Question

14£®Ã̵Ļ¯ºÏÎïÊÇÓÅÁ¼µÄ´ß»¯¼Á£¬¿ÉÓÃÓڸɵç³ØÔ­ÁÏÉú²úµÈ£®
£¨1£©Ð¿Ã̸ɵç³ØµÄ·´Ó¦Îª2MnO₂+Zn+2NH₄Cl=2MnO£¨OH£©+Zn£¨NH₃£©₂Cl₂£¬MnO£¨OH£©ÖÐÃÌÔªËصĻ¯ºÏ¼ÛΪ+3£®
£¨2£©Ïò·Ïµç³Ø»¹Ô­ºóµÄ·ÏÒº£¨º¬ÓÐMn²⁺¡¢Fe²⁺¡¢Zn²⁺µÈ£©ÖÐÖðµÎµÎ¼ÓNa₂SÈÜÒº£¬×îÏÈÉú³ÉµÄ³ÁµíΪZnS£¨Ìѧʽ£©£®[ÒÑÖªKsp£¨MnS£©=1.4¡Á10^-15£¬Ksp£¨ZnS£©=2.9¡Á10 ^-25£¬Ksp£¨FeS£©=6.0¡Á10^-18]
£¨3£©Mn²⁺´ß»¯H₂O₂·Ö½â£º2H₂O₂£¨l£©=2H₂O£¨l£©+O₂£¨g£©¡÷H₁£¬Æä·´Ó¦»úÀíÈçͼ£º
¢ÙÒÑÖª·´Ó¦¢òΪMnO₂£¨s£©+H₂O₂£¨1£©+2H⁺ £¨ aq£©=Mn²⁺ £¨aq£©+O₂£¨g£©+2H₂O£¨1£©¡÷H₂£®Ð´³ö·´Ó¦ IµÄÈÈ»¯Ñ§·½³Ìʽ£¨ìʱäÓá÷H₁ºÍ¡÷H₂±íʾ£©£ºH₂O₂£¨1£©+Mn²⁺£¨aq£©=2H⁺£¨aq£©+MnO₂£¨s£©¡÷H=¡÷H₁-¡÷H₂£®
¢ÚijζÈʱ£¬Ïò10mL0.4mol/L H₂O₂ÒºÖеÎÈë1µÎMnSO₄·¢Éú·Ö½â£º2H₂O₂=2H₂O+O₂£¬²âµÃ²»Í¬Ê±¿ÌÉú³ÉO₂µÄÌå»ý£¨ÒÑÕÛËãΪ±ê×¼×´¿öϵÄÌå»ý£©Èç±í£º

t/min	0	2	4	6
V£¨O2£©mL	0	9.9	17.2	22.4

0¡«2minʱ·´Ó¦ËÙÂʱÈ2¡«4minʱµÄ¿ì£¬ÆäÔ­ÒòÊÇËæ×Å·´Ó¦µÄ½øÐУ¬H₂O₂Ũ¶È²»¶Ï¼õС£¬·´Ó¦ËÙÂʲ»¶Ï¼õÂý£¬0¡«6minµÄƽ¾ù·´Ó¦ËÙÂÊv£¨H₂O₂£©=3.3¡Á10^-2mol/£¨L•min£©£¨ºöÂÔÈÜÒºÌå»ýµÄ±ä»¯£©£®
£¨4£©ÃÌ»ù´ß»¯¼ÁÊǺϳɼ״¼¡¢¶þ¼×ÃѵĴ߻¯¼Á£®ÒÑÖª£º

·´Ó¦	ƽºâ³£ÊýKP
	773K	873K
¢ñ£®CO2£¨g£©+4H2£¨g£©?CH4£¨g£©+2H2£¨g£©	19.4	0.803
¢ò£®CO2£¨g£©+3H2£¨g£©?CH3OH£¨g£©+H2O£¨g£©	6.07¡Á10-9	3.65¡Á10-9

¢Ù·´Ó¦IµÄÕý·´Ó¦ÊÇ·ÅÈÈ£¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©·´Ó¦£®
¢Ú·´Ó¦¢òµÄƽºâ³£Êý±í´ïʽΪK=$\frac{c£¨C{H}_{3}OH£©¡Ác£¨{H}_{2}O£©}{c£¨C{O}_{2}£©¡Á{c}^{3}£¨{H}_{2}£©}$£®

Answer 1

·ÖÎö £¨1£©MnO£¨OH£©ÖÐOÔªËصĻ¯ºÏ¼ÛΪ-2¼Û£¬HÔªËصĻ¯ºÏ¼ÛΪ+1¼Û£¬½áºÏ»¯ºÏÎïÖÐÕý¸º»¯ºÏ¼ÛµÄ´úÊýºÍΪ0¼ÆË㣻
£¨2£©ÒòΪKsp£¨MnS£©=1.4¡Á10^-15£¾Ksp£¨FeS£©=6.0¡Á10^-18£¾Ksp£¨ZnS£©=2.9¡Á10 ^-25£¬Ïò·Ïµç³Ø»¹Ô­ºóµÄ·ÏÒº£¨º¬ÓÐMn²⁺¡¢Fe²⁺¡¢Zn²⁺µÈ£©ÖÐÖðµÎµÎ¼ÓNa₂SÈÜÒº£¬×îÏÈÉú³ÉÈܽâ¶È×îСµÄÎïÖÊ£»
£¨3£©¢Ùת»¯¹Øϵ¿ÉÖª£¬¹ýÑõ»¯ÇâºÍÃÌÀë×Ó·´Ó¦µÃµ½¶þÑõ»¯Ã̺ÍÇâÀë×Ó£¬
2H₂O₂£¨l£©¨T2H₂O£¨l£©+O₂£¨g£©¡÷H₁¢Ù£¬
MnO₂£¨s£©+H₂O₂£¨l£©+2H⁺¨TMn²⁺£¨aq£©+O₂£¨g£©+2H₂O£¨l£©¡÷H₂£¬¢Ú
ÓɸÇ˹¶¨ÂÉ¢Ù-¢ÚµÃµ½H₂O₂£¨l£©+Mn²⁺£¨aq£©¨TMnO₂£¨s£©+2H⁺£¨aq£©¡÷H=¡÷H₁-¡÷H₂£¬ÒÔ´ËÊéдÈÈ»¯Ñ§·½³Ìʽ£»
¢Ú0¡«2min²úÉúµÄÑõÆøµÄÎïÖʵÄÁ¿n£¨O₂£©=$\frac{9.9¡Á1{0}^{-3}L}{22.4L/mol}$=0.00044mol£¬n£¨H₂0₂£©=2n£¨O₂£©=0.00088mol£¬v£¨H₂O₂£©=$\frac{{\frac{0.00088mol}{0.01L}}}{2min}$¡Ö4.4¡Á10^-2mol/£¨L•min£©£¬4¡«6min²úÉúµÄÑõÆøµÄÎïÖʵÄÁ¿n£¨O₂£©=$\frac{£¨22.4-17.2£©¡Á1{0}^{-3}L}{22.4L/mol}$=2.3¡Á10^-4mol£¬n£¨H₂0₂£©=2n£¨O₂£©=4.6¡Á10^-4mol£¬v£¨H₂0₂£©=$\frac{{\frac{{4.6¡Á{{10}^{-4}}mol}}{0.01L}}}{2min}$=2.3¡Á10^-2mol/L•min£¬0¡«2min H₂O₂ƽ¾ù·´Ó¦ËÙÂʱÈ4¡«6min¿ì£¬Ô­ÒòΪËæ×Å·´Ó¦µÄ½øÐУ¬H₂O₂Ũ¶È²»¶Ï¼õС£¬·´Ó¦ËÙÂʲ»¶Ï¼õÂý£»0¡«6min²úÉúµÄÑõÆøµÄÎïÖʵÄÁ¿n£¨O₂£©=$\frac{{22.4¡Á{{10}^{-3}}L}}{22.4L/mol}$=0.001mol£¬n£¨H₂0₂£©=2n£¨O₂£©=0.002mol£¬v£¨H₂0₂£©=$\frac{{\frac{0.002}{0.01}}}{6}$¡Ö3.3¡Á10^-2mol/£¨L•min£©£»
£¨4£©¢ÙζÈÓÐ773KÉý¸ßµ½873K£¬Æ½ºâ³£Êý¼õС£¬±íÃ÷Éý¸ßζȣ¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£»
¢Úƽºâ³£ÊýΪÉú³ÉÎïŨ¶ÈÃÝÖ®»ýÓë·´Ó¦ÎïŨ¶ÈÃÝÖ®»ýµÄ±È£®

½â´ð ½â£º£¨1£©MnO£¨OH£©ÖÐOÔªËصĻ¯ºÏ¼ÛΪ-2¼Û£¬HÔªËصĻ¯ºÏ¼ÛΪ+1¼Û£¬¸ù¾ÝÕý¸º»¯ºÏ¼ÛµÄ´úÊýºÍΪ0£¬ÃÌÔªËصĻ¯ºÏ¼ÛΪ+3¼Û£¬¹Ê´ð°¸Îª£º+3£»
£¨2£©ÒòΪKsp£¨MnS£©=1.4¡Á10^-15£¾Ksp£¨FeS£©=6.0¡Á10^-18£¾Ksp£¨ZnS£©=2.9¡Á10 ^-25£¬Ïò·Ïµç³Ø»¹Ô­ºóµÄ·ÏÒº£¨º¬ÓÐMn²⁺¡¢Fe²⁺¡¢Zn²⁺µÈ£©ÖÐÖðµÎµÎ¼ÓNa₂SÈÜÒº£¬×îÏÈÉú³ÉÈܽâ¶È×îСµÄÎïÖÊ£¬×îÏÈÉú³ÉµÄ³ÁµíΪZnS£¬¹Ê´ð°¸Îª£ºZnS£»
£¨3£©¢Ùת»¯¹Øϵ¿ÉÖª£¬¹ýÑõ»¯ÇâºÍÃÌÀë×Ó·´Ó¦µÃµ½¶þÑõ»¯Ã̺ÍÇâÀë×Ó£¬
2H₂O₂£¨l£©¨T2H₂O£¨l£©+O₂£¨g£©¡÷H₁¢Ù£¬
MnO₂£¨s£©+H₂O₂£¨l£©+2H⁺¨TMn²⁺£¨aq£©+O₂£¨g£©+2H₂O£¨l£©¡÷H₂£¬¢Ú
ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù-¢ÚµÃµ½H₂O₂£¨l£©+Mn²⁺£¨aq£©¨TMnO₂£¨s£©+2H⁺£¨aq£©¡÷H=¡÷H₁-¡÷H₂
¹Ê´ð°¸Îª£ºH₂O₂£¨l£©+Mn²⁺£¨aq£©¨TMnO₂£¨s£©+2H⁺£¨aq£©¡÷H=¡÷H₁-¡÷H₂
¢Ú0¡«2min²úÉúµÄÑõÆøµÄÎïÖʵÄÁ¿n£¨O₂£©=$\frac{9.9¡Á1{0}^{-3}L}{22.4L/mol}$=0.00044mol£¬n£¨H₂0₂£©=2n£¨O₂£©=0.00088mol£¬v£¨H₂O₂£©=$\frac{{\frac{0.00088mol}{0.01L}}}{2min}$¡Ö4.4¡Á10^-2mol/£¨L•min£©£¬4¡«6min²úÉúµÄÑõÆøµÄÎïÖʵÄÁ¿n£¨O₂£©=$\frac{£¨22.4-17.2£©¡Á1{0}^{-3}L}{22.4L/mol}$=2.3¡Á10^-4mol£¬n£¨H₂0₂£©=2n£¨O₂£©=4.6¡Á10^-4mol£¬v£¨H₂0₂£©=$\frac{{\frac{{4.6¡Á{{10}^{-4}}mol}}{0.01L}}}{2min}$=2.3¡Á10^-2mol/L•min£¬0¡«2min H₂O₂ƽ¾ù·´Ó¦ËÙÂʱÈ4¡«6min¿ì£¬Ô­ÒòΪËæ×Å·´Ó¦µÄ½øÐУ¬H₂O₂Ũ¶È²»¶Ï¼õС£¬·´Ó¦ËÙÂʲ»¶Ï¼õÂý£»0¡«6min²úÉúµÄÑõÆøµÄÎïÖʵÄÁ¿n£¨O₂£©=$\frac{{22.4¡Á{{10}^{-3}}L}}{22.4L/mol}$=0.001mol£¬n£¨H₂0₂£©=2n£¨O₂£©=0.002mol£¬v£¨H₂0₂£©=$\frac{{\frac{0.002}{0.01}}}{6}$¡Ö3.3¡Á10^-2mol/£¨L•min£©£¬
¹Ê´ð°¸Îª£º¢ÚËæ×Å·´Ó¦µÄ½øÐУ¬H₂O₂Ũ¶È²»¶Ï¼õС£¬·´Ó¦ËÙÂʲ»¶Ï¼õÂý£»3.3¡Á10^-2mol/£¨L•min£©£»
£¨4£©¢ÙζÈÓÐ773KÉý¸ßµ½873K£¬Æ½ºâ³£Êý¼õС£¬±íÃ÷Éý¸ßζȣ¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬ËµÃ÷Õý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬¹Ê´ð°¸Îª£º·ÅÈÈ£»
¢Ú·´Ó¦¢òΪCO₂£¨g£©+3H₂£¨g£©?CH₃OH£¨g£©+H₂O£¨g£©£¬Æäƽºâ³£Êý±í´ïʽΪK=$\frac{c£¨C{H}_{3}OH£©¡Ác£¨{H}_{2}O£©}{c£¨C{O}_{2}£©¡Á{c}^{3}£¨{H}_{2}£©}$£¬¹Ê´ð°¸Îª£ºK=$\frac{c£¨C{H}_{3}OH£©¡Ác£¨{H}_{2}O£©}{c£¨C{O}_{2}£©¡Á{c}^{3}£¨{H}_{2}£©}$£®

µãÆÀ ±¾Ì⿼²é»¯Ñ§Æ½ºâ¼ÆËã¡¢ÄÑÈܵç½âÖʼ°Æ½ºâÒƶ¯µÈ£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕÈÈ»¯Ñ§·½³ÌʽµÄÊéд¡¢»¯Ñ§Æ½ºâµÄÒƶ¯¡¢»¯Ñ§·´Ó¦ËÙÂʵļÆËãµÄÏà¹Ø֪ʶµÈΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëÓ¦ÓÃÄÜÁ¦µÄ¿¼²é£¬ÌâÄ¿ÄѶȲ»´ó£®