H2ºÍI2ÔÚÒ»¶¨Ìõ¼þÏÂÄÜ·¢Éú·´Ó¦£ºH2(g)£«I2(g)2HI(g)¡¡¦¤H£½£­a kJ¡¤mol£­1

ÒÑÖª£º

(a¡¢b¡¢c¾ù´óÓÚÁã)

ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®·´Ó¦ÎïµÄ×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿

B£®¶Ï¿ª1 mol H¡ªH¼üºÍ1 mol I¡ªI¼üËùÐèÄÜÁ¿´óÓڶϿª2 mol H¡ªI¼üËùÐèÄÜÁ¿

C£®¶Ï¿ª2 mol H¡ªI¼üËùÐèÄÜÁ¿Ô¼Îª(c£«b£«a) kJ

D£®ÏòÃܱÕÈÝÆ÷ÖмÓÈë2 mol H2ºÍ2 mol I2£¬³ä·Ö·´Ó¦ºó·Å³öµÄÈÈÁ¿Ð¡ÓÚ2a kJ


½âÎö¡¡¸Ã·´Ó¦µÄ¦¤H<0£¬Îª·ÅÈÈ·´Ó¦£¬¹Ê·´Ó¦ÎïµÄ×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬AÏîÕýÈ·£»¸Ã·´Ó¦Îª·ÅÈÈ·´Ó¦£¬Ôò¶Ï¿ª1 mol H¡ªH¼üºÍ1 mol I¡ªI¼üËùÐèµÄÄÜÁ¿Ð¡ÓڶϿª2 mol H¡ªI¼üËùÐèµÄÄÜÁ¿£¬BÏî´íÎ󣻦¤H£½E(H¡ªH)£«E(I¡ªI)£­2E(H¡ªI)£½£­a kJ¡¤mol£­1£¬¹Ê¶Ï¿ª2 mol H¡ªI¼üËùÐèµÄÄÜÁ¿Ô¼Îª(c£«b£«a)kJ£¬CÏîÕýÈ·£»H2ºÍI2µÄ·´Ó¦Îª¿ÉÄæ·´Ó¦£¬2 mol H2ºÍ2 mol I2·´Ó¦²»¿ÉÄÜÉú³É2 mol HI£¬Òò´Ë·Å³öµÄÈÈÁ¿Ð¡ÓÚ2a kJ£¬DÏîÕýÈ·¡£

´ð°¸¡¡B


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑÖªH¡ªH¼üÄÜΪ436 KJ/mol£¬H¡ªN¼üÄÜΪ391KJ/mol£¬¸ù¾Ý»¯Ñ§·½³Ìʽ£ºN2 + 3H2 = 2NH3

 ¦¤H=¡ª92.4 KJ/mol£¬ÔòN¡ÔN¼üµÄ¼üÄÜÊÇ£¨   £©

A£®431 KJ/mol      B£®946 KJ/mol       C£®649 KJ/mol    D£®869 KJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)

¢ÙάÉúËØC¾ßÓл¹Ô­ÐÔ£¬ÔÚÈËÌåÄÚÆð¿¹Ñõ»¯×÷ÓÃ

¢ÚNO2ÈÜÓÚˮʱ·¢ÉúÑõ»¯»¹Ô­·´Ó¦¡¡¢Û1 mol Cl2²Î¼Ó·´Ó¦×ªÒƵç×ÓÊýÒ»¶¨Îª2NA¡¡¢ÜÒõÀë×Ó¶¼Ö»Óл¹Ô­ÐÔ

A£®¢Ù¢Ú                                 B£®¢Ú¢Û

C£®¢Û¢Ü                                 D£®¢Ù¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÏÂÁÐÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ(¡¡¡¡)

A£®ÓÃPtµç¼«µç½âÉÙÁ¿µÄMgCl2ÈÜÒº£º2H2O£«2Cl£­H2¡ü£«Cl2¡ü£«2OH£­

B£®ÏòÇâÑõ»¯ÑÇÌúÖмÓÈë×ãÁ¿µÄÏ¡ÏõË᣺Fe(OH)2£«2H£«===Fe2£«£«2H2O

C£®ÏòNaAlO2ÈÜÒºÖÐͨÈë¹ýÁ¿CO2ÖÆAl(OH)3£ºAlO£«CO2£«2H2O===Al(OH)3¡ý£«HCO

D£®ÓÃʳ´×³ýȥˮƿÖеÄË®¹¸£ºCO£«2CH3COOH===2CH3COO£­£«CO2¡ü£«H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ij¹¤Òµ·ÏË®½öº¬Ï±íÖеÄijЩÀë×Ó£¬ÇÒ¸÷ÖÖÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÏàµÈ£¬¾ùΪ0.1 mol/L(´ËÊýÖµºöÂÔË®µÄµçÀë¼°Àë×ÓµÄË®½â)¡£

ÑôÀë×Ó

K£«

Ag£«

Mg2£«

Cu2£«

Al3£«

NH

ÒõÀë×Ó

Cl£­

CO

NO

SO

SiO

I£­

¼×ͬѧÓû̽¾¿·ÏË®µÄ×é³É£¬½øÐÐÁËÈçÏÂʵÑ飺

¢ñ.È¡¸ÃÎÞÉ«ÈÜÒº5 mL£¬µÎ¼ÓÒ»µÎ°±Ë®ÓгÁµíÉú³É£¬ÇÒÀë×ÓÖÖÀàÔö¼Ó¡£

¢ò.Óò¬Ë¿ÕºÈ¡ÈÜÒº£¬ÔÚ»ðÑæÉÏ×ÆÉÕ£¬Í¸¹ýÀ¶É«îܲ£Á§¹Û²ì£¬ÎÞ×ÏÉ«»ðÑæ¡£

¢ó.ÁíÈ¡ÈÜÒº¼ÓÈë¹ýÁ¿ÑÎËᣬÓÐÎÞÉ«ÆøÌåÉú³É£¬¸ÃÎÞÉ«ÆøÌåÓö¿ÕÆø±ä³Éºì×ØÉ«¡£

¢ô.Ïò¢óÖÐËùµÃµÄÈÜÒºÖмÓÈëBaCl2ÈÜÒº£¬Óа×É«³ÁµíÉú³É¡£

ÇëÍƶϣº

(1)ÓÉ¢ñ¡¢¢òÅжϣ¬ÈÜÒºÖÐÒ»¶¨²»º¬ÓеÄÑôÀë×ÓÊÇ______¡£

(2)¢óÖмÓÈëÑÎËáÉú³ÉÎÞÉ«ÆøÌåµÄÀë×Ó·½³ÌʽÊÇ________________¡£

(3)¼×ͬѧ×îÖÕÈ·¶¨Ô­ÈÜÒºÖÐËùº¬ÑôÀë×ÓÓÐ________£¬ÒõÀë×ÓÓÐ________£»²¢¾Ý´ËÍƲâÔ­ÈÜÒºÓ¦¸Ã³Ê________ÐÔ£¬Ô­ÒòÊÇ________________________(ÇëÓÃÀë×Ó·½³Ìʽ˵Ã÷)¡£

(4)ÁíÈ¡100 mLÔ­ÈÜÒº£¬¼ÓÈë×ãÁ¿µÄNaOHÈÜÒº£¬´Ë¹ý³ÌÖÐÉæ¼°µÄÀë×Ó·½³ÌʽΪ_____________________________________________________

___________________________________________________________¡£

³ä·Ö·´Ó¦ºó¹ýÂË£¬Ï´µÓ£¬×ÆÉÕ³ÁµíÖÁºãÖØ£¬µÃµ½µÄ¹ÌÌåÖÊÁ¿Îª________g¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÒÑÖªÒÒȲÓë±½ÕôÆøÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³ÌʽÈçÏ£º

¢ÙC2H2(g)£«5/2O2(g)¨D¡ú2CO2(g)£«H2O(l)

¦¤H£½£­1 300 kJ/mol

¢ÚC6H6(g)£«15/2O2(g)¨D¡ú6CO2(g)£«3H2O(l)¡¡¦¤H£½£­3 295 kJ/mol

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡¡¡)

A£®1 mol C2H2(g)ÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·ÅÈÈ´óÓÚ1 300 kJ

B£®1 mol C6H6(l)ÍêȫȼÉÕÉú³ÉҺ̬ˮʱ·ÅÈÈ´óÓÚ3 295 kJ

C£®ÏàͬÌõ¼þÏ£¬µÈÖÊÁ¿µÄC2H2(g)ÓëC6H6(g)ÍêȫȼÉÕ£¬C6H6(g)·ÅÈȸü¶à

D£®C2H2(g)Èý¾ÛÉú³ÉC6H6(g)µÄ¹ý³ÌÊôÓÚ·ÅÈÈ·´Ó¦

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¢ÙCaCO3(s)===CaO£«CO2(g)¡¡¦¤H£½177.7 kJ

¢ÚC(s)£«H2O(s)===CO(g)£«H2(g)

¦¤H£½£­131.3 kJ/mol

¢ÛH2SO4(l)£«NaOH(l)===Na2SO4(l)£«H2O(l)

¦¤H£½£­57.3 kJ/mol

¢ÜC(s)£«O2(g)===CO2(g)¡¡¦¤H£½£­393.5 kJ/mol

¢ÝCO(g)£«O2(g)===CO2(g)¡¡¦¤H£½£­283 kJ/mol

¢ÞHNO3(aq)£«NaOH(aq)===NaNO3(aq)£«H2O(l)

¦¤H£½£­57.3 kJ/mol

¢ß2H2(g)£«O2(g)===2H2O(l)

¦¤H£½£­517.6 kJ/mol

(1)ÉÏÊöÈÈ»¯Ñ§·½³ÌʽÖУ¬²»ÕýÈ·µÄÓÐ________£¬²»ÕýÈ·µÄÀíÓÉ·Ö±ðÊÇ______________________________________________________________________________________________________________________________¡£

(2)¸ù¾ÝÉÏÊöÐÅÏ¢£¬Ð´³öCת»¯ÎªCOµÄÈÈ»¯Ñ§·½³Ìʽ

_________________________________________________________¡£

(3)ÉÏÊö·´Ó¦ÖУ¬±íʾȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽÓÐ______________£»±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³ÌʽÓÐ__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¹èÊÇÖØÒªµÄ°ëµ¼Ìå²ÄÁÏ£¬¹¹³ÉÁËÏÖ´úµç×Ó¹¤ÒµµÄ»ù´¡¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)»ù̬SiÔ­×ÓÖУ¬µç×ÓÕ¼¾ÝµÄ×î¸ßÄܲã·ûºÅΪ______£¬¸ÃÄܲã¾ßÓеÄÔ­×Ó¹ìµÀÊýΪ______£¬µç×ÓÊýΪ________¡£

(2)¹èÖ÷ÒªÒÔ¹èËáÑΡ¢________µÈ»¯ºÏÎïµÄÐÎʽ´æÔÚÓڵؿÇÖС£

(3)µ¥Öʹè´æÔÚÓë½ð¸Õʯ½á¹¹ÀàËƵľ§Ì壬ÆäÖÐÔ­×ÓÓëÔ­×ÓÖ®¼äÒÔ________Ïà½áºÏ£¬Æ侧°ûÖй²ÓÐ8¸öÔ­×Ó£¬ÆäÖÐÔÚÃæÐÄλÖù±Ï×________¸öÔ­×Ó¡£

(4)µ¥Öʹè¿Éͨ¹ý¼×¹èÍé(SiH4)·Ö½â·´Ó¦À´ÖƱ¸¡£¹¤ÒµÉϲÉÓÃMg2SiºÍNH4ClÔÚÒº°±½éÖÊÖз´Ó¦ÖƵÃSiH4£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________________________________________¡£

(5)̼ºÍ¹èµÄÓйػ¯Ñ§¼ü¼üÄÜÈçÏÂËùʾ£¬¼òÒª·ÖÎöºÍ½âÊÍÏÂÁÐÓйØÊÂʵ£º

»¯Ñ§¼ü

C¡ªC

C¡ªH

C¡ªO

Si¡ªSi

Si¡ªH

Si¡ªO

¼üÄÜ/ (kJ¡¤mol£­1)

356

413

336

226

318

452

¢Ù¹èÓë̼ͬ×壬ҲÓÐϵÁÐÇ⻯Îµ«¹èÍéÔÚÖÖÀàºÍÊýÁ¿É϶¼Ô¶²»ÈçÍéÌþ¶à£¬Ô­ÒòÊÇ___________________________________________________¡£

¢ÚSiH4µÄÎȶ¨ÐÔСÓÚCH4£¬¸üÒ×Éú³ÉÑõ»¯ÎԭÒòÊÇ

_______________________________________________________¡£

(6)ÔÚ¹èËáÑÎÖУ¬SiOËÄÃæÌå(Èçͼa)ͨ¹ý¹²Óö¥½ÇÑõÀë×Ó¿ÉÐγɵº×´¡¢Á´×´¡¢²ã×´¡¢¹Ç¼ÜÍø×´ËÄ´óÀà½á¹¹ÐÍʽ¡£Í¼bΪһÖÖÎÞÏÞ³¤µ¥Á´½á¹¹µÄ¶à¹èËá¸ù£¬ÆäÖÐSiÔ­×ÓµÄÔÓ»¯ÐÎʽΪ________£¬SiÓëOµÄÔ­×ÓÊýÖ®±ÈΪ________£¬»¯Ñ§Ê½Îª________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


ÓÐÒ»ÖÖÓлúÎïµÄ½á¹¹¼òʽΪ£º

ÊԻشðÏÂÁÐÎÊÌ⣺

(1)¸Ã»¯ºÏÎïÊÇ________(ÌîÑ¡Ïî×Öĸ£¬ÏÂͬ)¡£

A£®Ï©Ìþ                                B£®ÓÍÖ¬

C£®µ°°×ÖÊ                              D£®ÌÇÀà

(2)¸Ã»¯ºÏÎïµÄÃܶÈ________¡£

A£®±ÈË®´ó                              B£®±ÈˮС

C£®ÓëË®Ïàͬ

(3)³£ÎÂϸû¯ºÏÎï³Ê________¡£

A£®ÒºÌ¬                                B£®¹Ì̬

C£®Æø̬

(4)ÏÂÁÐÎïÖÊÖУ¬ÄÜÓë¸ÃÎïÖÊ·´Ó¦µÄÓÐ________¡£

A£®NaOH(aq)                            B£®äåË®

C£®ÒÒ´¼                                D£®ÒÒËá

E£®H2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸