³¤Õ÷¶þºÅÀ¦°óʽ»ð¼ýÍÆ½ø¼ÁÒÔëÂ(N2H4)×÷ΪȼÁÏ£¬NO2×÷ΪÑõ»¯¼Á£¬·´Ó¦Éú³ÉN2ºÍË®ÕôÆø¡£ÒÑÖª£ºN2(g) + 2O2(g) £½2NO2(g)£»      ¡÷H£½£«67.7 kJ/mol
N2H4(g) + O2(g) £½ N2(g) + 2H2O(g)£»       ¡÷H£½£­534 kJ/mol
ÏÂÁйØÓÚëºÍNO2·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÖУ¬ÕýÈ·µÄÊÇ

A£®2N2H4(g) + 2NO2(g) £½ 3N2(g) + 4H2O(l)£»¡÷H£½£­1135.7 kJ/mol
B£®N2H4(g) + NO2(g) £½ 3/2N2(g) + 2H2O(g)£»¡÷H£½£­567.85 kJ/mol
C£®N2H4(g) + NO2(g) £½ 3/2N2(g) + 2H2O(l)£»¡÷H£½£­1135.7 kJ/mol
D£®2N2H4(g) + 2NO2(g) = 3N2(g) + 4H2O(g)£»¡÷H£½+1135.7 kJ/mol

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¾«Ó¢¼Ò½ÌÍø½üÄ꣬ÎÒ¹úÔÚº½ÌìÊÂÒµÉÏÈ¡µÃÁËÁîÊÀ½çÖõÄ¿µÄ³É¾Í£¬ÉñÖÛ·É´¬¶à´Î±»³¤Õ÷ϵÁлð¼ýËÍÈëÌ«¿Õ£®
£¨1£©³¤Õ÷¶þºÅÀ¦°óʽ»ð¼ýÍÆ½ø¼ÁÒÔÁª°±£¨N2H4£©×÷ΪȼÁÏ£¬N204×÷ÎªÍÆ½ø¼Á£®
¢ÙN204µÄÖ÷Òª×÷ÓÃÊÇÖúȼ£¬µ«ÆäÔÚ¹¤×÷ʱ»á²úÉúºì×ØÉ«ÆøÌåN02£¬¶Ô»·¾³»áÔì³ÉÎÛȾ£¬Îª±ÜÃâÎÛȾ¿ÉʹÓÃÏÂÁÐ
 
£¨Ìî×Öĸ£©´úÌæÖ®£®
A£®ÒºÌ¬°±       B£®H202         C£®KMn04       D£®ÒºÌ¬Ñõ
¢ÚÔÚ¼îÐÔÌõ¼þÏ£¬ÓÃÄòËØ[CO£¨NH2£©2]ºÍ´ÎÂÈËáÄÆ·´Ó¦¿ÉÒÔÖÆÈ¡Áª°±£¨²úÎïÖÐͬʱÓÐÁ½ÖÖÕýÑΣ©£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
 
£®
£¨2£©·É´¬×ù²ÕÄÚ¿ÕÆøµÄ¸üйý³ÌÈçͼËùʾ
¢Ù×ù²ÕÄÚ¿ÕÆø¸üйý³Ì¿ÉÒÔÑ­»·ÀûÓõÄÎïÖÊΪH2¡¢O2ºÍ
 
£»×°ÖâòÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
¢Ú´Ó×°Öâñ¡¢¢ò¡¢¢ó¿É¿´³öO2µÄÀ´Ô´£¬ÈôÓԱÿÌìÏûºÄ35mol02£¬Ã¿Ììºô³öµÄÆøÌåÖк¬18mol H20£¬Ôòºô³öµÄÆøÌåÖк¬C02
 
mol£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³¤Õ÷¶þºÅÀ¦°óʽ»ð¼ýÍÆ½ø¼ÁÒÔ루N2H4£©×÷ΪȼÁÏ£¬NO2×÷ΪÑõ»¯¼Á£¬·´Ó¦Éú³ÉN2ºÍË®ÕôÆø¡£ÒÑÖª£ºN2(g)+2O2(g)====2NO2(g);¦¤H=+67.7 kg¡¤mol-1

N2H4(g)+O2(g)====N2(g)+2H2O(g);¦¤H=-534 kJ¡¤mol-1

ÏÂÁйØÓÚëºÍNO2·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÖУ¬ÕýÈ·µÄÊÇ

A.2N2H4£¨g£©+2NO2(g) ====3N2(g)+4H2O(l);¦¤H=-1 135.7 kJ¡¤mol-1

B.2N2H4(g)+2NO2(g) ====3N2(g)+4H2O(g);¦¤H=-1 000.3 kJ¡¤mol-1

C.N2H4(g)+NO2(g) ====3/2N2(g)+2H2O(l);¦¤H=-1 135.7 kJ¡¤mol-1

D.2N2H4(g)+2NO2(g) ====3N2(g)+4H2O(g);¦¤H=-1 135.7 kJ¡¤mol-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê½­ËÕÊ¡ÐìÖÝÊиßÈýµÚÈý´Îµ÷Ñп¼ÊÔ£¨»¯Ñ§ÊÔÌ⣩ ÌâÐÍ£ºÌî¿ÕÌâ

½üÄ꣬ÎÒ¹úÔÚº½ÌìÊÂÒµÉÏÈ¡µÃÁËÁîÊÀ½çÖõÄ¿µÄ³É¾Í£¬ÉñÖÛ·É´¬¶à´Î±»³¤Õ÷ϵ
Áлð¼ýËÍÈëÌ«¿Õ¡£
(1)³¤Õ÷¶þºÅÀ¦°óʽ»ð¼ýÍÆ½ø¼ÁÒÔÁª°±()×÷ΪȼÁÏ£¬×÷ÎªÍÆ½ø¼Á¡£
¢ÙµÄÖ÷Òª×÷ÓÃÊÇÖúȼ£¬µ«ÆäÔÚ¹¤×÷ʱ»á²úÉúºì×ØÉ«ÆøÌ壬¶Ô»·¾³»áÔì
³ÉÎÛȾ£¬Îª±ÜÃâÎÛȾ¿ÉʹÓÃÏÂÁР  (Ìî×Öĸ)´úÌæÖ®¡£

A£®ÒºÌ¬°±B£®C£®D£®ÒºÌ¬Ñõ
¢ÚÔÚ¼îÐÔÌõ¼þÏ£¬ÓÃÄòËØºÍ´ÎÂÈËáÄÆ·´Ó¦¿ÉÒÔÖÆÈ¡Áª°±(²úÎïÖÐ
ͬʱÓÐÁ½ÖÖÕýÑÎ)£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ   ¡£
(2)·É´¬×ù²ÕÄÚ¿ÕÆøµÄ¸üйý³ÌÈçÏÂͼËùʾ£º

¢Ù×ù²ÕÄÚ¿ÕÆø¸üйý³Ì¿ÉÒÔÑ­»·ÀûÓõÄÎïÖÊΪ¡¢ºÍ   £»×°ÖâòÖÐ
·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ   ¡£
¢Ú´Ó×°ÖÃI¡¢¢ò¡¢¢ó¿É¿´³öµÄÀ´Ô´£¬ÈôÓԱÿÌìÏûºÄ35mol£¬Ã¿Ììºô³ö
µÄÆøÌåÖк¬18 mol£¬Ôòºô³öµÄÆøÌåÖк¬   mol£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄ긣½¨Ê¡ÏÃÃÅÊиßÈýÉÏѧÆÚ12ÔÂÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÑ¡ÔñÌâ

³¤Õ÷¶þºÅÀ¦°óʽ»ð¼ýÍÆ½ø¼ÁÒÔëÂ(N2H4)×÷ΪȼÁÏ£¬NO2×÷ΪÑõ»¯¼Á£¬·´Ó¦Éú³ÉN2ºÍË®ÕôÆø¡£ÒÑÖª£ºN2(g) + 2O2(g) £½2NO2(g)£»       ¡÷H£½£«67.7 kJ/mol

N2H4(g) + O2(g) £½ N2(g) + 2H2O(g)£»        ¡÷H£½£­534 kJ/mol

ÏÂÁйØÓÚëºÍNO2·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÖУ¬ÕýÈ·µÄÊÇ

A£®2N2H4(g) + 2NO2(g) £½ 3N2(g) + 4H2O(l)£»¡÷H£½£­1135.7 kJ/mol

B£®N2H4(g) + NO2(g) £½ 3/2N2(g) + 2H2O(g)£»¡÷H£½£­567.85 kJ/mol

C£®N2H4(g) + NO2(g) £½ 3/2N2(g) + 2H2O(l)£»¡÷H£½£­1135.7 kJ/mol

D£®2N2H4(g) + 2NO2(g) = 3N2(g) + 4H2O(g)£» ¡÷H£½+1135.7 kJ/mol

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

½üÄ꣬ÎÒ¹úÔÚº½ÌìÊÂÒµÉÏÈ¡µÃÁËÁîÊÀ½çÖõÄ¿µÄ³É¾Í£¬ÉñÖÛ·É´¬¶à´Î±»³¤Õ÷ϵ

  Áлð¼ýËÍÈëÌ«¿Õ¡£

  (1)³¤Õ÷¶þºÅÀ¦°óʽ»ð¼ýÍÆ½ø¼ÁÒÔÁª°±()×÷ΪȼÁÏ£¬×÷ÎªÍÆ½ø¼Á¡£

    ¢ÙµÄÖ÷Òª×÷ÓÃÊÇÖúȼ£¬µ«ÆäÔÚ¹¤×÷ʱ»á²úÉúºì×ØÉ«ÆøÌ壬¶Ô»·¾³»áÔì

    ³ÉÎÛȾ£¬Îª±ÜÃâÎÛȾ¿ÉʹÓÃÏÂÁР   (Ìî×Öĸ)´úÌæÖ®¡£

    A£®ÒºÌ¬°±    B£®    C£®    D£®ÒºÌ¬Ñõ

    ¢ÚÔÚ¼îÐÔÌõ¼þÏ£¬ÓÃÄòËØºÍ´ÎÂÈËáÄÆ·´Ó¦¿ÉÒÔÖÆÈ¡Áª°±(²úÎïÖÐ

    ͬʱÓÐÁ½ÖÖÕýÑÎ)£¬Ôò¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ    ¡£

(2)·É´¬×ù²ÕÄÚ¿ÕÆøµÄ¸üйý³ÌÈçÏÂͼËùʾ£º

    ¢Ù×ù²ÕÄÚ¿ÕÆø¸üйý³Ì¿ÉÒÔÑ­»·ÀûÓõÄÎïÖÊΪ¡¢ºÍ    £»×°ÖâòÖÐ

    ·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ    ¡£

    ¢Ú´Ó×°ÖÃI¡¢¢ò¡¢¢ó¿É¿´³öµÄÀ´Ô´£¬ÈôÓԱÿÌìÏûºÄ35mol£¬Ã¿Ììºô³ö

    µÄÆøÌåÖк¬18 mol£¬Ôòºô³öµÄÆøÌåÖк¬    mol£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸