ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÒÑÖª£º3CH4(g) + 2N2(g)3C(s) + 4NH3(g) ¦¤H£¾0£¬ÔÚ700¡æ£¬CH4ÓëN2ÔÚ

²»Í¬ÎïÖʵÄÁ¿Ö®±È[n(CH4)/n(N2)]ʱCH4µÄƽºâת»¯ÂÊÈçÏÂͼËùʾ£º

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A. n(CH4)/n(N2)Ô½´ó£¬CH4µÄת»¯ÂÊÔ½¸ß

B. n(CH4)/n(N2)²»±äʱ£¬ÈôÉýΣ¬NH3µÄÌå»ý·ÖÊý»áÔö´ó

C. bµã¶ÔÓ¦µÄƽºâ³£Êý±ÈaµãµÄ´ó

D. aµã¶ÔÓ¦µÄNH3µÄÌå»ý·ÖÊýԼΪ26%

¡¾´ð°¸¡¿B

¡¾½âÎö¡¿ÊÔÌâ·ÖÎö£ºA£®ÓÉͼÏó¿´³ö£¬CH4µÄת»¯ÂÊËæ×ŵÄÔö´ó¶ø½µµÍ£¬¹ÊA´íÎó£»B£®¡÷H£¾0£¬¸Ã·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬Éý¸ßζÈƽºâÕýÏòÒƶ¯£¬NH3µÄÌå»ý·ÖÊý»áÔö´ó£¬¹ÊBÕýÈ·£»C£®abÁ½µãµÄζÈÏàͬ£¬Æ½ºâ³£ÊýÖ»ÓëζÈÓйأ¬Ôòƽºâ³£Êý²»±ä£¬¹ÊC´íÎó£»D£®aµã¼×Íéת»¯ÂÊΪ22%£¬ =0.75£¬ÔòÉè¼×ÍéΪ3mol£¬µªÆøΪ4mol£¬

3CH4(g)+2N2(g)3C(s)+4NH3(g)¡÷H£¾0

¿ªÊ¼ 3 4 0

ת»¯ 0.66 0.44 0.88

ƽºâ 2.34 3.56 0.88

ÔòNH3µÄÌå»ý·ÖÊýԼΪ¡Á100%=13%£¬¹ÊD´íÎó£»¹ÊÑ¡B¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Îª¼õСCO2¶Ô»·¾³µÄÓ°Ï죬ÔÚÏÞÖÆÆäÅÅ·ÅÁ¿µÄͬʱ£¬Ó¦¼ÓÇ¿¶ÔCO2´´ÐÂÀûÓõÄÑо¿¡£

(1)¢Ù °Ñº¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøͨÈë±¥ºÍK2CO3ÈÜÒº¡£

¢Ú ÔÚ¢ÙµÄÎüÊÕÒºÖÐͨ¸ßÎÂË®ÕôÆøµÃµ½¸ßŨ¶ÈµÄCO2ÆøÌå¡£

д³ö¢ÚÖз´Ó¦µÄ»¯Ñ§·½³Ìʽ________¡£

(2)È罫CO2ÓëH2 ÒÔ1:3µÄÌå»ý±È»ìºÏ¡£

¢ÙÊʵ±Ìõ¼þϺϳÉijÌþºÍË®£¬¸ÃÌþÊÇ_______(ÌîÐòºÅ)¡£

A£®ÍéÌþ B£®Ï©Ìþ C£®È²Ìþ D£®±½µÄͬϵÎï

¢Ú Êʵ±Ìõ¼þϺϳÉȼÁϼ״¼ºÍË®¡£ÔÚÌå»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë2 mol CO2ºÍ6 mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g) ¡÷H£½£­49.0 kJ/mol¡£

²âµÃCO2(g)ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£

´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬v(H2)£½______£»ÇâÆøµÄת»¯ÂÊ£½_______£»ÄÜʹƽºâÌåϵÖÐn(CH3OH)Ôö´óµÄ´ëÊ©ÓÐ______¡£

(3)È罫CO2ÓëH2 ÒÔ1:4µÄÌå»ý±È»ìºÏ£¬ÔÚÊʵ±µÄÌõ¼þÏ¿ÉÖƵÃCH4¡£

ÒÑÖª£º

CH4 (g) + 2O2(g) CO2(g)+ 2H2O(l) ¦¤H1£½¨D 890.3 kJ/mol

H2(g) + 1/2O2(g) H2O(l) ¦¤H2£½£­285.8 kJ/mol

ÔòCO2(g)ÓëH2(g)·´Ó¦Éú³ÉCH4(g)ÓëҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽÊÇ________¡£

(4)ijͬѧÓóÁµí·¨²â¶¨º¬ÓнϸßŨ¶ÈCO2µÄ¿ÕÆøÖÐCO2µÄº¬Á¿£¬¾­²éµÃһЩÎïÖÊÔÚ20¡æµÄÊý¾ÝÈçÏÂ±í¡£

Èܽâ¶È(S)/g

ÈܶȻý(Ksp)

Ca(OH)2

Ba(OH)2

CaCO3

BaCO3

0.16

3.89

2.9¡Á10-9

2.6¡Á10-9

(˵Ã÷£ºKspԽС£¬±íʾ¸ÃÎïÖÊÔÚË®ÈÜÒºÖÐÔ½Ò׳Áµí)

ÎüÊÕCO2×îºÏÊʵÄÊÔ¼ÁÊÇ_________[Ìî¡°Ca(OH)2¡±»ò¡°Ba(OH)2¡±]ÈÜÒº£¬ÊµÑéʱ³ýÐèÒª²â¶¨¹¤Òµ·ÏÆøµÄÌå»ý(ÕÛËã³É±ê×¼×´¿ö)Í⣬»¹ÐèÒª²â¶¨__________¡£

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

¾«Ó¢¼Ò½ÌÍø