÃµ½µç×Ó£¬¢Û²»ÕýÈ·£»·´Ó¦ÊÇÏûºÄÇâÑõ»¯¼ØµÄ£¬ËùÒÔÈÜÒºµÄ¼îÐÔ½µµÍ£¬pH¼õС£¬¢Ü²»ÕýÈ·£¬´ð°¸Ñ¡A¡£
µãÆÀ£º¸ÃÌâÊǸ߿¼Öеij£¼ûÌâÐÍ£¬ÊôÓÚÖеÈÄѶȵÄÊÔÌâ¡£ÊÔÌâ×ÛºÏÐÔÇ¿£¬²àÖØ¶ÔѧÉú»ù´¡ÖªÊ¶µÄ¹®¹ÌºÍѵÁ·£¬ÓÐÀûÓÚÅàÑøÑ§ÉúµÄÂß¼­ÍÆÀíÄÜÁ¦£¬Ìá¸ßѧÉúÁé»îÔËÓûù´¡ÖªÊ¶½â¾öʵ¼ÊÎÊÌâµÄÄÜÁ¦¡£¸ÃÌâµÄ¹Ø¼üÊÇÃ÷È·Ô­µç³ØµÄ¹¤×÷Ô­Àí£¬È»ºóÁé»îÔËÓü´¿É¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ȼÁÏµç³ØÊÇÀûÓÃȼÁÏ£¨ÈçCO¡¢H2¡¢CH4µÈ£©ÓëÑõÆø·´Ó¦£¬½«·´Ó¦²úÉúµÄ»¯Ñ§ÄÜת±äΪµçÄܵÄ×°Öã¬Í¨³£ÓÃÇâÑõ»¯¼Ø×÷Ϊµç½âÖÊÈÜÒº¡£
£¨1£©Íê³ÉÏÂÁйØÓÚ¼×Í飨CH4£©È¼ÁÏµç³ØµÄÌî¿Õ£º
¢Ù¼×ÍéÓëÑõÆø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                          
¢ÚÒÑ֪ȼÁÏµç³ØµÄ×Ü·´Ó¦Ê½ÎªCH4+2O2+2KOH==K2CO3+3H2O£¬µç³ØÖÐÓÐÒ»¼«µÄµç¼«·´Ó¦ÎªCH4+10OH--8e-="=" CO32-+7H2O£¬Õâ¸öµç¼«ÊÇȼÁÏµç³ØµÄ     £¨Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±£©£¬ÁíÒ»¸öµç¼«Éϵĵ缫·´Ó¦Ê½Îª£º             
¢ÛËæ×Åµç³Ø²»¶Ï·Åµç£¬µç½âÖÊÈÜÒºµÄ¼îÐÔ         £¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©
¢Üͨ³£Çé¿öÏ£¬¼×ÍéȼÁÏµç³ØµÄÄÜÁ¿ÀûÓÃÂÊ        £¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©¼×ÍéȼÉÕµÄÄÜÁ¿ÀûÓÃÂÊ¡£
£¨2£©¾Ý±¨µÀ£¬×î½üĦÍÐÂÞÀ­£¨MOTOROLA£©¹«Ë¾Ñз¢ÁËÒ»ÖÖÓɼ״¼ºÍÑõÆøÒÔ¼°Ç¿¼î×öµç½âÖÊÈÜÒºµÄÐÂÐÍÊÖ»úµç³Ø£¬µçÁ¿ÊÇÏÖÓÃÄøÇâµç³ØºÍï®µç³ØµÄ10±¶£¬¿ÉÁ¬ÐøÊ¹ÓÃ1¸öÔ³äµçÒ»´Î¡£¼Ù¶¨·Åµç¹ý³ÌÖУ¬¼×´¼ÍêÈ«Ñõ»¯²úÉúµÄCO2±»³ä·ÖÎüÊÕÉú³ÉCO32£­¡£
¢Ù¸Ãµç³Ø·´Ó¦µÄ×ÜÀë×Ó·½³ÌʽΪ____________________________
¢Ú¼×´¼ÔÚ____¼«·¢Éú·´Ó¦£¨ÌîÕý»ò¸º£©£¬µç³ØÔڷŵç¹ý³ÌÖÐÈÜÒºµÄpH½«____£¨Ìî½µµÍ»òÉÏÉý¡¢²»±ä£©¡£
£¨15·Ö£©¿ÆÑ§¼ÒÈÏΪ£¬ÇâÆøÊÇÒ»ÖÖ¸ßЧ¶øÎÞÎÛȾµÄÀíÏëÄÜÔ´£¬½ü20ÄêÀ´£¬¶ÔÒÔÇâÆø×÷ΪδÀ´µÄ¶¯Á¦È¼ÁÏÇâÄÜÔ´µÄÑо¿»ñµÃÁËѸËÙ·¢Õ¹¡£
£¨1£©ÎªÁËÓÐЧ·¢Õ¹ÃñÓÃÇâÄÜÔ´£¬Ê×ÏȱØÐëÖÆµÃÁ®¼ÛµÄÇâÆø£¬ÏÂÁпɹ©¿ª·¢ÓֽϾ­¼ÃÇÒ×ÊÔ´¿É³ÖÐøÀûÓõÄÖÆÇâÆøµÄ·½·¨ÊÇ         ¡££¨Ñ¡Ìî×Öĸ£©
A£®µç½âË®B£®Ð¿ºÍÏ¡ÁòËá·´Ó¦
C£®¹â½âº£Ë®D£®·Ö½âÌìÈ»Æø
£¨2£©ÓÃË®·Ö½â»ñµÃÇâÆøµÄÄÜÁ¿±ä»¯ÈçÓÒͼËùʾ£¬±íʾʹÓô߻¯¼ÁÊÇÇúÏß         ¡£¸Ã·´Ó¦Îª         (·ÅÈÈ»¹ÊÇÎüÈÈ)·´Ó¦

£¨3£©1gµÄÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮÊͷųö142.9kJµÄÈÈÁ¿Ð´³öÆäÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º                               ¡£
£¨4£©ÀûÓÃÇâÆøºÍCOºÏ³É¶þ¼×ÃѵÄÈý²½·´Ó¦ÈçÏ£º
¢Ù 2H2(g) + CO(g)  CH3OH(g)£»¦¤H £½£­90.8 kJ¡¤mol£­1
¢Ú 2CH3OH(g)  CH3OCH3(g) + H2O(g)£»¦¤H£½£­23.5 kJ¡¤mol£­1
¢Û CO(g) + H2O(g)  CO2(g) + H2(g)£»¦¤H£½£­41.3 kJ¡¤mol£­1
×Ü·´Ó¦£º3H2(g) + 3CO(g)  CH3OCH3(g) + CO2 (g)µÄ¦¤H£½         
£¨5£©ÇâÑõȼÁÏµç³ØÄÜÁ¿×ª»¯Âʸߣ¬¾ßÓйãÀ«µÄ·¢Õ¹Ç°¾°¡£ÏÖÓÃÇâÑõȼÁÏµç³Ø½øÐÐÏÂͼËùʾʵÑ飺

¢ÙÇâÑõȼÁÏµç³ØÖУ¬Õý¼«µÄµç¼«·´Ó¦Ê½Îª                 ¡£
¢ÚÉÏͼװÖÃÖУ¬Ä³Ò»Í­µç¼«µÄÖÊÁ¿¼õÇá3.2g£¬Ôò a ¼«ÉÏÏûºÄµÄO2ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ
         L¡£
£¨6£©ÓÐÈËÉèÏëѰÇóºÏÊʵĴ߻¯¼ÁºÍµç¼«²ÄÁÏ£¬ÒÔN2¡¢H2Ϊµç¼«·´Ó¦ÎÒÔHCl£­NH4ClΪµç½âÖÊÈÜÒºÖÆÈ¡ÐÂÐÍȼÁÏµç³Ø¡£ÔòÕý¼«µç¼«·½³Ìʽ         ¡£

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

¾«Ó¢¼Ò½ÌÍø