计算：（1）在25℃、101kPa下，1g甲醇燃烧生成CO2和液态水时放热22.68kJ．则表示甲醇燃烧热的热化学方程 在25℃，101kpa下，1g甲醇燃烧生成二氧化碳和液态水时...
\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\uff09CH3OH(l)+3/2O2(g)=CO2(g)+2H2O(l) \u0394H\uff1d-725.76kJ\u00b7mol-1
\uff082\uff09N2(g)+2O2(g)=2NO2(g) \u0394H\uff1d+67.8kJ\u00b7mol-1
\uff083\uff09N2(g)+3H2(g)=2NH3(g) \u0394H\uff1d-92kJ\u00b7mol-1
\uff084\uff09CO(g)+FeO(s)=CO2(g)+Fe(s) \u0394H\uff1d-11kJ\u00b7mol-1
| 3 |
| 2 |
故答案为：CH3OH（l）+
| 3 |
| 2 |
（2）在反应N2+3H2?2NH3中，断裂3molH-H键，1mol N三N键共吸收的能量为：3×436kJ+946kJ=2254kJ，生成2mol NH3，共形成6mol N-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；
故答案为：N2（g）+3H2（g）?2NH3（g）△H=-92kJ?mol-1；
（3）①pH=12 的NaOH溶液中氢氧根离子浓度是0.01mol/L，pH=11的氢氧化钠溶液中氢氧根离子浓度是0.001mol/L，设加入水的体积是V2，C1V1=C2（V1+V2）=0.01mol/L×0.1L=（0.1+V2）L×0.001mol/L，V2=
| 0.01mol/L×0.1L |
| 0.001mol/L |
故答案为：900；
②pH=12 的NaOH溶液中氢氧根离子浓度是0.01mol/L，pH=11的氢氧化钠溶液中氢氧根离子浓度是0.001mol/L，pH=10的氢氧化钠溶液中氢氧根离子浓度是0.0001mol/L，
设加入pH=10的NaOH溶液体积是V2，C1V1+C2V2=C3（V1+V2）=0.01mol/L×0.1L+0.0001mol/L×V2=0.001mol/L（0.1+V2），V2=1L=1000mL，
故答案为：1000；
③盐酸的浓度是0.01mol/L，设加入盐酸的体积是V，C（OH-）=
| n(碱)?n(酸) |
| V(酸)+V(碱) |
| 0.01mol/L×0.1L?0.01mol/L×VL |
| (0.1+V)L |
故答案为：81.8．
绛旓細锛1锛25鈩銆101KPa鏃锛1molH2瀹屽叏鐕冪儳鐢熸垚娑叉佹按閲婃斁鍑285.8KJ鐨勮兘閲忥紝鍒欒〃绀篐2鐕冪儳鐑殑鍖栧鏂圭▼寮忎负锛欻2锛坓锛+12O2锛坓锛=H2O锛坙锛夆柍H=-285.8kJ/mol锛屾晠绛旀涓猴細H2锛坓锛+12O2锛坓锛=H2O锛坙锛夆柍H=-285.8kJ/mol锛涳紙2锛夎瑁呯疆鏄妸鍖栧鐗╄川涓殑鍖栧鑳借浆鍖栦负鐢佃兘锛屾墍浠ユ槸鍖栧鑳借浆鍙樹负鐢佃兘...
绛旓細1 g姘㈡皵瀹屽叏鐕冪儳鐢熸垚娑叉佹按鏃舵斁鍑142.9kJ鐑噺锛屽垯2g姘㈡皵鍗1mol姘㈡皵瀹屽叏鐕冪儳鐢熸垚娑叉佹按鏃舵斁鍑142.9kJ鐑噺脳2锛285.8kJ銆傜敱浜庣噧鐑х儹鏄湪涓瀹氭潯浠朵笅锛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忥紝鍒欒〃绀烘阿姘旂噧鐑х儹鐨勭儹鍖栧鏂圭▼寮忎负H 2 (g)+ 1/2O 2 (g)=H 2 O(l) 鈻矵锛-285.8...
绛旓細锛1锛塇 2 锛坓锛+1/2O 2 锛坓锛==H 2 O锛坓锛 鈻矵=锛241.8 kJ/mol 锛 +131.3锛2锛夆憼K= 锛涒憽50% 锛> 锛= 鈶 闄嶄綆
绛旓細锛1锛1 molCH4瀹屽叏鐕冪儳鐢熸垚娑叉佹按鍜孋O2锛屾斁鍑890.3kJ鐨勭儹閲忥紝鍒欑儹鍖栧鏂圭▼寮忎负锛欳H4锛坓锛+2O2锛坓锛夆晲CO2锛坓锛+2H2O锛坙锛夆柍H=-890.3 kJ?mol锛涳紙2锛夆憼C锛坰锛+O2锛坓锛夆晲CO2锛坓锛夆柍H1=-437.3kJ?mol-1 鈶2锛坓锛+12O2锛坓锛夆晲H2O锛坓锛夆柍H3=-285.8kJ?mol-1鈶O锛坓锛+12O2锛坓...
绛旓細锛1锛塴g鐢茬兎瀹屽叏鐕冪儳鐢熸垚CO2鍜屾恫鎬佹按鏃舵斁鐑55.6kJ鐑噺锛16g鐢茬兎鐕冪儳鏀剧儹889.6KJ锛屼緷鎹儹鍖栧鏂圭▼寮忎功鍐欐柟娉曟爣娉ㄨ仛闆嗙姸鎬佸拰瀵瑰簲鐒撳彉鍐欏嚭鐑寲瀛︽柟绋嬪紡涓猴細CH4锛坓锛+2O2锛坓锛夆晲CO2锛坓锛+2H2O锛坙锛夆柍H=-889.6 kJ/mol锛屾晠绛旀涓猴細CH4锛坓锛+2O2锛坓锛夆晲CO2锛坓锛+2H2O锛坙锛夆柍H=-889.6 kJ/mol锛...
绛旓細骞宠 鍚戞鍙嶅簲鏂瑰悜绉诲姩 锛3鍒嗭級 璇曢鍒嗘瀽锛氾紙1锛鍦25鈩銆101kPa鏃讹紝H 2 涓嶰 2 鍖栧悎鐢熸垚1molH 2 O(g)鏀惧嚭241.8kJ鐨勭儹閲忥紝鎵浠ュ叾鐑寲瀛︽柟绋嬪紡涓篐 2 (g)+ O 2 (g)鈺恖H 2 O(g) 鈻矵锛濓紞241.8kJ/mol銆傚凡鐭モ憼C(s)锛婳 2 (g)鈺怌O 2 (g) 鈻矵锛濓紞393.5kJ/mol鈶O(...
绛旓細1.姒傚康:25 鈩,101 kPa鏃,1 mol绾墿璐ㄥ畬鍏ㄧ噧鐑х敓鎴愮ǔ瀹氱殑鍖栧悎鐗╂椂鎵鏀惧嚭鐨勭儹閲忋傜噧鐑х儹鐨勫崟浣嶇敤kJ/mol琛ㄧず銆傗绘敞鎰忎互涓嬪嚑鐐: 鈶犵爺绌舵潯浠:101 kPa鈶″弽搴旂▼搴:瀹屽叏鐕冪儳,浜х墿鏄ǔ瀹氱殑姘у寲鐗┿ 鈶㈢噧鐑х墿鐨勭墿璐ㄧ殑閲:1 mol 鈶ｇ爺绌跺唴瀹:鏀惧嚭鐨勭儹閲忋(螖H<0,鍗曚綅kJ/mol) 鍥涖佷腑鍜岀儹1.姒傚康:鍦ㄧ█婧舵恫涓,閰歌窡纰...
绛旓細锛1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725.8KJ锛岀噧鐑х儹鐑寲瀛︽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锛鏁呯瓟妗堜负锛欳H3OH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...
绛旓細CH4锛坓)+2O2(g锛=CO2(g)+2H2O(l) 鈻矵=362.88mol/KJ
绛旓細锛1锛鍦25鈩銆101kPa涓嬶紝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锛鏁呯瓟妗堜负锛欳H3OH锛坙锛+32O2锛坓锛夆晲CO2锛...
