依据事实，写出下列反应的热化学方程式．（1）在250C、101KPa时，1克甲醇（CH30H）完全燃烧生成CO2和液态 依据事实，写出下列反应的热化学方程式。(1)在25℃、101...
\uff081\uff09\u71c3\u70e7\u9700\u8981\u7684\u4e09\u4e2a\u6761\u4ef6\u662f\uff1a\u2460\u53ef\u71c3\u7269\uff1b \u2461______\uff1b\u2462______\uff0e\uff082\uff09\u5199\u51fa\u4e0b\u5217\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\uff0e\u2460\u71c3\u70e7\u7ea2\uff081\uff09\u71c3\u70e7\u7684\u4e09\u4e2a\u6761\u4ef6\u662f\uff1a\u53ef\u71c3\u7269\u3001\u4e0e\u6c27\u6c14\u63a5\u89e6\u3001\u6e29\u5ea6\u8981\u8fbe\u5230\u7740\u706b\u70b9\uff0e\uff082\uff09\u2460\u7ea2\u78f7\u71c3\u70e7\u4ea7\u751f\u5927\u91cf\u7684\u767d\u70df\uff0c\u53ef\u7528\u4e8e\u5236\u9020\u70df\u5e55\uff0c\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1a4P+5O2 \u70b9\u71c3 . 2P2O5 \uff0e\u2461\u8fc7\u6c27\u5316\u6c22\u5728\u4e8c\u6c27\u5316\u9530\u7684\u50ac\u5316\u4f5c\u7528\u4e0b\u751f\u6210\u6c34\u548c\u6c27\u6c14\uff0c\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1a2H2O2 MnO2 . 2H2O+O2\u2191\uff0e\u2462\u94dd\u7c89\u4e0e\u6c27\u5316\u94c1\u9ad8\u6e29\u4e0b\u751f\u6210\u94c1\u548c\u6c27\u5316\u94dd\uff0c\u53cd\u5e94\u7684\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3a\uff1a2Al+Fe2O3 \u9ad8\u6e29 . 2Fe+Al2O3\uff0e\u6545\u7b54\u6848\u4e3a\uff1a\uff081\uff09\u4e0e\u6c27\u6c14\u63a5\u89e6\uff1b\u6e29\u5ea6\u8981\u8fbe\u5230\u7740\u706b\u70b9\uff1b\uff082\uff09\u24604P+5O2 \u70b9\u71c3 . 2P2O5 \uff1b\u24612H2O2 MnO2 . 2H2O+O2\u2191\uff1b\u24622Al+Fe2O3 \u9ad8\u6e29 . 2Fe+Al2O3\uff0e
(1)CH 3 OH(l)\uff0b O 2 (g)===CO 2 (g)\uff0b2H 2 O(l) \u0394H\uff1d\uff0d725.76 kJ\u00b7mol \uff0d 1 (2)N 2 (g)\uff0b2O 2 (g)===2NO 2 (g) \u0394H\uff1d\uff0b67.8 kJ\u00b7mol \uff0d 1 \u8bd5\u9898\u5206\u6790\uff1a\uff081\uff09\u71c3\u70e7\u70ed\u662f\u5728\u4e00\u5b9a\u6761\u4ef6\u4e0b\uff0c1mol\u53ef\u71c3\u7269\u5b8c\u5168\u71c3\u70e7\u751f\u6210\u7a33\u5b9a\u7684\u6c27\u5316\u7269\u65f6\u6240\u653e\u51fa\u7684\u70ed\u91cf\uff0c\u6240\u4ee51mol\u7532\u9187\u5b8c\u5168\u71c3\u70e7\u653e\u51fa\u7684\u70ed\u91cf\u662f22.68 kJ\u00d732\uff1d725.76 kJ\uff0c\u5219\u793a\u7532\u9187\u71c3\u70e7\u70ed\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3aCH 3 OH(l)\uff0b O 2 (g)===CO 2 (g)\uff0b2H 2 O(l) \u0394H\uff1d\uff0d725.76 kJ\u00b7mol \uff0d 1 \u3002\uff082\uff09\uff0c\u6bcf\u751f\u621023 g NO 2 \u9700\u8981\u5438\u653616.95 kJ\u70ed\u91cf\uff0c\u5219\u8868\u793a\u8be5\u53cd\u5e94\u7684\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u4e3aN 2 (g)\uff0b2O 2 (g)===2NO 2 (g) \u0394H\uff1d\uff0b67.8 kJ\u00b7mol \uff0d 1 \u3002\u70b9\u8bc4\uff1a\u8be5\u9898\u662f\u57fa\u7840\u6027\u8bd5\u9898\u7684\u8003\u67e5\uff0c\u4fa7\u91cd\u5bf9\u5b66\u751f\u57fa\u7840\u77e5\u8bc6\u7684\u8bad\u7ec3\u3002\u8be5\u9898\u7684\u5173\u952e\u662f\u660e\u786e\u70ed\u5316\u5b66\u65b9\u7a0b\u5f0f\u7684\u542b\u4e49\u4ee5\u53ca\u4e66\u5199\u6ce8\u610f\u4e8b\u9879\uff0c\u6709\u5229\u4e8e\u57f9\u517b\u5b66\u751f\u89c4\u8303\u7684\u7b54\u9898\u80fd\u529b\uff0c\u63d0\u9ad8\u5b66\u751f\u7684\u5e94\u8bd5\u80fd\u529b\u548c\u5b66\u4e60\u6548\u7387\u3002
（1）在25℃、101kPa下，1g甲醇（CH3OH）燃烧生成CO2和液态水时放热22.68kJ，1mol甲醇完全燃烧生成二氧化碳和液态水放热为22.68kJ×32=725.76KJ，所以甲醇燃烧热的热化学方程式为CH3OH（l）+| 3 |
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故答案为：CH3OH（l）+
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（2）1.00L 1.00mol/L H2SO4溶液与2.00L 1.00mol/L NaOH溶液完全反应，放出114.6kJ的热量，即生成2mol水放出114.6kJ的热量，反应的反应热为-11.46kJ/mol，中和热为-57.3kJ/mol，则中和热的热化学方程式：NaOH（aq）+
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故答案为：NaOH（aq）+
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绛旓細mol-1锛涙晠绛旀涓猴細C2H5OH锛坙锛+3O2锛坓锛夆晲2CO2锛坓锛+3H2O锛坙锛夆柍H=-10588 kJ?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16.95 kJ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锛屾晠绛旀涓猴細N2锛坓...
绛旓細锛1锛塁(g)+O 2 (g)=CO 2 (g) 鈻矵= -393.6 kJ/mol 锛2锛塁H 3 OH(l)+O 2 (g)鈫扖O 2 (g)+2H 2 O(l) 螖H=-725.76kJ路mol -1
绛旓細鐑寲瀛鏂圭▼寮忔槸琛ㄧず鍖栧鍙嶅簲涓鍙嶅簲鐑鍏崇郴鐨勬柟绋嬪紡銆備功鍐欏拰搴旂敤鐑寲瀛︽柟绋嬪紡鏃跺繀椤绘敞鎰浠ヤ笅鍑犵偣锛氣憼鏄庣‘鍐欏嚭鍙嶅簲鐨璁￠噺鏂圭▼寮忥紝鍚勭墿璐ㄥ寲瀛﹀紡鍓嶇殑鍖栧璁￠噺绯绘暟鍙互鏄暣鏁帮紝涔熷彲浠ユ槸鍒嗘暟銆傗憽鍚勭墿璐ㄥ寲瀛﹀紡鍙充晶鐢ㄥ渾鎷姬锛堬級琛ㄦ槑鐗╄川鐨勮仛闆嗙姸鎬併傚彲浠ョ敤g銆乴銆乻鍒嗗埆浠ｈ〃姘旀併佹恫鎬併佸浐鎬併傚浐浣撴湁涓嶅悓鏅舵佹椂锛岃繕闇灏...
绛旓細锛1锛塁H 3 OH锛坙锛+O 2 锛坓锛夆啋CO 2 锛坓锛+2H 2 O 螖H锛-725锛76kJ路mol -1 锛2锛57.3kJ/mol锛 鈻矵锛濓紞57.3kJ/mol 锛3锛塏 2 锛坓锛+3H 2 锛坓锛=2NH 3 锛坓锛壩擧锛-92kJ路mol -1
绛旓細锛1锛塁H 3 OH(l)+3/2O 2 (g)鈫扖O 2 (g)+2H 2 O(l)锛浳擧=-725锛76kJ路mol -1 锛2锛塏 2 (g)+2O 2 (g)==2NO 2 (g)锛浳擧=67锛8kJ路mol -1 锛3锛塁 2 H 2 (g)锛5/2O 2 (g)鈫2CO 2 (g)+H 2 O(l)锛浳擧=-1300kJ路mol -1 锛4锛塏 2 (g)+3H...
绛旓細3kJ/mol銆傚洜姝よ〃绀哄叾涓拰鐑殑鐑寲瀛︽柟绋嬪紡涓 鈻矵锛濓紞57锛3kJ/mol銆傦紙3锛鍙嶅簲鐑灏辨槸鏂敭鍚告敹鐨勮兘閲忓拰褰㈡垚鍖栧閿墍鏀惧嚭鐨勮兘閲忕殑宸硷紝鎵浠ヨ鍙嶅簲鐨勫弽搴旂儹鏄436kJ/mol脳3锛946kJ/mol锛2脳3脳391kJ/mol锛濓紞92kJ/mol锛屾墍浠ヨ鍙嶅簲鐨勭儹鍖栧鏂圭▼寮忔槸N 2 锛坓锛+3H 2 锛坓锛=2NH 3 锛坓锛...
绛旓細1mol鍙噧鐗╁畬鍏ㄧ噧鐑х敓鎴愮ǔ瀹氱殑姘у寲鐗╂椂鎵鏀惧嚭鐨勭儹閲锛屾牴鎹棰樻剰1mol涔欓唶瀹屽叏鐕冪儳鏀惧嚭鐨勭儹閲忔槸29.7kJ脳46锛1366.8kJ锛屾墍浠鐑寲瀛鏂圭▼寮忎负C 2 H 5 OH(l)+3 O 2 (g)=2CO 2 (g)+3H 2 O(l) 螖H锛濃1366.8kJ路mol 锛 1 銆傦紙2锛23g NO 2 闇瑕佸惛鏀16.95kJ鐑噺锛屽垯鐢熸垚2molNO 2 ...
绛旓細N2(g)+3H2(g)==NH3(g) 螖H=-92.9 kj /mol
绛旓細NO2鍗充负0.5molNO2锛屾墍浠2鍜孫2瀹屽叏鍙嶅簲鐢熸垚NO2鐨勭儹鍖栧鏂圭▼寮忎负 N2(g)锛2O2(g)===2NO2(g)螖H锛濓紜67.8kJ/mol锛(3)C2H2(姘旀)瀹屽叏鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鐨勫弽搴旀柟绋嬪紡涓篊2H2锛婳2===2CO2锛婬2O锛岃鍙嶅簲杩囩▼杞Щ鐨勭數瀛愭暟涓10锛岀敱浜庢瘡5NA涓數瀛愯浆绉绘椂锛屾斁鍑650kJ鐨勭儹閲忥紝鎵浠ヤ竴鎽╁皵C2H2瀹屽叏鐕冪儳...
绛旓細锛堝叡5鍒嗭級锛1锛塁H 3 OH(l)+3/2 O 2 (g)=CO 2 (g) +2H 2 O(l) 鈻矵="-725.76kJ/mol " 锛2鍒嗭級(2) CO (g)+ FeO (s)=CO 2 (g) +Fe (s) 锛3鍒嗭級 鐣
