依据事实，写出下列反应的热化学方程式．（1）在25℃、101kPa下，1g液态甲醇（CH3OH）完全燃烧生成CO2和 依据事实，写出下列反应的热化学方程式．（1）在25℃、101...
\u4f9d\u636e\u4e8b\u5b9e\uff0c\u5199\u51fa\u4e0b\u5217\u53cd\u5e94\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\uff0e\uff081\uff09\u572825\u2103\u3001101kPa\u4e0b\uff0c1g\u7532\u9187\uff08CH3OH\uff09\u71c3\u70e7\u751f\u6210CO2\u548c\u6db2\u6001\u6c34\u65f6\uff081\uff09\u572825\u2103\u3001101kPa\u4e0b\uff0c1g\u7532\u9187\uff08CH3OH\uff09\u71c3\u70e7\u751f\u6210CO2\u548c\u6db2\u6001\u6c34\u65f6\u653e\u70ed22.68kJ\uff0e32g\u7532\u9187\u71c3\u70e7\u751f\u6210\u4e8c\u6c27\u5316\u78b3\u548c\u6db2\u6001\u6c34\u653e\u51fa\u70ed\u91cf\u4e3a725.76KJ\uff1b\u5219\u8868\u793a\u7532\u9187\u71c3\u70e7\u70ed\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aCH3OH\uff08l\uff09+32O2\uff08g\uff09=CO2\uff08g\uff09+2H2O\uff08l\uff09\u25b3H=-725.76kJ?mol-1\uff0c\u6545\u7b54\u6848\u4e3a\uff1aCH3OH\uff08l\uff09+32O2\uff08g\uff09=CO2\uff08g\uff09+2H2O\uff08l\uff09\u25b3H=-725.76kJ?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\u751f\u621046g\u4e8c\u6c27\u5316\u6c2e\u53cd\u5e94\u653e\u51fa\u70ed\u91cf67.8KJ\uff0c\u53cd\u5e94\u8272\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\u5728C2H2\uff08\u6c14\u6001\uff09\u5b8c\u5168\u71c3\u70e7\u751f\u6210CO2\u548c\u6db2\u6001\u6c34\u7684\u53cd\u5e94\u4e2d\uff0c\u6bcf\u67095NA\u4e2a\u7535\u5b50\u8f6c\u79fb\u65f6\uff0c\u653e\u51fa650kJ\u7684\u70ed\u91cf\uff0c\u53cd\u5e94C2H2\uff08g\uff09+52O2\uff08g\uff09=2CO2\uff08g\uff09+H2O\uff08l\uff09\u4e2d\u7535\u5b50\u8f6c\u79fb\u4e3a10mol\uff0c\u5219\u8f6c\u79fb10mol\u7535\u5b50\u653e\u51fa\u70ed\u91cf1300KJ\uff0c\u53cd\u5e94\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1aC2H2\uff08g\uff09+52O2\uff08g\uff09=2CO2\uff08g\uff09+H2O\uff08l\uff09\u25b3H=-1300kJ?mol-1\uff0c\u6545\u7b54\u6848\u4e3a\uff1aC2H2\uff08g\uff09+52O2\uff08g\uff09=2CO2\uff08g\uff09+H2O\uff08l\uff09\u25b3H=-1300kJ?mol-1\uff0e
\uff081\uff09\u71c3\u70e7\u70ed\u662f1mol\u53ef\u71c3\u7269\u5b8c\u5168\u71c3\u70e7\u751f\u6210\u7a33\u5b9a\u6c27\u5316\u7269\u65f6\u653e\u51fa\u7684\u70ed\u91cf\uff1b\u572825\u2103\u3001101kPa\u4e0b\uff0c1g\u7532\u9187\uff08CH3OH\uff09\u71c3\u70e7\u751f\u6210CO2\u548c\u6db2\u6001\u6c34\u65f6\u653e\u70ed22.68kJ\uff0c1mol\u7532\u9187\u537332g\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\uff09Fe2O3\uff08s\uff09+3CO\uff08g\uff09=2Fe\uff08s\uff09+3CO2\uff08g\uff09\u25b3H=-24.8kJ/mol \u2460 3Fe2O3\uff08s\uff09+CO\uff08g\uff09=2Fe3O4\uff08s\uff09+CO2\uff08g\uff09\u25b3H=-47.2kJ/mol \u2461 Fe3O4\uff08s\uff09+CO\uff08g\uff09=3FeO\uff08s\uff09+CO2\uff08g\uff09\u25b3H=+640.5kJ/mol \u2462\u7531\u2460\u00d73-\u2461-\u2462\u00d72\u5f97 6CO\uff08g\uff09+6FeO\uff08s\uff09=6Fe\uff08s\uff09+6CO2\uff08g\uff09\u25b3H=\uff08-24.8kJ/mol\uff09\u00d73-\uff08-47.2kJ/mol\uff09-\uff08+640.5kJ/mol\uff09\u00d72=-1308.0kJ/mol\uff0c\u5373 CO\uff08g\uff09+FeO\uff08s\uff09=Fe\uff08s\uff09+CO2\uff08g\uff09\u25b3H=-218.0kJ/mol \u6545\u7b54\u6848\u4e3a\uff1aCO\uff08g\uff09+FeO\uff08s\uff09=Fe\uff08s\uff09+CO2\uff08g\uff09\u25b3H=-218.0kJ/mol\uff0e
（1）在25℃、101kPa下，1g甲醇（CH3OH）燃烧生成CO2和液态水时放热22.68kJ．32g甲醇燃烧生成二氧化碳和液态水放出热量为725.76KJ；则表示甲醇燃烧热的热化学方程式为：CH3OH（l）+| 3 |
| 2 |
故答案为：CH3OH（l）+
| 3 |
| 2 |
（2）适量的N2和O2完全反应，每生成23克NO2需要吸收16.95kJ热量，生成46g二氧化氮反应放出热量67.8KJ，反应色热化学方程式为：N2（g）+2O2（g）=2NO2（g）△H=67.8kJ?mol-1，
故答案为：N2（g）+2O2（g）=2NO2（g）△H=67.8kJ?mol-1；
（3）在反应N2+3H2?2NH3中，断裂3molH-H键，1molN三N键共吸收的能量为3×436kJ+946kJ=2254kJ，生成2molNH3，共形成6molN-H键，放出的能量为6×391kJ=2346kJ，吸收的能量少，放出的能量多，该反应为放热反应，放出的热量为2346kJ-2254kJ=92kJ，N2与H2反应生成NH3的热化学方程式为，N2（g）+3H2（g）?2NH3（g）△H=-92kJ?mol-1；
故答案为：N2（g）+3H2（g）?2NH3（g）△H=-92kJ?mol-1；
（4）①Fe（s）+
| 1 |
| 2 |
②2Al（s）+
| 3 |
| 2 |
将方程式②-①×3得2Al（s）+3FeO（s）═Al2O3（s）+3Fe（s）△H=-859.7 kJ?mol-1，
故答案为：2Al（s）+3FeO（s）═Al2O3（s）+3Fe（s）△H=-859.7 kJ?mol-1．
绛旓細鍙嶅簲鐨勭儹鍖栧鏂圭▼寮忎负锛歂2锛坓锛+3H2锛坓锛=2NH3锛坓锛夆柍H=-92.2kJ/mol锛涙晠绛旀涓猴細N2锛坓锛+3H2锛坓锛=2NH3锛坓锛夆柍H=-92.2kJ/mol锛涳紙2锛1molN2锛坓锛変笌閫傞噺O2锛坓锛夎捣鍙嶅簲鐢熸垚NO锛坓锛夛紝鍚告敹68kJ鐑噺锛屽弽搴旂殑鐑寲瀛︽柟绋嬪紡涓猴細N2锛坓锛+O2锛坓锛=NO锛坓锛夆柍H=+68KJ/mol锛...
绛旓細mol-1锛屾晠绛旀涓猴細CH3OH锛坙锛+32O2锛坓锛=CO2锛坓锛+2H2O锛坙锛夆柍H=-725.76kJ?mol-1锛涳紙2锛夐傞噺鐨凬2鍜孫2瀹屽叏鍙嶅簲锛屾瘡鐢熸垚23鍏婲O2闇瑕佸惛鏀16.95kJ鐑噺锛岀敓鎴46g浜屾哀鍖栨爱鍙嶅簲鏀惧嚭鐑噺67.8KJ锛屽弽搴旇壊鐑寲瀛︽柟绋嬪紡涓猴細N2锛坓锛+2O2锛坓锛=2NO2锛坓锛夆柍H=67.8kJ?mol-1锛屾晠绛旀涓猴細N2锛坓锛...
绛旓細mol -1 锛2锛塁 2 H 5 OH(l)+3O 2 (g)=2CO 2 (g)+3H 2 O(l) 鈻矵锛濃1366锛8kJ?mol -1 锛3锛堿l(s)+ O 2 (g)=Al 2 O 3 (s) 鈻矵锛濃1669锛8kJ?mol -1 璇曢鍒嗘瀽锛氳棰樿冩煡鐨勬槸鐑寲瀛︽柟绋嬪紡鐨勪功鍐欙紝鍏充簬鐑寲瀛︽柟绋嬪紡鐨勪功鍐欒娉ㄦ剰浠ヤ笅鍑犵偣锛氾紙1锛夆柍H鍙兘鍐...
绛旓細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锛涚噧鐑х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+32 O2锛坓锛=CO2锛坓锛+2H2O锛坙锛夆柍H=-725.76 kJ/mol锛涳紙3锛0.4mol娑叉佽偧鍜岃冻閲廐2O2鍙嶅簲锛岀敓鎴愭爱姘斿拰姘磋捀姘旓紝鏀惧嚭...
绛旓細锛1锛1molN 2 锛坓锛変笌閫傞噺O 2 锛坓锛夎捣鍙嶅簲锛岀敓鎴怤O 2 锛坓锛夛紝鍚告敹68kJ鐑噺锛屽弽搴斿惛鐑椂鐒撳彉鍊间负姝ｅ锛屾墍浠ヨ鍙嶅簲鐨勭儹鍖栧鏂圭▼寮忎负N 2 锛坓锛+2O 2 锛坓锛=2NO 2 锛坓锛夆柍H=+68kJ?mol -1 锛屾晠绛旀涓猴細N 2 锛坓锛+2O 2 锛坓锛=2NO 2 锛坓锛夆柍H=+68kJ?mol -1 锛涳紙2锛夋斁鐑...
绛旓細=2NO2锛坓锛夆柍H=+68kJ/mol锛屾晠绛旀涓猴細N2锛坓锛+2O2锛坓锛夆晲2NO2锛坓锛夆柍H=+68kJ/mol锛涳紙3锛2mol涔欑倲鍙嶅簲杞Щ20NA涓數瀛愶紝鍦–2H2锛堟皵鎬侊級瀹屽叏鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鐨勫弽搴斾腑锛屾瘡鏈5NA涓數瀛愯浆绉绘椂锛屾斁鍑650kJ鐨勭儹閲忥紝鎵浠ユ湁20NA涓數瀛愯浆绉绘椂锛屾斁鍑2600kJ鐨勭儹閲忥紝鍒鐑寲瀛鏂圭▼寮忎负锛2C2H2锛...
绛旓細锛1锛1molC涓1mol姘磋捀姘鍙嶅簲鐢熸垚1mol CO鍜1mol H2锛屽惛鏀131.5kJ鐨勭儹閲忥紝鍒鐑寲瀛鏂圭▼寮忎负锛欳 锛圫锛+H2O 锛坓锛 CO 锛坓锛+H2 锛坓锛夆柍H=+131.5kJ?mol-1锛屾晠绛旀涓猴細C 锛圫锛+H2O 锛坓锛 CO 锛坓锛+H2 锛坓锛夆柍H=+131.5kJ?mol-1锛涳紙2锛夊湪鍙嶅簲N2+3H2?2NH3涓紝鏂3molH-H閿紝...
绛旓細2NH 3 锛屼緷鎹寲瀛鏂圭▼寮忚绠鍙嶅簲鐑=946kJ/mol+3脳436kJ/mol-6脳391kJ/mol=-92KJ/mol锛鍙嶅簲鐨勭儹鍖栧鏂圭▼寮忥細N 2 锛坓锛+3H 2 锛坓锛?2NH 3 锛坓锛夆柍H=-92 kJ?mol -1 锛屾晠绛旀涓猴細N 2 锛坓锛+3H 2 锛坓锛?2NH 3 锛坓锛夆柍H=-92 kJ?mol -1 锛涳紙3锛夆憼CH 4 锛坓锛+H 2 O锛...
绛旓細mol-1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=-725.76kJ?mol-1锛涳紙2锛夐傞噺鐨凬2鍜孫2瀹屽叏鍙嶅簲锛姣忕敓鎴23鍏婲O2闇瑕佸惛鏀16.95kJ鐑噺锛屾墍浠ユ瘡鐢熸垚92鍏婲O2闇瑕佸惛鏀67.8kJ鐑噺锛屽垯鐑寲瀛鏂圭▼寮忎负锛歂2锛坓锛+2O2锛坓锛=2NO2锛坓锛夆柍H=67.8kJ?mol-1锛屾晠绛旀涓猴細N2锛坓锛...
绛旓細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锛坙锛夆柍H=-725.76 kJ?mol-1锛涳紙2锛1.00L 1.00mol/L H2SO4婧舵恫涓2.00L 1.00mol/L NaOH婧舵恫瀹屽叏鍙嶅簲锛屾斁鍑114.6kJ鐨勭儹閲忥紝鍗崇敓鎴2mol姘存斁鍑114.6kJ鐨勭儹閲锛屽弽搴旂殑鍙嶅簲鐑涓-11.46kJ/mol锛屼腑鍜岀儹涓-57.3kJ/mol锛屽垯涓拰...
