»¯Ñ§·´Ó¦¹ý³ÌÖз¢ÉúÎïÖʱ仯µÄͬʱ£¬³£³£°éÓÐÄÜÁ¿µÄ±ä»¯£®ÕâÖÖÄÜÁ¿µÄ±ä»¯³£ÒÔÈÈÁ¿µÄÐÎʽ±íÏÖ³öÀ´£¬½Ð×ö·´Ó¦ÈÈ£®ÓÉÓÚ·´Ó¦µÄÇé¿ö²»Í¬£¬·´Ó¦ÈÈ¿ÉÒÔ·ÖΪÐí¶àÖÖ£¬ÈçȼÉÕÈȺÍÖкÍÈȵȣ®
£¨1£©ÏÂÁС÷H±íʾÎïÖÊȼÉÕÈȵÄÊÇ
¡÷H4¡¢¡÷H5
¡÷H4¡¢¡÷H5
£»±íʾÎïÖÊÖкÍÈȵÄÊÇ
¡÷H6¡¢¡÷H8
¡÷H6¡¢¡÷H8
£¨Ìî¡°¡÷H1¡±¡¢¡°¡÷H2¡±¡¢¡°¡÷H3¡±µÈ£©£®
A£®2H2£¨g£©+O2£¨g£©¨T2H2O£¨l£©¡÷H1
B£®C£¨s£©+1/2O2£¨g£©¨TCO£¨g£©¡÷H2
C£®CH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨g£©¡÷H3
D£®C£¨s£©+O2£¨g£©¨TCO2£¨g£©¡÷H4
E£®C6H12O6£¨s£©+6O2£¨g£©¨T6CO2£¨g£©+6H2O£¨l£©¡÷H5
F£®NaOH£¨aq£©+HCl£¨aq£©¨TNaCl£¨aq£©+H2O£¨l£©¡÷H6
G£®2NaOH£¨aq£©+H2SO4£¨aq£©¨TNa2SO4£¨aq£©+2H2O£¨l£©¡÷H7
H£®CH3COOH£¨aq£©+NaOH£¨aq£©¨TCH3COONa£¨aq£©+H2O£¨l£©¡÷H8
£¨2£©2.00g C2H2ÆøÌåÍêȫȼÉÕÉú³ÉҺ̬ˮºÍCO2ÆøÌ壬·Å³ö99.6kJµÄÈÈÁ¿£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
C2H2£¨g£©+
5
2
O2£¨g£©¨T2CO2£¨g£©+H2O£¨l£©¡÷H=-1294.8kJ/mol
C2H2£¨g£©+
5
2
O2£¨g£©¨T2CO2£¨g£©+H2O£¨l£©¡÷H=-1294.8kJ/mol
£®
£¨3£©³ä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡Íé·Å³öµÄÈÈÁ¿´óСΪQ£¬Éú³ÉµÄCO2Ç¡ºÃÓë100mLŨ¶ÈΪ5mol?L-1µÄKOHÈÜÒºÍêÈ«·´Ó¦Éú³ÉÒ»ÖÖÕýÑΣ®ÔòȼÉÕ1mol¶¡Íé·Å³öµÄÈÈÁ¿Îª
16QkJ
16QkJ
£®
·ÖÎö£º£¨1£©È¼ÉÕÈÈÖ¸1molÎïÖÊÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯ÎïµÄ·´Ó¦ÈÈ£¬Ò»°ãH¡úH2O£¨l£©¡¢C¡úCO2£¨g£©¡¢S¡úSO2£¨g£©£¬ÖкÍÈÈָϡÈÜÒºÖÐËáÓë¼î·´Ó¦Éú³É1molH2OµÄ·´Ó¦ÈÈÈÈÁ¿£»
£¨2£©CO2Ç¡ºÃÓëKOHÈÜÒºÍêÈ«·´Ó¦Éú³ÉÕýÑÎΪK2CO3£¬¸ù¾Ýn=cV¼ÆËãKOHµÄÎïÖʵÄÁ¿£¬¸ù¾Ý¼ØÔªËØÊغã¼ÆËãn£¨K2CO3£©£¬¸ù¾Ý̼ԪËØÊغãÓÉn£¨CO2£©=n£¨K2CO3£©£¬½ø¶ø¼ÆË㶡ÍéµÄÎïÖʵÄÁ¿£¬¾Ý´Ë¼ÆË㣮
½â´ð£º½â£º£¨1£©A£®ÇâÆøµÄÎïÖʵÄÁ¿Îª2mol£¬·´Ó¦ÈÈ¡÷H1²»ÄܱíʾȼÉÕÈÈ£¬
B£®Ì¼È¼ÉÕµÄÉú³ÉÎïΪCO£¬±íʾζȵÄÑõ»¯ÎïCO2£¬·´Ó¦ÈÈ¡÷H2²»ÄܱíʾȼÉÕÈÈ£¬
C£®1mol¼×ÍéÍêȫȼÉÕ£¬Éú³ÉµÄË®ÊÇÆø̬£¬²»ÊÇÎȶ¨µÄ״̬£¬Ó¦ÎªÒºÌ¬Ë®£¬¹Ê·´Ó¦ÈÈ¡÷H3²»ÄܱíʾȼÉÕÈÈ£¬
D£®C£¨s£©+O2£¨g£©¨TCO2£¨g£©¡÷H4ÖÐ1molCÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼£¬·ûºÏȼÉÕÈȸÅÄ·´Ó¦ÈÈ¡÷H4ÄܱíʾȼÉÕÈÈ£¬
E£®C6H12O6£¨s£©+6O2£¨g£©¨T6CO2£¨g£©+6H2O£¨l£©¡÷H5ÖÐ1molC6H12O6ÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼ÓëҺ̬ˮ£¬·ûºÏȼÉÕÈȸÅÄ·´Ó¦ÈÈ¡÷H5ÄܱíʾȼÉÕÈÈ£¬
F£®NaOH£¨aq£©+HCl£¨aq£©¨TNaCl£¨aq£©+H2O£¨l£©¡÷H6ÖÐʵÖÊÊÇ1molÇâÀë×ÓÓë1molÇâÑõ¸ùÀë×Ó·´Ó¦Éú³É1molH2O£¬·ûºÏÖкÍÈȸÅÄ·´Ó¦ÈÈ¡÷H6ÄܱíʾÖкÍÈÈ£¬
G£®·´Ó¦Éú³ÉµÄˮΪ2mol£¬·´Ó¦ÈÈ¡÷H7²»ÄܱíʾÖкÍÈÈ£¬
H£®CH3COOH£¨aq£©+NaOH£¨aq£©¨TCH3COONa£¨aq£©+H2O£¨l£©¡÷H8ÊÇËáÓë¼î·´Ó¦Éú³É1molH2O£¬·ûºÏÖкÍÈȸÅÄ·´Ó¦ÈÈ¡÷H8ÄܱíʾÖкÍÈÈ£¬
¹Ê´ð°¸Îª£º¡÷H4¡¢¡÷H5£¬¡÷H6¡¢¡÷H8£»
£¨2£©2.00g C2H2ÆøÌåÍêȫȼÉÕÉú³ÉҺ̬ˮºÍCO2ÆøÌ壬·Å³ö99.6kJµÄÈÈÁ¿£¬Ôò1molC2H2ÆøÌåÆøÌåÍêȫȼÉշųöµÄÈÈÁ¿Îª
1mol¡Á26g/mol
2g
¡Á99.6kJ=1294.8kJ£¬¹Ê¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºC2H2£¨g£©+
5
2
O2£¨g£©¨T2CO2£¨g£©+H2O£¨l£©¡÷H=-1294.8kJ/mol£¬
¹Ê´ð°¸Îª£ºC2H2£¨g£©+
5
2
O2£¨g£©¨T2CO2£¨g£©+H2O£¨l£©¡÷H=-1294.8kJ/mol£»
£¨3£©KOHµÄÎïÖʵÄÁ¿Îª0.1L¡Á5mol/L=0.5mol£¬¸ù¾Ý¼ØÀë×ÓÊغ㣬¹Ên£¨K2CO3£©=0.5mol¡Á
1
2
=0.25mol£¬¸ù¾Ý̼ԪËØÊغãÓÉn£¨CO2£©=n£¨K2CO3£©=0.25mol£¬¸ù¾Ý̼ԪËØÊغã¿ÉÖª£¬¶¡ÍéµÄÎïÖʵÄÁ¿Îª0.25mol¡Á
1
4
=
1
16
mol£¬¼´
1
16
mol¶¡Íé·Å³öµÄÈÈÁ¿´óСΪQkJ£¬¹Ê1mol¶¡ÍéÍêȫȼÉշųöµÄÈÈÁ¿ÎªQkJ¡Á
1mol
1
16
mol
=16QkJ£¬
¹Ê´ð°¸Îª£º16QkJ£®
µãÆÀ£º¿¼²éȼÉÕÈÈÓëÖкÍÈȵĸÅÄî¡¢·´Ó¦ÈȵÄÓйؼÆËã¡¢ÈÈ»¯Ñ§·½³ÌʽµÄÒâÒåµÈ£¬×¢ÒâÖкÍÈÈÓëȼÉÕÈȸÅÄîµÄ°ÑÎÕ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄê°²»ÕÊ¡Êæ³ÇÏþÌìÖÐѧ¸ß¶þÏÂѧÆÚÖÊÁ¿²âÊÔ3»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö)(2011¡¤Òø´¨¸ß¶þ¼ì²â)»¯Ñ§·´Ó¦¹ý³ÌÖз¢ÉúÎïÖʱ仯µÄͬʱ£¬³£³£°éÓÐÄÜÁ¿µÄ±ä»¯£¬ÕâÖÖÄÜÁ¿µÄ±ä»¯³£ÒÔÈÈÄܵÄÐÎʽ±íÏÖ³öÀ´£¬½Ð×ö·´Ó¦ÈÈ¡£ÓÉÓÚ·´Ó¦µÄÇé¿ö²»Í¬£¬·´Ó¦ÈÈ¿ÉÒÔ·ÖΪÐí¶àÖÖ£¬ÈçȼÉÕÈȺÍÖкÍÈȵȡ£
(1)ÏÂÁЦ¤H±íʾÎïÖÊȼÉÕÈȵÄÊÇ________£»±íʾÎïÖÊÖкÍÈȵÄÊÇ________¡£(ÌH1¡¢¦¤H2ºÍ¦¤H3µÈ)

A£®2H2(g)£«O2(g)===2H2O(l)¡¡¦¤H1
B£®C(s)£«O2(g)===CO(g)¡¡¦¤H2
C£®CH4(g)£«2O2(g)===CO2(g)£«2H2O(g)¡¡¦¤H3
D£®C(s)£«O2(g)===CO2(g)¡¡¦¤H4
E£®C6H12O6(s)£«6O2(g)===6CO2(g)£«6H2O(l)¡¡¦¤H5
F£®NaOH(aq)£«HCl(aq)===NaCl(aq)£«H2O(l)¡¡¦¤H6
G£®2NaOH(aq)£«H2SO4(aq)===Na2SO4(aq)£«2H2O(l)¡¡¦¤H7
(2)½øÒ»²½Ñо¿±íÃ÷£¬»¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯(¦¤H)Óë·´Ó¦ÎïºÍÉú³ÉÎïµÄ¼üÄÜÓйء£¼üÄÜ¿ÉÒÔ¼òµ¥µØÀí½âΪ¶Ï¿ª1 mol»¯Ñ§¼üʱËùÐèÎüÊÕµÄÄÜÁ¿¡£Ï±íÊDz¿·Ö»¯Ñ§¼üµÄ¼üÄÜÊý¾Ý£º
»¯Ñ§¼ü
P¡ªP
P¡ªO
O===O
P===O
¼üÄÜkJ/mol
197
360
499
x
ÒÑÖª°×Á×(P4)µÄȼÉÕÈÈΪ2 378.0 kJ/mol£¬°×Á×ÍêȫȼÉյIJúÎï(P4O10)µÄ½á¹¹ÈçͼËùʾ£¬ÔòÉϱíÖÐx£½________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄêÖØÇìÊеÚÆßÖÐѧ¸ß¶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö)»¯Ñ§·´Ó¦¹ý³ÌÖз¢ÉúÎïÖʱ仯µÄͬʱ£¬³£³£°éÓÐÄÜÁ¿µÄ±ä»¯£¬ÕâÖÖÄÜÁ¿µÄ±ä»¯³£ÒÔÈÈÄܵÄÐÎʽ±íÏÖ³öÀ´£¬½Ð×ö·´Ó¦ÈÈ¡£ÓÉÓÚ·´Ó¦µÄÇé¿ö²»Í¬£¬·´Ó¦ÈÈ¿ÉÒÔ·ÖΪÐí¶àÖÖ£¬ÈçȼÉÕÈȺÍÖкÍÈȵȡ£
£¨1£©ÏÂÁЦ¤H±íʾÎïÖÊȼÉÕÈȵÄÊÇ           £»±íʾÎïÖÊÖкÍÈȵÄÊÇ            ¡£
(Ìî¡°¦¤H1¡±¡¢¡°¦¤H2¡±ºÍ¡°¦¤H3¡±µÈ)
A£®2H2(g)+O2(g) = 2H2O(l) ¡¡¦¤H1                
B£®C(s)+1/2O2(g) = CO(g)¡¡¦¤H2
C£®CH4(g)+2O2(g) = CO2(g)+2H2O(g)¡¡¦¤H3       
D£®C(s)+O2(g)= CO2(g) ¡¡¦¤H4
E£®C6H12O6(s)+6O2(g) = 6CO2(g)+6H2O(l)¡¡¦¤H5
F£®NaOH(aq)+HCl(aq) = NaCl(aq)+H2O(l)¡¡¦¤H6
G£®2NaOH(aq)+H2SO4(aq) = Na2SO4(aq)+2H2O(l)   ¦¤H7
£¨2£©ÒÑÖªÔÚ101kPa¡¢273Kʱ£¬15gÒÒÍéȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³öakJµÄÈÈÁ¿£¬ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ                ¡£
A£®C2H6(g)+7/2O2(g) = 2CO2(g)+3H2O(l)  ¦¤H= +2akJ/mol
B£®C2H6(g)+7/2O2(g) = 2CO2(g)+3H2O(g)   ¦¤H= -2akJ/mol
C£®2C2H6(g)+7O2(g) = 4CO2(g)+6H2O(l)   ¦¤H= -4akJ/mol
D£®2C2H6(g)+7O2(g) = 4CO2(g)+6H2O(g)  ¦¤H= -4akJ/mol
£¨3£©½øÒ»²½Ñо¿±íÃ÷£¬»¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯(¦¤H)Óë·´Ó¦ÎïºÍÉú³ÉÎïµÄ¼üÄÜÓйأ¨¼üÄÜ¿ÉÒÔ¼òµ¥µØÀí½âΪ¶Ï¿ª1mol»¯Ñ§¼üʱËùÐèÎüÊÕµÄÄÜÁ¿£©£¬Ï±íÊDz¿·Ö»¯Ñ§¼üµÄ¼üÄÜÊý¾Ý£º

ÒÑÖª°×Á×(P4)µÄȼÉÕÈÈΪ2378.0kJ/mol£¬°×Á×ÍêȫȼÉյIJúÎï(P4O10)µÄ½á¹¹ÈçÏÂͼËùʾ£¬ÔòÉϱíÖÐa=             ¡£

£¨4£©ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×´¼È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22.68kJ¡£Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ                                               ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2014½ìÖØÇìÊи߶þÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö)»¯Ñ§·´Ó¦¹ý³ÌÖз¢ÉúÎïÖʱ仯µÄͬʱ£¬³£³£°éÓÐÄÜÁ¿µÄ±ä»¯£¬ÕâÖÖÄÜÁ¿µÄ±ä»¯³£ÒÔÈÈÄܵÄÐÎʽ±íÏÖ³öÀ´£¬½Ð×ö·´Ó¦ÈÈ¡£ÓÉÓÚ·´Ó¦µÄÇé¿ö²»Í¬£¬·´Ó¦ÈÈ¿ÉÒÔ·ÖΪÐí¶àÖÖ£¬ÈçȼÉÕÈȺÍÖкÍÈȵȡ£

£¨1£©ÏÂÁЦ¤H±íʾÎïÖÊȼÉÕÈȵÄÊÇ           £»±íʾÎïÖÊÖкÍÈȵÄÊÇ            ¡£

(Ìî¡°¦¤H1¡±¡¢¡°¦¤H2¡±ºÍ¡°¦¤H3¡±µÈ)

A£®2H2(g)+O2(g) = 2H2O(l) ¡¡¦¤H1                

B£®C(s)+1/2O2(g) = CO(g)¡¡¦¤H2

C£®CH4(g)+2O2(g) = CO2(g)+2H2O(g)¡¡¦¤H3       

D£®C(s)+O2(g)= CO2(g) ¡¡¦¤H4

E£®C6H12O6(s)+6O2(g) = 6CO2(g)+6H2O(l)¡¡¦¤H5

F£®NaOH(aq)+HCl(aq) = NaCl(aq)+H2O(l)¡¡¦¤H6

G£®2NaOH(aq)+H2SO4(aq) = Na2SO4(aq)+2H2O(l)   ¦¤H7

£¨2£©ÒÑÖªÔÚ101kPa¡¢273Kʱ£¬15gÒÒÍéȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³öakJµÄÈÈÁ¿£¬ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ                ¡£

A£®C2H6(g)+7/2O2(g) = 2CO2(g)+3H2O(l)  ¦¤H= +2akJ/mol

B£®C2H6(g)+7/2O2(g) = 2CO2(g)+3H2O(g)   ¦¤H= -2akJ/mol

C£®2C2H6(g)+7O2(g) = 4CO2(g)+6H2O(l)   ¦¤H= -4akJ/mol

D£®2C2H6(g)+7O2(g) = 4CO2(g)+6H2O(g)  ¦¤H= -4akJ/mol

£¨3£©½øÒ»²½Ñо¿±íÃ÷£¬»¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯(¦¤H)Óë·´Ó¦ÎïºÍÉú³ÉÎïµÄ¼üÄÜÓйأ¨¼üÄÜ¿ÉÒÔ¼òµ¥µØÀí½âΪ¶Ï¿ª1mol»¯Ñ§¼üʱËùÐèÎüÊÕµÄÄÜÁ¿£©£¬Ï±íÊDz¿·Ö»¯Ñ§¼üµÄ¼üÄÜÊý¾Ý£º

ÒÑÖª°×Á×(P4)µÄȼÉÕÈÈΪ2378.0kJ/mol£¬°×Á×ÍêȫȼÉյIJúÎï(P4O10)µÄ½á¹¹ÈçÏÂͼËùʾ£¬ÔòÉϱíÖÐa=             ¡£

£¨4£©ÔÚ25¡æ¡¢101kPaÏ£¬1g¼×´¼È¼ÉÕÉú³ÉCO2ºÍҺ̬ˮʱ·ÅÈÈ22.68kJ¡£Ôò±íʾ¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ                                               ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì°²»ÕÊ¡¸ß¶þÏÂѧÆÚÖÊÁ¿²âÊÔ3»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö)(2011¡¤Òø´¨¸ß¶þ¼ì²â)»¯Ñ§·´Ó¦¹ý³ÌÖз¢ÉúÎïÖʱ仯µÄͬʱ£¬³£³£°éÓÐÄÜÁ¿µÄ±ä»¯£¬ÕâÖÖÄÜÁ¿µÄ±ä»¯³£ÒÔÈÈÄܵÄÐÎʽ±íÏÖ³öÀ´£¬½Ð×ö·´Ó¦ÈÈ¡£ÓÉÓÚ·´Ó¦µÄÇé¿ö²»Í¬£¬·´Ó¦ÈÈ¿ÉÒÔ·ÖΪÐí¶àÖÖ£¬ÈçȼÉÕÈȺÍÖкÍÈȵȡ£

(1)ÏÂÁЦ¤H±íʾÎïÖÊȼÉÕÈȵÄÊÇ________£»±íʾÎïÖÊÖкÍÈȵÄÊÇ________¡£(ÌH1¡¢¦¤H2ºÍ¦¤H3µÈ)

A£®2H2(g)£«O2(g)===2H2O(l)¡¡¦¤H1

B£®C(s)£«O2(g)===CO(g)¡¡¦¤H2

C£®CH4(g)£«2O2(g)===CO2(g)£«2H2O(g)¡¡¦¤H3

D£®C(s)£«O2(g)===CO2(g)¡¡¦¤H4

E£®C6H12O6(s)£«6O2(g)===6CO2(g)£«6H2O(l)¡¡¦¤H5

F£®NaOH(aq)£«HCl(aq)===NaCl(aq)£«H2O(l)¡¡¦¤H6

G£®2NaOH(aq)£«H2SO4(aq)===Na2SO4(aq)£«2H2O(l)¡¡¦¤H7

(2)½øÒ»²½Ñо¿±íÃ÷£¬»¯Ñ§·´Ó¦µÄÄÜÁ¿±ä»¯(¦¤H)Óë·´Ó¦ÎïºÍÉú³ÉÎïµÄ¼üÄÜÓйء£¼üÄÜ¿ÉÒÔ¼òµ¥µØÀí½âΪ¶Ï¿ª1 mol»¯Ñ§¼üʱËùÐèÎüÊÕµÄÄÜÁ¿¡£Ï±íÊDz¿·Ö»¯Ñ§¼üµÄ¼üÄÜÊý¾Ý£º

»¯Ñ§¼ü

P¡ªP

P¡ªO

O===O

P===O

¼üÄÜkJ/mol

197

360

499

x

ÒÑÖª°×Á×(P4)µÄȼÉÕÈÈΪ2 378.0 kJ/mol£¬°×Á×ÍêȫȼÉյIJúÎï(P4O10)µÄ½á¹¹ÈçͼËùʾ£¬ÔòÉϱíÖÐx£½________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸