（1）在25℃、101kPa下，1g甲醇（CH3OH）燃烧生成CO2和液态水时放热22.68kJ，则该反应的热化学方程式应为 依据叙述，完成下列三个小题．（1）在25℃、101kPa下，...

\u6309\u8981\u6c42\u5199\u51fa\u4e0b\u5217\u65b9\u7a0b\u5f0f\uff081\uff09\u572825\u2103\u3001101kPa\u4e0b\uff0c1g\u7532\u9187\u71c3\u70e7\u751f\u6210CO2\u548c\u6db2\u6001\u6c34\u65f6\u653e\u70ed22.68kJ\uff0e\u5219\u8868\u793a\u7532\u9187\u71c3\u70e7

\uff081\uff091mol\u7532\u9187\u5b8c\u5168\u71c3\u70e7\u751f\u6210\u4e8c\u6c27\u5316\u78b3\u548c\u6db2\u6001\u6c34\u653e\u70ed725.8KJ\uff1b\u71c3\u70e7\u70ed\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aCH3OH\uff08l\uff09+32O2\uff08g\uff09\u2550CO2\uff08g\uff09+2H2O\uff08l\uff09\u25b3H=-725.8 kJ?mol-1\uff0c\u6545\u7b54\u6848\u4e3a\uff1aCH3OH\uff08l\uff09+32O2\uff08g\uff09\u2550CO2\uff08g\uff09+2H2O\uff08l\uff09\u25b3H=-725.8 kJ?mol-1\uff1b\uff082\uff09CO\u7684\u71c3\u70e7\u70ed\u4e3a283kJ/mol\uff0e\u76f8\u540c\u6761\u4ef6\u4e0b\uff0c2molCH4\u5b8c\u5168\u71c3\u70e7\u751f\u6210\u6db2\u6001\u6c34\uff0c\u6240\u653e\u51fa\u7684\u70ed\u91cf\u4e3a283kJ\u00d76.3=1782.9kJ\uff0c\u6545\u7532\u70f7\u5b8c\u5168\u71c3\u70e7\u751f\u6210\u6db2\u6001\u6c34\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aCH4\uff08g\uff09+2O2 \uff08g\uff09=CO2\uff08g\uff09+2H2O\uff081\uff09\uff1b\u25b3H=-891.45kJ/mol\uff0c\u6545\u7b54\u6848\u4e3a\uff1aCH4\uff08g\uff09+2O2 \uff08g\uff09=CO2\uff08g\uff09+2H2O\uff081\uff09\uff1b\u25b3H=-891.45kJ/mol\uff1b\uff083\uff09Na2S\u662f\u4e8c\u5143\u5f31\u9178\u6c22\u786b\u9178\u7684\u76d0\uff0c\u6c34\u89e3\u79bb\u5b50\u65b9\u7a0b\u5f0f\u4e3a\uff1aS2-+H2O?HS-+OH-\u3001HS-+H2O?H2S+OH-\uff1b\u6545\u7b54\u6848\u4e3a\uff1aS2-+H2O?HS-+OH-\u3001HS-+H2O?H2S+OH-\uff1b\uff084\uff09\u786b\u5316\u94a0\u4e3a\u79bb\u5b50\u5316\u5408\u7269\uff0c\u7528\u7535\u5b50\u5f0f\u8868\u793aNa2S\u7684\u5f62\u6210\u8fc7\u7a0b\u4e3a\uff1a\u6545\u7b54\u6848\u4e3a\uff1a\uff0e

\uff081\uff09\u572825\u2103\u3001101kPa\u4e0b\uff0c1g\u7532\u9187\uff08CH3OH\uff09\u71c3\u70e7\u751f\u6210CO2\u548c\u6db2\u6001\u6c34\u65f6\u653e\u70ed22.68kJ\uff0c1mol\u7532\u9187\u5b8c\u5168\u71c3\u70e7\u751f\u6210\u4e8c\u6c27\u5316\u78b3\u548c\u6db2\u6001\u6c34\u653e\u70ed725.76KJ\uff0c\u5219\u7532\u9187\u71c3\u70e7\u70ed\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aCH3OH\uff08l\uff09+32O2\uff08g\uff09\u2550CO2\uff08g\uff09+2H2O\uff08l\uff09\u25b3H=-725.76 kJ?mol-1\uff0c\u6545\u7b54\u6848\u4e3a\uff1aCH3OH\uff08l\uff09+32O2\uff08g\uff09\u2550CO2\uff08g\uff09+2H2O\uff08l\uff09\u25b3H=-725.76 kJ?mol-1\uff1b\uff082\uff09\u9002\u91cf\u7684N2\u548cO2\u5b8c\u5168\u53cd\u5e94\uff0c\u6bcf\u751f\u621023\u514bNO2\u9700\u8981\u5438\u653616.95kJ\u70ed\u91cf\uff0c\u6240\u4ee5\u6bcf\u751f\u621092\u514b\u53732molNO2\u9700\u8981\u5438\u653667.8kJ\u70ed\u91cf\uff0c\u5219\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aN2\uff08g\uff09+2O2\uff08g\uff09=2NO2\uff08g\uff09\u25b3H=+67.8kJ?mol-1\uff0c\u6545\u7b54\u6848\u4e3a\uff1aN2\uff08g\uff09+2O2\uff08g\uff09=2NO2\uff08g\uff09\u25b3H=+67.8kJ?mol-1\uff1b\uff083\uff09\u8bbe0.2mol\u7684\u70ad\u7c89\u548c\u6c22\u6c14\u4e2d\u78b3\u7684\u7269\u8d28\u7684\u91cf\u4e3axmol\uff0c\u6c22\u6c14\u7684\u7269\u8d28\u7684\u91cf\u4e3aymol\uff0c\u4f9d\u636eC\uff08s\uff09+O2\uff08g\uff09=CO2\uff08g\uff09\u25b3H=-393.5kJ?mol-1\uff0cxmol\u78b3\u71c3\u70e7\u653e\u51fa\u7684\u70ed\u91cf\u4e3a393.5xkJ\uff1b\u4f9d\u636e2H2\uff08g\uff09+O2\uff08g\uff09=2H2O\uff08g\uff09\u25b3H=-483.6kJ?mol-1\u73b0\uff0cymol\u6c22\u6c14\u71c3\u70e7\u653e\u51fa\u7684\u70ed\u91cf\u4e3a483.62\u00d7yKJ\uff0c\u7531\u9898\u610f\u5f97\uff1a393.5xkJ+483.62\u00d7yKJ=63.53kJ\uff0cx+y=0.2mol\uff0c\u89e3\u5f97\uff1ax=0.1mol\uff0cy=0.1mol \u6240\u4ee5\u6df7\u5408\u7269\u4e2dC\u4e0eH2\u7684\u7269\u8d28\u7684\u91cf\u4e4b\u6bd4\u4e3a0.1mol\uff1a0.1mol=1\uff1a1\uff0c\u6545\u7b54\u6848\u4e3a\uff1a1\uff1a1\uff0e

（1）在25℃、101kPa下，1g甲醇（CH₃OH）燃烧生成CO₂和液态水时放热22.68kJ，则32g甲醇即1mol甲醇完全燃烧生成二氧化碳和液态水放出的热量=

22.68kJ

1g

×32g=725.8kJ，其热化学反应方程式为CH₃OH（l）+

3

2

O₂（g）=CO₂（g）+2H₂O（l）△H=-725.8KJ/mol，
故答案为：CH₃OH（l）+

3

2

O₂（g）=CO₂（g）+2H₂O（l）△H=-725.8KJ/mol；
（2）由氢气和氧气反应生成1mol水蒸气放热241.8kJ，生成2mol水蒸气放出热量为483.6kJ，该反应热化学反应方程式为2H₂（g）+O₂（g）=2H₂O（g）△H=-483.6kJ/mol，若1g水蒸气转化成液态水放热2.444kJ，则1mol水蒸气转化为液态水放出热量=

2.444kJ

1g 18g/mol

=43.992kJ/mol，根据盖斯定律知其焓变=
-483.6kJ/mol+（-43.992kJ/mol）×2=-571.58kJ/mol，则氢气的燃烧热=

571.58kJ/mol

2

=285.8kJ/mol，
故答案为：2H₂（g）+O₂（g）=2H₂O（g）△H=-483.6kJ/mol；-571.58kJ/mol；285.8kJ/mol；
（3）0.5mol甲烷与0.5mol水蒸气在t℃．p kPa时，完全反应生成一氧化碳和氢气，吸收了a kJ热量，则1mol甲烷和1mol水蒸气在t℃．p kPa时，完全反应生成一氧化碳和氢气，吸收了2a kJ热量，所以其热化学反应方程式为CH₄（g）+H₂O（g）=CO（g）+3H₂（g）△H=-2akJ/mol，
故答案为：CH₄（g）+H₂O（g）=CO（g）+3H₂（g）△H=-2akJ/mol．

(1)鍦25鈩銆101kPa涓,1g涔欑兎鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鏃舵斁鐑璔kJ.鍒欒〃绀轰箼鐑...
绛旓細锛1锛夊湪25鈩冦101kPa涓锛1g涔欑兎鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鏃舵斁鐑璔kJ锛1mol涔欑兎锛堣川閲忎负30g锛夌噧鐑х敓鎴愪簩姘у寲纰冲拰娑叉佹按鏀剧儹30QKJ鐑噺锛屽垯琛ㄧず涔欑兎鐕冪儳鐑殑鐑寲瀛︽柟绋嬪紡涓篊₂H₆锛坓锛+7/2O₂锛坓锛夆晲2CO₂锛坓锛+3H₂O锛坙锛夆柍H=-30QkJ/mol锛屾晠绛旀涓猴細C₂H̀...

...鍐欏嚭涓嬪垪鍙嶅簲鐨勭儹鍖栧鏂圭▼寮.(1)鍦25鈩銆101kPa涓,1g鐢查唶鐕冪儳鐢熸垚CO2...
绛旓細锛1锛夊湪25鈩銆101kPa涓锛1g鐢查唶锛圕H3OH锛夌噧鐑х敓鎴怌O2鍜屾恫鎬佹按鏃舵斁鐑22.68kJ锛1mol鐢查唶瀹屽叏鐕冪儳鐢熸垚浜屾哀鍖栫⒊鍜屾恫鎬佹按鏀剧儹涓22.68kJ脳32=725.8KJ锛屾墍浠ョ敳閱囩噧鐑х儹鐨勭儹鍖栧鏂圭▼寮忎负CH3OH锛坙锛+32O2锛坓锛夆晲CO2锛坓锛+2H2O锛坙锛夆柍H=-725.8 kJ?mol-1锛屾晠绛旀涓猴細CH3OH锛坙锛+32O2锛坓锛夆晲CO2锛...

璁＄畻:(1)鍦25鈩銆101kPa涓,1g鐢查唶鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鏃舵斁鐑22.68kJ...
绛旓細锛1锛夊湪25鈩銆101kPa涓锛1g鐢查唶锛圕H3OH锛夌噧鐑х敓鎴怌O2鍜屾恫鎬佹按鏃舵斁鐑22.68kJ锛32g鐢查唶鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鏃舵斁鐑22.68kJ脳32=725.76kJ锛1mol鐢查唶璐ㄩ噺涓32鍏嬶紝鎵浠ュ畬鍏ㄧ噧鐑1mol鐢查唶鐢熸垚浜屾哀鍖栫⒊鍜屾恫鎬佹按鏀剧儹725.8KJ锛岀噧鐑х儹鐨勭儹鍖栧鏂圭▼寮忎负锛欳H3OH锛坙锛+32O2锛坓锛夆晲CO2锛坓锛+2H2O锛坙锛夆柍H=-7...

渚濇嵁鍙欒堪,瀹屾垚涓嬪垪涓変釜灏忛.(1)鍦25鈩銆101kPa涓,1g鐢查唶(CH3OH)鐕冪儳鐢...
绛旓細锛1锛夊湪25鈩銆101kPa涓锛1g鐢查唶锛圕H3OH锛夌噧鐑х敓鎴怌O2鍜屾恫鎬佹按鏃舵斁鐑22.68kJ锛1mol鐢查唶瀹屽叏鐕冪儳鐢熸垚浜屾哀鍖栫⒊鍜屾恫鎬佹按鏀剧儹725.76KJ锛屽垯鐢查唶鐕冪儳鐑殑鐑寲瀛︽柟绋嬪紡涓猴細CH3OH锛坙锛+32O2锛坓锛夆晲CO2锛坓锛+2H2O锛坙锛夆柍H=-725.76 kJ?mol-1锛屾晠绛旀涓猴細CH3OH锛坙锛+32O2锛坓锛夆晲CO2锛坓锛+2H2O锛坙...

渚濇嵁浜嬪疄,鎸夎姹備綔绛.(1)鍦25鈩銆101kPa涓,1g鐢查唶鐕冪儳鐢熸垚CO2鍜屾恫鎬...
绛旓細锛1锛鏍规嵁鏉′欢寰楋細 CH3OH锛坙锛+32O2锛坓锛夆啋CO2锛坓锛+2H2O锛坙锛 鏀惧嚭鐑噺 1g 22.68kJ 32g 725.76 kJ鐢查唶鐕冪儳鐑殑鍖栧鏂圭▼寮忎负锛欳H3OH锛坙锛+32O2锛坓锛夆啋CO2锛坓锛+2H2O锛坙锛夆柍H=-725.76 kJ?mol-1鏁呯瓟妗堜负锛欳H3OH锛坙锛+32O2锛坓锛夆啋CO2锛坓锛+2H2O锛坙锛夆柍H=-725...

姹備綔涓氱瓟妗:鍦25鈩 101kPa涓,鐧
绛旓細D

姹備綔涓氱瓟妗:(1)鍦25鈩銆101kPa
绛旓細锛1锛CH 4 锛坓锛+ 2O 2 锛坓锛夛紳 CO 2 锛坓锛+ 2H 2 O锛坙锛 螖H锛-880kJ锛弇ol 锛2锛壩擧3锛+260kJ锛弇ol 锛3锛塏O 2 锛坓锛+CO锛坓锛夛紳CO 2 锛坓锛+NO锛坓锛 螖H锛-234kJ锛弇ol

渚濇嵁浜嬪疄,濉┖:(1)鍦25鈩銆101kPa涓,1g鐢查唶娑蹭綋瀹屽叏鐕冪儳鐢熸垚CO2鍜屾恫鎬...
绛旓細锛1锛1g鐢查唶鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鏃舵斁鐑22.68kJ锛屽垯64g鐢查唶鍗2mol鐢查唶鐕冪儳鏀剧殑鐑噺涓1452kJ锛岃岀噧鐑х儹鏄寚1mol鍙噧鐗╁畬鍏ㄧ噧鐑х敓鎴愮ǔ瀹氱殑姘у寲鐗╂斁鍑虹殑鐑噺锛屽垯鐢查唶鐕冪儳鐑殑鐑寲瀛︽柟绋嬪紡涓篊H3OH锛坙锛+32O2锛坓锛夆啋CO2锛坓锛+2H2O锛屾晠绛旀涓猴細CH3OH锛坙锛+32O2锛坓锛夆啋CO2锛坓锛+2H2O锛坙锛夆柍H=-...

渚濇嵁鍙欒堪,鍐欏嚭涓嬪垪鍙嶅簲鐨勭儹鍖栧鏂圭▼寮忋
绛旓細銆愮瓟妗堣В鏋愩戣瘯棰樺垎鏋愶細(1)鍦25鈩銆101kPa涓锛1g娑叉佺敳閱囷紙CH3OH锛夌噧鐑х敓鎴怌O2鍜屾恫鎬佹按鏃舵斁鐑22.68kJ锛屾墍浠1mol娑叉佺敳閱囷紙CH3OH锛夌噧鐑х敓鎴怌O2鍜屾恫鎬佹按鏃舵斁鐑负32*22.68=725.76kJ锛屾墍浠ョ敳閱囩噧鐑х儹鐨勭儹鍖栧鏂圭▼寮忎负CH3OH(l)锛婳2(g)===CO2(g)锛2H2O(l)螖H锛濓紞725.76kJ/mol锛23g NO2鍗充负0...

(1)鍦25鈩銆101kPa鐘跺喌涓,4g姘㈡皵鍜岄傞噺姘ф皵鍙嶅簲鐢熸垚1mol姘磋捀姘旀斁鐑241.8...
绛旓細锛1锛鐢熸垚1mol姘磋捀姘旀斁鐑241.8kJ锛屽湪鍙嶅簲2H2锛坓锛+O2锛坓锛夆晲2H2O锛坓锛変腑锛岀敓鎴2molHCl鏀惧嚭2脳241.8kJ=483.6kJ锛屽垯鐑寲瀛︽柟绋嬪紡涓2H2锛坓锛+O2锛坓锛夆晲2H2O锛坓锛夆柍H=-483.6kJ/mol锛屾晠绛旀涓猴細2H2锛坓锛+O2锛坓锛夆晲2H2O锛坓锛夆柍H=-483.6kJ/mol锛涳紙2锛夊悇鍖栧閿敭鑳戒负P-P 198kJ?

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