ÅжÏÕýÎó£¬ÕýÈ·µÄ»®¡°¡Ì¡±£¬´íÎóµÄ»®¡°¡Á¡±

(1)S(s)£«O2(g)===SO3(g)¡¡¦¤H£½£­315 kJ·mol£­1(ȼÉÕÈÈ)¡¡(¦¤HµÄÊýÖµÕýÈ·)(¡¡¡¡)

(2)NaOH(aq)£«HCl(aq)===NaCl(aq)£«H2O(l)

¦¤H£½£­57.3 kJ·mol£­1(ÖкÍÈÈ)¡¡(¦¤HµÄÊýÖµÕýÈ·)(¡¡¡¡)

(3)ÒÑÖªH£«(aq)£«OH£­(aq)===H2O(l)¡¡¦¤H£½£­57.3 kJ·mol£­1£¬ÔòH2SO4ºÍBa(OH)2·´Ó¦µÄ·´Ó¦ÈȦ¤H£½2¡Á(£­57.3) kJ·mol£­1(¡¡¡¡)

(4)ȼÁϵç³ØÖн«¼×´¼ÕôÆøת»¯ÎªÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽÊÇCH3OH(g)£«O2(g)===CO2(g)£«2H2(g)

¦¤H£½£­192.9 kJ·mol£­1£¬ÔòCH3OH(g)µÄȼÉÕÈÈΪ192.9 kJ·mol£­1(¡¡¡¡)

(5)H2(g)µÄȼÉÕÈÈÊÇ285.8 kJ·mol£­1£¬Ôò2H2O(g)===2H2(g)£«O2(g)¡¡¦¤H£½571.6 kJ·mol£­1(¡¡¡¡)

(6)ÆÏÌÑÌǵÄȼÉÕÈÈÊÇ2 800 kJ·mol£­1£¬ÔòC6H12O6(s)£«3O2(g)===3CO2(g)£«3H2O(l)¡¡¦¤H£½£­1 400 kJ·mol£­1(¡¡¡¡)

(7)ÒÑÖª101 kPaʱ£¬2C(s)£«O2(g)===2CO(g)

¦¤H£½£­221 kJ·mol£­1£¬Ôò¸Ã·´Ó¦µÄ·´Ó¦ÈÈΪ221 kJ·mol£­1(¡¡¡¡)

(8)ÒÑ֪ϡÈÜÒºÖУ¬H£«(aq)£«OH£­(aq)===H2O(l)

¦¤H£½£­57.3 kJ·mol£­1£¬ÔòÏ¡´×ËáÓëÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³É1 molˮʱ·Å³ö57.3 kJµÄÈÈÁ¿(¡¡¡¡)

(9)ÒÑÖªHClºÍNaOH·´Ó¦µÄÖкÍÈȦ¤H£½£­57.3 kJ·mol£­1£¬Ôò98%µÄŨÁòËáÓëÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦Éú³É1 molË®µÄÖкÍÈÈΪ£­57.3 kJ·mol£­1(¡¡¡¡)

(10)CO(g)µÄȼÉÕÈÈÊÇ283.0 kJ·mol£­1£¬Ôò2CO2(g)===2CO(g)£«O2(g)·´Ó¦µÄ¦¤H£½2¡Á283.0 kJ·mol£­1(¡¡¡¡)

(11)ÇâÆøµÄȼÉÕÈÈΪ285.5 kJ·mol£­1£¬Ôòµç½âË®µÄÈÈ»¯Ñ§·½³ÌʽΪ2H2O(l)2H2(g)£«O2(g)¡¡¦¤H£½285.5 kJ·mol£­1(¡¡¡¡)


´ð°¸¡¡(1)¡Á¡¡(2)¡Ì¡¡(3)¡Á¡¡(4)¡Á¡¡(5)¡Á¡¡(6)¡Ì

(7)¡Á¡¡(8)¡Á¡¡(9)¡Á¡¡(10)¡Ì¡¡(11)¡Á


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÖÜÆÚ±íÖеÄÖ÷×嶼ÓзǽðÊôÔªËØ

B£®ÖÜÆÚ±íÖеÄÖ÷×嶼ÓнðÊôÔªËØ

C£®ÖÜÆÚ±íÖеķǽðÊôÔªËض¼Î»ÓÚ¶ÌÖÜÆÚ

D£®ÖÜÆÚ±íÖеĹý¶ÉÔªËض¼ÊǽðÊôÔªËØ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¹ØÓÚ1s¡¢2s¡¢3s¡¢4sÔ­×Ó¹ìµÀµÄ˵·¨£¬ÕýÈ·µÄÊÇ (¡¡¡¡)

A£®µç×ÓÖ»ÄÜÔÚµç×ÓÔÆÂÖÀªÍ¼ÖÐÔ˶¯

B£®¹ìµÀ²»Í¬£¬µç×ÓÔÆÂÖÀªÍ¼ÐÎ×´Ïàͬ

C£®¹ìµÀÊýÄ¿Ïàͬ£¬µç×ÓÔÆÂÖÀªÍ¼ÐÎ×´¡¢´óСÍêÈ«Ïàͬ

D£®Äܲ㲻ͬ£¬µç×ÓÔÆÂÖÀªÍ¼ÐÎ×´Ò²²»Ïàͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÔõÑùÀí½â¿ÉÄæ·´Ó¦Öеķ´Ó¦ÈÈ£¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ʵÑé²âµÃ£º101 kPaʱ£¬1 mol H2ÍêȫȼÉÕÉú³ÉҺ̬ˮ£¬·Å³ö285.8 kJµÄÈÈÁ¿£»1 mol CH4ÍêȫȼÉÕÉú³ÉҺ̬ˮºÍCO2£¬·Å³ö890.3 kJµÄÈÈÁ¿¡£ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽµÄÊéдÕýÈ·µÄÊÇ(¡¡¡¡)

¢ÙCH4(g)£«2O2(g)===CO2(g)£«2H2O(l)

¦¤H£½890.3 kJ·mol£­1

¢ÚCH4(g)£«2O2(g)===CO2(g)£«2H2O(l)

¦¤H£½£­890.3 kJ·mol£­1

¢ÛCH4(g)£«2O2(g)===CO2(g)£«2H2O(g)

¦¤H£½£­890.3 kJ·mol£­1

¢Ü2H2(g)£«O2(g)===2H2O(l)

¦¤H£½£­571.6 kJ·mol£­1

A£®½öÓТڠ B£®½öÓТڢܠ       C£®½öÓТڢۢܠ D£®È«²¿·ûºÏÒªÇó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


LiH¿É×÷·É´¬µÄȼÁÏ£¬ÒÑÖªÏÂÁз´Ó¦£º

¢Ù2Li(s)£«H2(g)===2LiH(s)¡¡¦¤H£½£­182 kJ·mol£­1

¢Ú2H2(g)£«O2(g)===2H2O(l)¡¡¦¤H£½£­572 kJ·mol£­1

¢Û4Li(s)£«O2(g)===2Li2O(s)¡¡¦¤H£½£­1 196 kJ·mol£­1

ÊÔд³öLiHÔÚO2ÖÐȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


Ϊ»º½âÄÜÔ´½ôÕÅ£¬Ô½À´Ô½¶àµÄ¹ú¼Ò¿ªÊ¼ÖØÊÓÉúÎïÖÊÄÜÔ´(ÀûÓÃÄÜÔ´×÷ÎïºÍÓлú·ÏÁÏ£¬¾­¹ý¼Ó¹¤×ª±äΪÉúÎïȼÁϵÄÒ»ÖÖÄÜÔ´)µÄ¿ª·¢ÀûÓá£

(1)ÈçͼÊÇij¹úÄÜÔ´½á¹¹±ÈÀýͼ£¬ÆäÖÐÉúÎïÖÊÄÜÔ´ËùÕ¼µÄ±ÈÀýÊÇ______¡£

 (2)ÉúÎï²ñÓÍÊÇÓɶ¯Ö²ÎïÓÍ֬ת»¯¶øÀ´£¬ÆäÖ÷Òª³É·ÖΪ֬·¾Ëáõ¥£¬¼¸ºõ²»º¬Áò£¬ÉúÎï½µ½âÐԺã¬Ò»Ð©¹ú¼ÒÒѽ«ÆäÌí¼ÓÔÚÆÕͨ²ñÓÍÖÐʹÓ᣹ØÓÚÉúÎï²ñÓͼ°ÆäʹÓã¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ________¡£

¢ÙÉúÎï²ñÓÍÊÇ¿ÉÔÙÉú×ÊÔ´¡¡¢Ú¿É¼õÉÙ¶þÑõ»¯ÁòµÄÅÅ·Å

¢ÛÓëÆÕͨ²ñÓÍÏà±ÈÒ׷ֽ⡡¢ÜÓëÆÕͨ²ñÓÍÖÆÈ¡·½·¨Ïàͬ

A£®¢Ù¢Ú¢Û  B£®¢Ù¢Ú¢Ü  C£®¢Ù¢Û¢Ü  D£®¢Ú¢Û¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÀûÓú¬Ì¼»¯ºÏÎïºÏ³ÉȼÁÏÊǽâ¾öÄÜԴΣ»úµÄÖØÒª·½·¨£¬ÒÑÖªCO(g)£«2H2(g)CH3OH(g)·´Ó¦¹ý³ÌÖеÄÄÜÁ¿±ä»¯Çé¿öÈçͼËùʾ£¬ÇúÏߢñºÍÇúÏߢò·Ö±ð±íʾ²»Ê¹Óô߻¯¼ÁºÍʹÓô߻¯¼ÁµÄÁ½ÖÖÇé¿ö¡£ÏÂÁÐÅжÏÕýÈ·µÄÊÇ(¡¡¡¡)

A£®¸Ã·´Ó¦µÄ¦¤H£½91 kJ·mol£­1

B£®¼ÓÈë´ß»¯¼Á£¬¸Ã·´Ó¦µÄ¦¤H±äС

C£®·´Ó¦ÎïµÄ×ÜÄÜÁ¿´óÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿

D£®Èç¹û¸Ã·´Ó¦Éú³ÉҺ̬CH3OH£¬Ôò¦¤HÔö´ó

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ(¡¡¡¡)

A£®½«AlÌõͶÈëNaOHÈÜÒºÖУºAl£«OH£­£«H2O===AlO£«H2¡ü

B£®Í­ÈÜÓÚÏ¡ÏõËáÖУºCu£«4H£«£«2NO===Cu2£«£«2NO2¡ü£«2H2O

C£®Ì¼ËáÇâ¸ÆÈÜÒºÖмÓÈë¹ýÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£ºHCO£«OH£­===CO£«H2O

D£®Ïò̼ËáÄÆÈÜÒºÖÐÖðµÎ¼ÓÈëÓëÖ®µÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄÏ¡´×Ë᣺CO£«CH3COOH===CH3COO£­£«HCO

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸