2021年高考化學(xué)真題和模擬題分類匯編專題09化學(xué)能與熱能【含答案】

專題09化學(xué)能與熱能/r/n2021年化學(xué)高考題/r/n一、單選題/r/n1．（山東高考真題試卷）/r/n18/r/nO/r/n標記的乙酸甲酯在足量/r/nNaOH/r/n溶液中發(fā)生水解，部分反應(yīng)歷程可表示為：/r/n+OH/r/n-/r/n+CH/r/n3/r/nO/r/n-/r/n能量變化如圖所示。已知/r/n為快速平衡，下列說法正確的是/r/nA．/r/n反應(yīng)/r/nⅡ/r/n、/r/nⅢ/r/n為決速步/r/nB．/r/n反應(yīng)結(jié)束后，溶液中存在/r/n18/r/nOH/r/n-/r/nC．/r/n反應(yīng)結(jié)束后，溶液中存在/r/nCH/r/n3/r/n18/r/nOH/r/nD．/r/n反應(yīng)/r/nⅠ/r/n與反應(yīng)/r/nⅣ/r/n活化能的差值等于圖示總反應(yīng)的焓變/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．一般來說，反應(yīng)的活化能越高，反應(yīng)速率越慢，由圖可知，反應(yīng)/r/nI/r/n和反應(yīng)/r/nIV/r/n的活化能較高，因此反應(yīng)的決速步為反應(yīng)/r/nI/r/n、/r/nIV/r/n，故/r/nA/r/n錯誤；/r/nB/r/n．反應(yīng)/r/nI/r/n為加成反應(yīng)，而/r/n與/r/n為快速平衡，反應(yīng)/r/nII/r/n的成鍵和斷鍵方式為/r/n或/r/n，后者能生成/r/n18/r/nOH/r/n-/r/n，因此反應(yīng)結(jié)束后，溶液中存在/r/n18/r/nOH/r/n-/r/n，故/r/nB/r/n正確；/r/nC/r/n．反應(yīng)/r/nIII/r/n的成鍵和斷鍵方式為/r/n或/r/n，因此反應(yīng)結(jié)束后溶液中不會存在/r/nCH/r/n3/r/n18/r/nH/r/n，故/r/nC/r/n錯誤；/r/nD/r/n．該總反應(yīng)對應(yīng)反應(yīng)物的總能量高于生成物總能量，總反應(yīng)為放熱反應(yīng)，因此/r/n和/r/nCH/r/n3/r/nO/r/n-/r/n的總能量與/r/n和/r/nOH/r/n-/r/n的總能量之差等于圖示總反應(yīng)的焓變，故/r/nD/r/n錯誤；/r/n綜上所述，正確的是/r/nB/r/n項，故答案為/r/nB/r/n。/r/n2．（浙江）/r/n相同溫度和壓強下，關(guān)于反應(yīng)的/r/n，下列判斷正確的是/r/nA．/r/n B．/r/nC．/r/n D．/r/n【KS5U答案】/r/nC/r/n【分析】/r/n一般的烯烴與氫氣發(fā)生的加成反應(yīng)為放熱反應(yīng)，但是，由于苯環(huán)結(jié)構(gòu)的特殊性決定了苯環(huán)結(jié)構(gòu)的穩(wěn)定性，苯與氫氣發(fā)生加成反應(yīng)生成/r/n1/r/n，/r/n3-/r/n環(huán)己二烯時，破壞了苯環(huán)結(jié)構(gòu)的穩(wěn)定性，因此該反應(yīng)為吸熱反應(yīng)。/r/n【KS5U解析】/r/nA/r/n．環(huán)己烯、/r/n1/r/n，/r/n3-/r/n環(huán)己二烯分別與氫氣發(fā)生的加成反應(yīng)均為放熱反應(yīng)，因此，/r/n/r/n，/r/nA/r/n不正確；/r/n/r/nB/r/n．苯分子中沒有碳碳雙鍵，其中的碳碳鍵是介于單鍵和雙鍵之間的特殊的共價鍵，因此，其與氫氣完全加成的反應(yīng)熱不等于環(huán)己烯、/r/n1/r/n，/r/n3-/r/n環(huán)己二烯分別與氫氣發(fā)生的加成反應(yīng)的反應(yīng)熱之和，即/r/n，/r/nB/r/n不正確；/r/nC/r/n．環(huán)己烯、/r/n1/r/n，/r/n3-/r/n環(huán)己二烯分別與氫氣發(fā)生的加成反應(yīng)均為放反應(yīng)，/r/n，由于/r/n1mol1/r/n，/r/n3-/r/n環(huán)己二烯與氫氣完全加成后消耗的氫氣是等量環(huán)己烯的/r/n2/r/n倍，故其放出的熱量更多，其/r/n；苯與氫氣發(fā)生加成反應(yīng)生成/r/n1/r/n，/r/n3-/r/n環(huán)己二烯的反應(yīng)為吸熱反應(yīng)（/r/n），根據(jù)蓋斯定律可知，苯與氫氣完全加成的反應(yīng)熱/r/n，因此/r/n，/r/nC/r/n正確；/r/n/r/nD/r/n．根據(jù)蓋斯定律可知，苯與氫氣完全加成的反應(yīng)熱/r/n，因此/r/n，/r/nD/r/n不正確。/r/n綜上所述，本題選/r/nC/r/n。/r/n3．（廣東高考真題試卷）“/r/n天問一號/r/n”/r/n著陸火星，/r/n“/r/n嫦娥五號/r/n”/r/n采回月壤。騰飛中國離不開化學(xué)，長征系列運載火箭使用的燃料有液氫和煤油等化學(xué)品。下列有關(guān)說法正確的是/r/nA．/r/n煤油是可再生能源/r/nB．/r/n燃燒過程中熱能轉(zhuǎn)化為化學(xué)能/r/nC．/r/n火星隕石中的/r/n質(zhì)量數(shù)為/r/n20/r/nD．/r/n月壤中的/r/n與地球上的/r/n互為同位素/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．煤油來源于石油，屬于不可再生能源，故/r/nA/r/n錯誤；/r/nB/r/n．氫氣的燃燒過程放出熱量，將化學(xué)能變?yōu)闊崮?，?r/nB/r/n錯誤；/r/nC/r/n．元素符號左上角數(shù)字為質(zhì)量數(shù)，所以火星隕石中的/r/n20/r/nNe/r/n質(zhì)量數(shù)為/r/n20/r/n，故/r/nC/r/n正確；/r/nD/r/n．同位素須為同種元素，/r/n3/r/nHe和/r/n3/r/nH/r/n的質(zhì)子數(shù)不同，不可能為同位素關(guān)系，故/r/nD/r/n錯誤；/r/n故選/r/nC/r/n。/r/n4．（河北高考真題試卷）/r/n下列操作規(guī)范且能達到實驗?zāi)康牡氖?r/nA．/r/n圖甲測定醋酸濃度/r/n B．/r/n圖乙測定中和熱/r/nC．/r/n圖丙稀釋濃硫酸/r/n D．/r/n圖丁萃取分離碘水中的碘/r/n【KS5U答案】/r/nA/r/n【KS5U解析】/r/nA/r/n．氫氧化鈉溶液呈堿性，因此需裝于堿式滴定管，氫氧化鈉溶液與醋酸溶液恰好完全反應(yīng)后生成的醋酸鈉溶液呈堿性，因此滴定過程中選擇酚酞作指示劑，當溶液由無色變?yōu)榈t色時，達到滴定終點，故/r/nA/r/n選；/r/nB/r/n．測定中和熱實驗中溫度計用于測定溶液溫度，因此不能與燒杯內(nèi)壁接觸，并且大燒杯內(nèi)空隙需用硬紙板填充，防止熱量散失，故/r/nB/r/n不選；/r/nC/r/n．容量瓶為定容儀器，不能用于稀釋操作，故/r/nC/r/n不選；/r/nD/r/n．分液過程中長頸漏斗下方放液端的長斜面需緊貼燒杯內(nèi)壁，防止液體留下時飛濺，故/r/nD/r/n不選；/r/n綜上所述，操作規(guī)范且能達到實驗?zāi)康牡氖?r/nA/r/n項，故答案為/r/nA/r/n。/r/n5．（浙江高考真題試卷）/r/n已知共價鍵的鍵能與熱化學(xué)方程式信息如下表：/r/n共價鍵/r/nH-H/r/nH-O/r/n鍵能/r/n/(kJ·mol/r/n-1/r/n)/r/n436/r/n463/r/n熱化學(xué)方程式/r/n2H/r/n2/r/n(g)+O/r/n2/r/n(g)=2H/r/n2/r/nO(g)Δ/r/nH/r/n=-482kJ·mol/r/n-1/r/n則/r/n2O(g)=O/r/n2/r/n(g)/r/n的/r/nΔ/r/nH/r/n為/r/nA．428kJ·mol/r/n-1/r/n B．-428kJ·mol/r/n-1/r/n C．498kJ·mol/r/n-1/r/n D．-498kJ·mol/r/n-1/r/n【KS5U答案】/r/nD/r/n【分析】/r/n根據(jù)Δ/r/nH/r/n=反應(yīng)物的鍵能總和-生成物的鍵能總和計算。/r/n【KS5U解析】/r/n反應(yīng)的/r/nΔ/r/nH/r/n=2(H-H)+(O-O)-4(H-O)/r/n；/r/n-482kJ/mol=2×436kJ/mol+(O-O)-4×463kJ/mol/r/n，解得/r/nO-O/r/n鍵的鍵能為/r/n498kJ/mol/r/n，/r/n2/r/n個氧原子結(jié)合生成氧氣的過程需要釋放能量，因此/r/n2O(g)=O/r/n2/r/n(g)/r/n的/r/nΔ/r/nH/r/n=-498kJ/mol/r/n。/r/n6．（浙江高考真題試卷）/r/n在/r/n298.15K/r/n、/r/n100kPa/r/n條件下，/r/nN/r/n2/r/n(g)+3H/r/n2/r/n(g)=2NH/r/n3/r/n(g)ΔH=-92.4kJ·mol/r/n-1/r/n，/r/nN/r/n2/r/n(g)/r/n、/r/nH/r/n2/r/n(g)/r/n和/r/nNH/r/n3/r/n(g)/r/n的比熱容分別為/r/n29.1/r/n、/r/n28.9/r/n和/r/n35.6J·K/r/n-1/r/n·mol/r/n-1/r/n。一定壓強下，/r/n1mol/r/n反應(yīng)中，反應(yīng)物/r/n[N/r/n2/r/n(g)+3H/r/n2/r/n(g)]/r/n、生成物/r/n[2NH/r/n3/r/n(g)]/r/n的能量隨溫度/r/nT/r/n的變化示意圖合理的是/r/nA．/r/n B．/r/nC．/r/n D．/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/n該反應(yīng)為放熱反應(yīng)，反應(yīng)物的總能量大于生成物的總能量，根據(jù)題目中給出的反應(yīng)物與生成物的比熱容可知，升高溫度反應(yīng)物能量升高較快，反應(yīng)結(jié)束后反應(yīng)放出的熱量也會增大，比較/r/n4/r/n個圖像/r/nB/r/n符合題意，故答案選/r/nB/r/n。/r/n二、原理綜合題/r/n7．（湖南高考真題試卷）/r/n氨氣中氫含量高，是一種優(yōu)良的小分子儲氫載體，且安全、易儲運，可通過下面兩種方法由氨氣得到氫氣。/r/n方法/r/nI/r/n：氨熱分解法制氫氣/r/n相關(guān)化學(xué)鍵的鍵能數(shù)據(jù)/r/n化學(xué)鍵/r/n鍵能/r/n946/r/n436.0/r/n390.8/r/n一定溫度下，利用催化劑將/r/n分解為/r/n和/r/n?；卮鹣铝袉栴}：/r/n(1)/r/n反應(yīng)/r/n_______/r/n；/r/n(2)/r/n已知該反應(yīng)的/r/n，在下列哪些溫度下反應(yīng)能自發(fā)進行？/r/n_______(/r/n填標號/r/n)/r/nA．25/r/n℃/r/nB．125/r/n℃/r/nC．225/r/n℃/r/nD．325/r/n℃/r/n(3)/r/n某興趣小組對該反應(yīng)進行了實驗探究。在一定溫度和催化劑的條件下，將/r/n通入/r/n3L/r/n的密閉容器中進行反應(yīng)/r/n(/r/n此時容器內(nèi)總壓為/r/n200kPa)/r/n，各物質(zhì)的分壓隨時間的變化曲線如圖所示。/r/n①若保持容器體積不變，/r/n時反應(yīng)達到平衡，用/r/n的濃度變化表示/r/n時間內(nèi)的反應(yīng)速率/r/n_______/r/n(/r/n用含/r/n的代數(shù)式表示/r/n)/r/n②/r/n時將容器體積迅速縮小至原來的一半并保持不變，圖中能正確表示壓縮后/r/n分壓變化趨勢的曲線是/r/n_______(/r/n用圖中/r/na/r/n、/r/nb/r/n、/r/nc/r/n、/r/nd/r/n表示/r/n)/r/n，理由是/r/n_______/r/n；/r/n③在該溫度下，反應(yīng)的標準平衡常數(shù)/r/n_______/r/n。/r/n(/r/n已知：分壓/r/n=/r/n總壓/r/n×/r/n該組分物質(zhì)的量分數(shù)，對于反應(yīng)/r/n，/r/n，其中/r/n，/r/n、/r/n、/r/n、/r/n為各組分的平衡分壓/r/n)/r/n。/r/n方法Ⅱ：氨電解法制氫氣/r/n利用電解原理，將氨轉(zhuǎn)化為高純氫氣，其裝置如圖所示。/r/n(4)/r/n電解過程中/r/n的移動方向為/r/n_______(/r/n填/r/n“/r/n從左往右/r/n”/r/n或/r/n“/r/n從右往左/r/n”)/r/n；/r/n(5)/r/n陽極的電極反應(yīng)式為/r/n_______/r/n。/r/nKOH/r/n溶液/r/nKOH/r/n溶液/r/n【KS5U答案】/r/n+90.8CD/r/nb/r/n

/r/n開始體積減半，/r/nN/r/n2/r/n分壓變?yōu)樵瓉淼?r/n2/r/n倍，隨后由于加壓平衡逆向移動，/r/nN/r/n2/r/n分壓比原來/r/n2/r/n倍要小/r/n0.48/r/n從右往左/r/n2NH/r/n3/r/n-6e/r/n-/r/n+6OH/r/n-/r/n=N/r/n2/r/n+6H/r/n2/r/nO/r/n【KS5U解析】/r/n(1)/r/n根據(jù)反應(yīng)熱/r/n=/r/n反應(yīng)物的總鍵能/r/n-/r/n生成物的總鍵能，/r/n2NH/r/n3/r/n(g)/r/nN/r/n2/r/n(g)+3H/r/n2/r/n(g)/r/n，/r/nH/r/n=390.8kJ/r/nmol/r/n-1/r/n-(946kJ/r/nmol/r/n-1/r/n+436.0kJ/r/nmol/r/n-1/r/n)=+90.8kJ/r/nmol/r/n-1/r/n，故/r/n+90.8/r/n；/r/n(2)/r/n若反應(yīng)自發(fā)進行，則需/r/nH/r/n-T/r/nS/r/n<0/r/n，/r/nT>/r/n=/r/n=456.5K/r/n，即溫度應(yīng)高于/r/n(456.5-273)/r/n℃/r/n=183.5/r/n℃，/r/nCD/r/n符合，故/r/nCD/r/n；/r/n(3)/r/n①設(shè)/r/nt/r/n1/r/n時達到平衡，轉(zhuǎn)化的/r/nNH/r/n3/r/n的物質(zhì)的量為/r/n2x/r/n，列出三段式：/r/n根據(jù)同溫同壓下，混合氣體的物質(zhì)的量等于體積之比，/r/n=/r/n，解得/r/nx=0.02mol/r/n，/r/n(H/r/n2/r/n)=/r/n=/r/nmol/r/nL/r/n-1/r/nmin/r/n-1/r/n，故/r/n；/r/n②/r/nt/r/n2/r/n時將容器體積壓縮到原來的一半，開始/r/nN/r/n2/r/n分壓變?yōu)樵瓉淼?r/n2/r/n倍，隨后由于加壓平衡逆向移動，/r/nN/r/n2/r/n分壓比原來/r/n2/r/n倍要小，故/r/nb/r/n曲線符合，故/r/nb/r/n；開始體積減半，/r/nN/r/n2/r/n分壓變?yōu)樵瓉淼?r/n2/r/n倍，隨后由于加壓平衡逆向移動，/r/nN/r/n2/r/n分壓比原來/r/n2/r/n倍要?。?r/n③由圖可知，平衡時，/r/nNH/r/n3/r/n、/r/nN/r/n2/r/n、/r/nH/r/n2/r/n的分壓分別為/r/n120kPa/r/n、/r/n40kPa/r/n、/r/n120kPa/r/n，反應(yīng)的標準平衡常數(shù)/r/n=/r/n=0.48/r/n，故/r/n0.48/r/n；/r/n(4)/r/n由圖可知，通/r/nNH/r/n3/r/n的一極氮元素化合價升高，發(fā)生氧化反應(yīng)，為電解池的陽極，則另一電極為陰極，電解過程中/r/nOH/r/n-/r/n移向陽極，則從右往左移動，故從右往左；/r/n(5)/r/n陽極/r/nNH/r/n3/r/n失電子發(fā)生氧化反應(yīng)生成/r/nN/r/n2/r/n，結(jié)合堿性條件，電極反應(yīng)式為：/r/n2NH/r/n3/r/n-6e/r/n-/r/n+6OH/r/n-/r/n=N/r/n2/r/n+6H/r/n2/r/nO/r/n，/r/n故/r/n2NH/r/n3/r/n-6e/r/n-/r/n+6OH/r/n-/r/n=N/r/n2/r/n+6H/r/n2/r/nO/r/n。/r/n2021年化學(xué)高考模擬題/r/n一、單選題/r/n1．（九龍坡區(qū)·重慶市育才中學(xué)高三三模）/r/n工業(yè)合成三氧化硫的反應(yīng)為/r/n2SO/r/n2/r/n(g)/r/n＋/r/nO/r/n2/r/n(g)/r/n2SO/r/n3/r/n(g)ΔH=-198kJ·mol/r/n-1/r/n，反應(yīng)過程可用下圖模擬/r/n(/r/n代表/r/nO/r/n2/r/n分子，/r/n代表/r/nSO/r/n2/r/n分子，/r/n代表催化劑/r/n)/r/n。下列說法不正確的是/r/nA．/r/n過程/r/nⅡ/r/n和過程/r/nⅢ/r/n決定了整個反應(yīng)進行的程度/r/nB．/r/n過程/r/nⅡ/r/n為吸熱過程，過程/r/nⅢ/r/n為放熱過程/r/nC．/r/n加入/r/nSO/r/n2/r/n和/r/nO/r/n2/r/n各/r/n1mol/r/n，充分反應(yīng)后放出的熱量小于/r/n99KJ/r/nD．/r/n催化劑可降低反應(yīng)的活化能，使/r/nΔH/r/n減小/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．過程/r/nI/r/n是吸附放熱過程，自發(fā)進行程度大，但過程/r/nII/r/n是共價鍵斷裂的過程，過程/r/nIV/r/n是生成物解吸過程，需要消耗能量，它們的活化能相對較大，決定了全部反應(yīng)進行的程度，/r/nA/r/n正確；/r/nB/r/n．由圖可知，過程/r/nII/r/n化學(xué)鍵斷裂，為吸熱過程，過程/r/nIII/r/n化學(xué)鍵形成，為放熱過程，/r/nB/r/n正確；/r/nC/r/n．反應(yīng)/r/n2SO/r/n2/r/n(g)+O/r/n2/r/n(g)/r/n?/r/n2SO/r/n3/r/n(g)△H=-198kJ/mol/r/n是可逆反應(yīng)，所以/r/n1molSO/r/n2/r/n和/r/n1molO/r/n2/r/n反應(yīng)時消耗/r/nSO/r/n2/r/n的物質(zhì)的量小于/r/n1mol/r/n，放熱小于/r/n99kJ/r/n，/r/nC/r/n正確；/r/nD/r/n．催化劑不能改變反應(yīng)的始終態(tài)，不能改變反應(yīng)物和生成物的內(nèi)能，所以不能改變反應(yīng)熱，/r/nD/r/n錯誤；/r/n故選：/r/nD/r/n。/r/n2．（重慶市第十一中學(xué)校高三二模）/r/n文獻報道：在/r/n45/r/n℃、/r/n0.1MPa/r/n時，科學(xué)家以鐵粉為催化劑，通過球磨法合成氨。部分反應(yīng)歷程如圖所示/r/n(/r/n吸附在催化劑表面的物種用/r/n*/r/n標注/r/n)/r/n，下列說法正確的是/r/nA．/r/n由此歷程可知：/r/nN*/r/n＋/r/n3H*/r/n＝/r/nNH*/r/n＋/r/n2H*ΔH>0/r/nB．/r/n鐵粉改變了合成氨的反應(yīng)歷程和反應(yīng)熱/r/nC．/r/n圖示過程中有極性共價鍵的生成/r/nD．/r/n用不同催化劑合成氨，反應(yīng)歷程均與上圖相同/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/nΔH/r/n與反應(yīng)過程無關(guān)，取決于反應(yīng)的起點和終點，由圖知，反應(yīng)物的總能量比生成物的總能量高，為放熱反應(yīng)/r/nΔH<0/r/n，則/r/nN*/r/n＋/r/n3H*/r/n＝/r/nNH*/r/n＋/r/n2H*ΔH<0/r/n，/r/nA/r/n錯誤；/r/nB/r/n．鐵粉做催化劑，只改變反應(yīng)歷程，不改變始終態(tài)，即不能改變反應(yīng)熱，/r/nB/r/n錯誤；/r/nC/r/n．過程中有/r/nN-H/r/n極性共價鍵的生成，/r/nC/r/n正確；/r/nD/r/n．不同的催化劑，改變反應(yīng)的歷程不一樣，/r/nD/r/n錯誤；/r/n故選：/r/nC/r/n。/r/n3．（青海高三三模）/r/n上海交通大學(xué)仇毅翔等研究了不同含金化合物催化乙烯加氫/r/n[C/r/n2/r/nH/r/n4/r/n(g)/r/n＋/r/nH/r/n2/r/n(g)=C/r/n2/r/nH/r/n6/r/n(g)/r/n△/r/nH/r/n=amol·L/r/n-1/r/n]/r/n的反應(yīng)歷程如下圖所示：/r/n下列說法正確的是/r/nA．/r/n該反應(yīng)為吸熱反應(yīng)/r/n B．a(chǎn)=-129.6/r/nC．/r/n催化乙烯加氫效果較好的催化劑是/r/nAuF/r/n D．/r/n兩種過渡態(tài)物質(zhì)中較穩(wěn)定的是過渡態(tài)/r/n1/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．/r/n/r/n該反應(yīng)生成物具有的能量低，為放熱反應(yīng)，故/r/nA/r/n錯誤；/r/nB/r/n．由反應(yīng)物、生成物的總能量可知/r/na=-129.6kJ/mol-0=-129.6kJ/mol/r/n，/r/na=-129.6/r/n，故/r/nB/r/n正確；/r/nC/r/n．/r/n/r/n由圖可知/r/nAuPF/r/n3/r/n+/r/n對應(yīng)的活化能小，則催化效果好，故/r/nC/r/n錯誤；/r/nD/r/n．/r/n/r/n過渡態(tài)/r/n1/r/n所處狀態(tài)能量高于狀態(tài)/r/n2/r/n，兩種過渡態(tài)物質(zhì)中較穩(wěn)定的是過渡態(tài)/r/n2/r/n，故/r/nD/r/n錯誤；/r/n故選/r/nB/r/n。/r/n4．（全國高三零模）/r/n科學(xué)家結(jié)合實驗和計算機模擬結(jié)果，研究了在貴重金屬催化劑表面上的氣態(tài)體系中，一個/r/n分子還原/r/n的能量變化與反應(yīng)歷程如圖所示。下列說法錯誤的是/r/nA．/r/n該反應(yīng)的熱化學(xué)方程式為/r/n/r/nB．/r/n決定整個反應(yīng)速率快慢的步驟是①/r/nC．/r/n反應(yīng)過程中斷裂與形成的化學(xué)鍵都包含/r/n鍵和/r/n鍵/r/nD．/r/n改變催化劑，不能使反應(yīng)的焓變發(fā)生改變/r/n【KS5U答案】/r/nA/r/n【KS5U解析】/r/nA/r/n．該過程表示一個/r/n分子還原/r/n的能量變化，熱化學(xué)方程式對應(yīng)的能量變化應(yīng)該是/r/n2mol/r/n分子被還原的能量，故/r/nA/r/n錯誤；/r/n/r/nB/r/n．活化能大速率慢，是決速步驟，①的活化能大于②，決定整個反應(yīng)速率快慢的步驟是①，故/r/nB/r/n正確；/r/nC/r/n．反應(yīng)物/r/n分子既含有/r/n鍵和/r/n鍵，生成物/r/nN/r/n2/r/n和/r/nCO/r/n2/r/n含有/r/n鍵和/r/n鍵，所以反應(yīng)過程中斷裂與形成的化學(xué)鍵都包含/r/n鍵和/r/n鍵，故/r/nC/r/n正確；/r/nD/r/n．催化劑只能改變反應(yīng)歷程，不能改變焓變，故/r/nD/r/n正確；/r/n故答案為/r/nA/r/n。/r/n5．（陜西寶雞市·）2007/r/n年諾貝爾化學(xué)獎授予埃特爾以表彰其對于合成氨反應(yīng)機理的/r/n研究，氮氣和氫氣分子在催化劑表面的部分變化過程如圖所示，下列說法不正確的是/r/nA．/r/n升高溫度不能提高一段時間內(nèi)/r/n的產(chǎn)率/r/nB．/r/n圖/r/n①→②/r/n過程吸熱，圖/r/n②→③/r/n過程放熱/r/nC．/r/n在反應(yīng)過程中三鍵均發(fā)生斷裂/r/nD．/r/n反應(yīng)過程中存在/r/n、/r/n等中間產(chǎn)物/r/n【KS5U答案】/r/nA/r/n【KS5U解析】/r/nA/r/n．升高溫度，化學(xué)反應(yīng)速率加快，可以提高一段時間內(nèi)/r/n的產(chǎn)率，/r/nA/r/n錯誤；/r/nB/r/n．圖/r/n①→②/r/n過程表示斷裂氮氣分子的化學(xué)鍵，斷鍵吸熱，圖/r/n②→③/r/n過程表示形成氮氫單鍵，成鍵放熱，/r/nB/r/n正確；/r/nC/r/n．由圖/r/n①→②/r/n可知，/r/n在反應(yīng)過程中三鍵均發(fā)生斷裂，/r/nC/r/n正確；/r/nD/r/n．圖③④表明反應(yīng)過程中存在/r/n、/r/n等中間產(chǎn)物，/r/nD/r/n正確；/r/nA/r/n。/r/n6．（安徽高三一模）/r/n馬里奧/r/n·/r/n莫利納等科學(xué)家通過研究揭示了大氣中臭氧層被破壞的機理，如圖所示。下列說法錯誤是/r/nA．CFCl/r/n3/r/n是四面體結(jié)構(gòu)/r/nB．/r/n過程Ⅲ中的/r/nO/r/n原子可能來自大氣中/r/nO/r/n2/r/n或/r/nO/r/n3/r/n的解離/r/nC．/r/n整個過程中，/r/nCFCl/r/n3/r/n是/r/nO/r/n3/r/n分解的催化劑/r/nD．/r/n紫外線輻射提供破壞化學(xué)鍵的能量/r/n【KS5U答案】/r/nC/r/n【分析】/r/n過程Ⅰ為：/r/nCFCl/r/n3/r/n→/r/nCl+/r/nCFCl/r/n2/r/n，過程Ⅱ為：/r/nCl+O/r/n3/r/n→O/r/n2/r/n+ClO/r/n，過程Ⅲ為：/r/nClO+O→O/r/n2/r/n+Cl/r/n【KS5U解析】/r/nA/r/n．/r/nCH/r/n4/r/n是正四面體結(jié)構(gòu)，/r/nCFCl/r/n3/r/n就相當于用/r/nF/r/n將一個/r/nH/r/n原子替代了，用/r/n3/r/n個/r/nCl/r/n將一個/r/n3/r/n個/r/nH/r/n原子替代了，雖然鍵長、鍵角有變化，但是還是四面體結(jié)構(gòu)，/r/nA/r/n正確；/r/nB/r/n．過程Ⅰ為：/r/nCFCl/r/n3/r/n→Cl+CFCl/r/n2/r/n，過程Ⅱ為：/r/nCl+O/r/n3/r/n→O/r/n2/r/n+ClO/r/n，過程Ⅲ為：/r/nClO+O→O/r/n2/r/n+Cl/r/n，故過程Ⅲ中的/r/nO/r/n原子可能來自大氣中/r/nO/r/n2/r/n或/r/nO/r/n3/r/n的解離/r/nB/r/n正確；/r/nC/r/n．過程Ⅰ為：/r/nCFCl/r/n3/r/n→Cl+CFCl/r/n2/r/n，過程Ⅱ為：/r/nCl+O/r/n3/r/n→O/r/n2/r/n+ClO/r/n，過程Ⅲ為：/r/nClO+O→O/r/n2/r/n+Cl/r/n，我們發(fā)現(xiàn)/r/nCFCl/r/n3/r/n作為反應(yīng)物被消耗，/r/nCFCl/r/n3不是催化劑/r/n，/r/nC/r/n錯誤；/r/nD/r/n．由圖可知，紫外線輻射提供破壞了/r/nC-Cl/r/n，斷鍵需要吸收能量，/r/nD/r/n正確；/r/n/r/n故選/r/nD/r/n。/r/n7．（浙江高三其他模擬）/r/n反應(yīng)/r/nA+B→C/r/n分兩步進行：反應(yīng)①/r/nA+B→X/r/n，反應(yīng)②/r/nX→C/r/n。反應(yīng)過程中能量變化如圖，下列說法正確的是/r/nA．/r/n該反應(yīng)為放熱反應(yīng)，/r/n△/r/nH/r/n=a-d/r/nB．/r/n催化劑通過降低化學(xué)反應(yīng)的焓變加快化學(xué)反應(yīng)速率/r/nC．/r/n升高溫度，/r/na/r/n、/r/nb/r/n、/r/nc/r/n、/r/nd/r/n的數(shù)值均會發(fā)生改變/r/nD．/r/n該反應(yīng)速率的快慢主要由反應(yīng)②決定/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．根據(jù)圖示可知：反應(yīng)物/r/nA/r/n和/r/nB/r/n的能量總和比生成物/r/nC/r/n的高，發(fā)生反應(yīng)放出熱量，故該反應(yīng)為放熱反應(yīng)，/r/n△/r/nH/r/n=(a+c)-(d+b)/r/n，/r/nA/r/n錯誤；/r/nB/r/n．催化劑通過改變反應(yīng)途徑，降低反應(yīng)的活化能來加快化學(xué)反應(yīng)速率，但該反應(yīng)的焓變不變，/r/nB/r/n錯誤；/r/nC/r/n．升高溫度，物質(zhì)的內(nèi)能發(fā)生改變，故/r/na/r/n、/r/nb/r/n、/r/nc/r/n、/r/nd/r/n的數(shù)值均會發(fā)生改變，/r/nC/r/n正確；/r/nD/r/n．化學(xué)反應(yīng)速率的快慢由反應(yīng)速率慢的決定。反應(yīng)的活化能越大，反應(yīng)需消耗的能量就越高，反應(yīng)就越難發(fā)生。根據(jù)上述圖示可知反應(yīng)①的活化能較大，故該反應(yīng)速率的快慢主要由反應(yīng)①決定，/r/nD/r/n錯誤；/r/n故合理選項是/r/nC/r/n。/r/n8．（廣西南寧市·南寧三中高三三模）/r/n為應(yīng)對全球氣候問題，中國政府承諾/r/n“2030/r/n年碳達峰/r/n”/r/n、/r/n“2060/r/n年碳中和/r/n”/r/n?？茖W(xué)家使用絡(luò)合物作催化劑，用多聚物來捕獲二氧化碳，反應(yīng)可能的過程如圖所示。下列敘述錯誤的是/r/nA．/r/n該反應(yīng)若得以推廣將有利于碳中和/r/nB．/r/n反應(yīng)過程中只有極性鍵的斷裂和形成/r/nC．/r/n總反應(yīng)方程式為/r/nCO/r/n2/r/n+3H/r/n2/r/nCH/r/n3/r/nOH+H/r/n2/r/nO/r/nD．/r/n開發(fā)太陽能、風(fēng)能等再生能源可降低/r/nCO/r/n2/r/n、/r/nCH/r/n4/r/n溫室氣體的碳排放/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．碳中和指將通過植樹造林、節(jié)能減排等方式，抵消自身產(chǎn)生的/r/nCO/r/n2/r/n等溫室氣體，實現(xiàn)相對碳的/r/n“/r/n零排放/r/n”/r/n，故/r/nA/r/n正確；/r/nB/r/n．第二、四步有/r/nH/r/n2/r/n參與反應(yīng)，存在非極性鍵的斷裂，故/r/nB/r/n錯誤；/r/nC/r/n．除去環(huán)上循環(huán)的物質(zhì)和催化劑，總反應(yīng)的反應(yīng)物為/r/nCO/r/n2/r/n、/r/nH/r/n2/r/n生成物為/r/nCH/r/n3/r/nOH/r/n與/r/nH/r/n2/r/nO/r/n，故/r/nC/r/n正確；/r/nD/r/n．碳排放是指/r/nCO/r/n2/r/n、/r/nCH/r/n4/r/n等溫室氣體排放的簡稱，故/r/nD/r/n正確；/r/n故選/r/nB/r/n。/r/n9．（長沙市明德中學(xué)高三三模）Ni/r/n可活化/r/nC/r/n2/r/nH/r/n6/r/n放出/r/nCH/r/n4/r/n，其反應(yīng)歷程如下圖所示：/r/n下列關(guān)于活化歷程的說法錯誤的是/r/nA．/r/n此反應(yīng)的決速步驟：中間體/r/n2→/r/n中間體/r/n3/r/nB．/r/n只涉及極性鍵的斷裂和生成/r/nC．/r/n在此反應(yīng)過程中/r/nNi/r/n的成鍵數(shù)目發(fā)生變化/r/nD．Ni(s)+C/r/n2/r/nH/r/n6/r/n(g)=NiCH/r/n2/r/n(s)+CH/r/n4/r/n(g)?/r/nH/r/n=?6.57kJ·mol/r/n-1/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．中間體/r/n2→/r/n中間體/r/n3/r/n能量差值最大，該步反應(yīng)的活化能最大，是化學(xué)反應(yīng)的決定速率的步驟，/r/nA/r/n正確；/r/nB/r/n．反應(yīng)過程涉及到/r/nC-C/r/n鍵斷裂和/r/nC-H/r/n鍵形成，涉及到非極性鍵的斷裂和極性鍵的形成，/r/nB/r/n錯誤；/r/nC/r/n．根據(jù)圖示可知：在此反應(yīng)過程中/r/nNi/r/n的成鍵數(shù)目在不斷發(fā)生變化，/r/nC/r/n正確；/r/nD/r/n．/r/nNi(s)+C/r/n2/r/nH/r/n6/r/n(g)=NiCH/r/n2/r/n(s)+CH/r/n4/r/n(g)?/r/nH/r/n=E/r/n生成物/r/n-E/r/n反應(yīng)物/r/n=-6.57kJ/mol-0kJ/mol=-6.57kJ/mo1/r/n，/r/nD/r/n正確；/r/n故合理選項是/r/nB/r/n。/r/n10．（北京高三其他模擬）Deacon/r/n催化氧化法將/r/nHCl/r/n轉(zhuǎn)化為/r/nCl/r/n2/r/n的反應(yīng)為：/r/n4HCl(g)+O/r/n2/r/n(g)=2Cl/r/n2/r/n(g)+2H/r/n2/r/nO(g)Δ/r/nH/r/n=-116kJ·mol/r/n-1/r/n研究發(fā)現(xiàn)/r/nCuCl/r/n2/r/n(s)/r/n催化反應(yīng)的過程如下：/r/n反應(yīng)/r/ni/r/n：/r/nCuCl/r/n2/r/n(s)=CuCl(s)+/r/nCl/r/n2/r/n(g)Δ/r/nH/r/n1/r/n=+83kJ·mol/r/n-1/r/n反應(yīng)/r/nii/r/n：/r/nCuCl(s)+/r/nO/r/n2/r/n(g)=CuO(s)+/r/nCl/r/n2/r/n(g)Δ/r/nH/r/n2/r/n=-20kJ·mol/r/n-1/r/n反應(yīng)/r/niii/r/n：/r/n……/r/n下列表述/r/n不正確/r/n的是/r/nA．/r/n反應(yīng)/r/ni/r/n中反應(yīng)物的總能量小于生成物的總能量/r/nB．/r/n反應(yīng)/r/nii/r/n中，/r/n1molCuCl(s)/r/n反應(yīng)時轉(zhuǎn)移/r/n2mole/r/n-/r/nC．/r/n推斷反應(yīng)/r/niii/r/n應(yīng)為/r/nCuO(s)+2HCl(g)=CuCl/r/n2/r/n(s)+H/r/n2/r/nO(g)Δ/r/nH/r/n3/r/n=-242kJ·mol/r/n-1/r/nD．/r/n由反應(yīng)過程可知催化劑參與反應(yīng)，通過改變反應(yīng)路徑提高反應(yīng)速率/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．反應(yīng)/r/ni/r/n為吸熱反應(yīng)，則反應(yīng)物的總能量小于生成物的總能量，/r/nA/r/n正確；/r/nB/r/n．/r/n/r/n反應(yīng)/r/nii/r/n中，銅從+1升高到+2、氯從-1升高到0，則/r/n1molCuCl(s)/r/n反應(yīng)時失去/r/n2mole/r/n-/r/n，氧氣中氧從/r/n0/r/n降低到/r/n-2/r/n，則/r/nmolO/r/n2/r/n得到/r/n2mole/r/n-/r/n，/r/nB/r/n正確；/r/nC/r/n．/r/n/r/n按蓋斯定律，反應(yīng)/r/n4HCl(g)+O/r/n2/r/n(g)=2Cl/r/n2/r/n(g)+2H/r/n2/r/nO(g)Δ/r/nH/r/n=-116kJ·mol/r/n-1/r/n減去/r/n2/r/n×/r/n(/r/n反應(yīng)/r/ni+/r/n反應(yīng)/r/nii)/r/n可得：2/r/nCuO(s)+4HCl(g)=2CuCl/r/n2/r/n(s)+2H/r/n2/r/nO(g)Δ/r/nH/r/n3/r/n=-242kJ·mol/r/n-1/r/n，則/r/nCuO(s)+2HCl(g)=CuCl/r/n2/r/n(s)+H/r/n2/r/nO(g)Δ/r/nH/r/n3/r/n=-121kJ·mol/r/n-1/r/n，/r/nC/r/n不正確；/r/nD/r/n．/r/n/r/n由反應(yīng)過程可知催化劑參與反應(yīng)后重新生成，但催化劑改變反應(yīng)路徑、通過降低反應(yīng)活化能提高反應(yīng)速率，/r/nD/r/n正確；/r/n答案選/r/nC/r/n。/r/n11．（河南新鄉(xiāng)市·新鄉(xiāng)縣一中高三其他模擬）CH/r/n3/r/n-CH=CH-COOCH/r/n3/r/n有兩種立體異構(gòu)體/r/n和/r/n，由/r/nCH/r/n3/r/nCHO/r/n和/r/nPh/r/n3/r/nP=CHCOOCH/r/n3/r/n反應(yīng)制取這兩種異構(gòu)體的歷程中能量變化如圖/r/n已知：在立體結(jié)構(gòu)中，實線表示該鍵在紙平面上，實楔形線表示該鍵在紙前方，虛線表示該鍵在紙后方。/r/n下列說法正確的是/r/nA．/r/n比/r/n穩(wěn)定/r/nB．/r/n溫度升高，/r/nCH/r/n3/r/nCHO/r/n的轉(zhuǎn)化率減小/r/nC．/r/n生成/r/n的過程中，速率最快的是由/r/n生成/r/n的反應(yīng)/r/nD．/r/n兩個反應(yīng)歷程中，中間產(chǎn)物相同/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．圖像中/r/n比/r/n能量高，/r/n穩(wěn)定，/r/nA/r/n錯誤；/r/nB/r/n．圖像該反應(yīng)歷程中生成兩種物質(zhì)都是吸熱反應(yīng)，溫度升高，/r/nCH/r/n3/r/nCHO/r/n的轉(zhuǎn)化率增大，/r/nB/r/n錯誤。/r/nC/r/n．由/r/n生成/r/n的過程活化能最小，反應(yīng)速率快，/r/nC/r/n正確；/r/nD/r/n．兩個反應(yīng)歷程中，/r/n是生成/r/n的中間產(chǎn)物，/r/n是生成/r/n的中間產(chǎn)物，兩個過程中間產(chǎn)物不同，/r/nD/r/n錯誤；/r/n故選/r/nC/r/n。/r/n12．（【一飛沖天】4.一模）CO/r/n與/r/nN/r/n2/r/nO/r/n在/r/nFe/r/n+/r/n作用下發(fā)生反應(yīng)的能量變化及反應(yīng)歷程如圖所示，兩步為①/r/nN/r/n2/r/nO+Fe/r/n+/r/n=N/r/n2/r/n+FeO/r/n+/r/n(/r/n慢/r/n)/r/n、②/r/nFeO/r/n+/r/n+CO=CO/r/n2/r/n+Fe/r/n+/r/n(/r/n快/r/n)/r/n。下列說法正確的是/r/n

/r/nA．/r/n分子構(gòu)型：/r/nCO/r/n2/r/n為直線型，/r/nN/r/n2/r/nO/r/n為/r/nV/r/n形/r/n(/r/n已知/r/nN/r/n2/r/nO/r/n中每個原子都滿足/r/n8/r/n電子穩(wěn)定結(jié)構(gòu)/r/n)/r/nB．/r/n反應(yīng)①的活化能比反應(yīng)②大/r/nC．/r/n反應(yīng)中每轉(zhuǎn)移/r/n1mol/r/n電子，生成/r/nN/r/n2/r/n體積為/r/n11.2L/r/nD．/r/n兩步反應(yīng)均為放熱反應(yīng)，總反應(yīng)的化學(xué)反應(yīng)速率由反應(yīng)②決定/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．/r/nCO/r/n2/r/n與/r/nN/r/n2/r/nO/r/n均為三原子分子，/r/n22/r/n個電子，互為等電子體，所以均為直線型分子，故/r/nA/r/n錯；/r/nB/r/n．反應(yīng)①/r/nN/r/n2/r/nO+Fe/r/n+/r/n=N/r/n2/r/n+FeO/r/n+/r/n(/r/n慢/r/n)/r/n，反應(yīng)②/r/nFeO/r/n+/r/n+CO=CO/r/n2/r/n+Fe/r/n+/r/n(/r/n快/r/n)/r/n，由圖示可知，則反應(yīng)①活化能較反應(yīng)②大，故/r/nB/r/n正確；/r/nC/r/n．選項中未告訴是否處于標況下，若標況下，則根據(jù)/r/n可知，每轉(zhuǎn)移/r/n1mol/r/n電子，生成/r/nN/r/n2/r/n體積為/r/n11.2L/r/n，故/r/nC/r/n錯；/r/nD/r/n．總反應(yīng)的化學(xué)反應(yīng)速率由反應(yīng)較慢一步所決定，即由反應(yīng)①決定，故/r/nD/r/n錯；/r/n答案選/r/nB/r/n。/r/n13．（阜新市第二高三其他模擬）CO/r/n2/r/n是廉價的碳資源，將其甲烷化具有重要意義。其原理是/r/nCO/r/n2/r/n(g)+4H/r/n2/r/n(g)=CH/r/n4/r/n(g)+2H/r/n2/r/nO(g)/r/nΔ/r/nH/r/n<0./r/n在某密閉容器中，充入/r/n1molCO/r/n2/r/n和/r/n4molH/r/n2/r/n發(fā)生上述反應(yīng)，下列敘述正確的是/r/nA．/r/n反應(yīng)物的總能量小于生成物的總能量/r/nB．/r/n升高溫度可增大活化分子的百分數(shù)及有效碰撞頻率，因而溫度越高越利于獲得甲烷/r/nC．/r/n恒溫、恒壓條件下，充入/r/nHe/r/n，平衡向正反應(yīng)方向移動/r/nD．/r/n在絕熱密閉容器中進行時，容器中溫度不再改變，說明已達到平衡/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．該反應(yīng)是放熱反應(yīng)，反應(yīng)物的總能量大于生成物的總能量，故/r/nA/r/n錯誤；/r/nB/r/n．該反應(yīng)是放熱反應(yīng)，升高溫度，化學(xué)平衡向逆反應(yīng)方向移動，則溫度越高越不利于獲得甲烷，故/r/nB/r/n錯誤；/r/nC/r/n．恒溫、恒壓條件下充入惰性氣體氦氣，容器的容積增大，相當于減小壓強，該反應(yīng)為氣體體積減小的反應(yīng)，減小壓強，平衡向逆反應(yīng)方向移動，故/r/nC/r/n錯誤；/r/nD/r/n．該反應(yīng)是放熱反應(yīng)，在絕熱密閉容器中進行時，反應(yīng)溫度會增大，則當容器中溫度不再改變時，說明正、逆反應(yīng)速率相等，反應(yīng)已達到平衡，故/r/nD/r/n正確；/r/n故選/r/nD/r/n。/r/n14．（阜新市第二高三其他模擬）/r/n在含有/r/nFe/r/n3+/r/n、/r/nS/r/n2/r/nO/r/n和/r/nI/r/n-/r/n的混合溶液中，反應(yīng)/r/nS/r/n2/r/nO/r/n(aq)+2I/r/n-/r/n(aq)=2SO/r/n(aq)+I/r/n2/r/n(aq)/r/n的分部機理如下，反應(yīng)進程中的能量變化如圖所示。/r/n步驟①：/r/n2Fe/r/n3+/r/n(aq)+2I/r/n-/r/n(aq)=2Fe/r/n2+/r/n(aq)+I/r/n2/r/n(aq)/r/n步驟②：/r/nS/r/n2/r/nO/r/n(aq)+2Fe/r/n2+/r/n(aq)=2SO/r/n(aq)+2Fe/r/n3+/r/n(aq)/r/n下列關(guān)于該反應(yīng)的說法錯誤的是/r/nA．Fe/r/n3+/r/n是該反應(yīng)的催化劑/r/nB．/r/n步驟②比步驟①速率快/r/nC．/r/n該反應(yīng)為放熱反應(yīng)/r/nD．/r/n若不加/r/nFe/r/n3+/r/n，則正反應(yīng)的活化能比逆反應(yīng)的大/r/n【KS5U答案】/r/nD/r/n【分析】/r/n步驟①：/r/n2Fe/r/n3+/r/n(aq)+2I/r/n-/r/n(aq)=2Fe/r/n2+/r/n(aq)+I/r/n2/r/n(aq)/r/n，步驟②：/r/nS/r/n2/r/nO/r/n(aq)+2Fe/r/n2+/r/n(aq)=2SO/r/n(aq)+2Fe/r/n3+/r/n(aq)/r/n，①/r/n+/r/n②得/r/nS/r/n2/r/nO/r/n(aq)+2I/r/n-/r/n(aq)=2SO/r/n(aq)+I/r/n2/r/n(aq)/r/n，/r/nFe/r/n3+/r/n作催化劑，/r/nFe/r/n2+/r/n為中間產(chǎn)物，/r/nS/r/n2/r/nO/r/n(aq)+2I/r/n-/r/n(aq)+2Fe/r/n3+/r/n(aq)/r/n的總能量高于/r/n2Fe/r/n3+/r/n(aq)+2SO/r/n(aq)+I/r/n2/r/n(aq)/r/n的總能量，反應(yīng)放熱。/r/n【KS5U解析】/r/nA/r/n．根據(jù)反應(yīng)的機理可知鐵離子參與了反應(yīng)，但是在反應(yīng)前后其物質(zhì)的量不變，則/r/nFe/r/n3+/r/n是該反應(yīng)的催化劑，故/r/nA/r/n正確；/r/nB/r/n．由圖象可知步驟①的活化能大于步驟②的，活化能越大，反應(yīng)速率越小，故步驟①反應(yīng)速率慢，故/r/nB/r/n正確；/r/nC/r/n．反應(yīng)物的總能量高于生成物的總能量，所以該反應(yīng)為放熱反應(yīng)，故/r/nC/r/n正確；/r/nD/r/n．/r/nΔ/r/nH/r/n=/r/n正反應(yīng)的活化能/r/n-/r/n逆反應(yīng)的活化能，該反應(yīng)為放熱反應(yīng)，不管加不加催化劑，正反應(yīng)活化能都低于逆反應(yīng)活化能，故/r/nD/r/n錯誤；/r/n故選/r/nD/r/n。/r/n15．（長沙市明德中學(xué)高三三模）/r/n我國研究人員研發(fā)了一種新型納米催化劑，實現(xiàn)/r/nCO/r/n2/r/n和/r/nH/r/n2/r/n反應(yīng)得到/r/nCH/r/n4/r/n，部分微粒轉(zhuǎn)化過程如圖/r/n(/r/n吸附在催化劑表面上的物種用/r/n*/r/n標注/r/n)/r/n。下列說法/r/n不正確/r/n的是/r/nA．/r/n過程②吸收熱量/r/nB．/r/n過程③涉及極性鍵的斷裂和形成/r/nC．/r/n結(jié)合過程③，過程④的方程式為/r/n*C+2*OH+6H→CH/r/n4/r/n+2H/r/n2/r/nO/r/nD．/r/n整個過程中制得/r/n1molCH/r/n4/r/n轉(zhuǎn)移電子的物質(zhì)的量為/r/n6mol/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．由圖示知，過程②涉及/r/n*CO/r/n2/r/n中化學(xué)鍵的斷裂，需要吸收能量，/r/nA/r/n正確；/r/nB/r/n．由圖示知，過程③涉及/r/n*CO/r/n中極性鍵斷裂和生成物/r/n*OH/r/n中極性鍵的形成，/r/nB/r/n正確；/r/nC/r/n．由圖示知，過程③對應(yīng)方程式為：/r/n*CO+*O+2H/r/n→/r/n*C+2*OH/r/n，過程④反應(yīng)物為/r/n*C/r/n、/r/n*OH/r/n、/r/nH/r/n，生成物為/r/nCH/r/n4/r/n和/r/nH/r/n2/r/nO/r/n，根據(jù)過程③知/r/n*C/r/n、/r/n*OH/r/n反應(yīng)比例為/r/n1:2/r/n，故過程④方程式為：/r/n*C+2*OH+6H/r/n→/r/nCH/r/n4/r/n+2H/r/n2/r/nO/r/n，/r/nC/r/n正確；/r/nD/r/n．/r/nCO/r/n2/r/n轉(zhuǎn)化為/r/nCH/r/n4/r/n，碳元素化合價由/r/n+4/r/n價降低為/r/n-4/r/n價，得關(guān)系式：/r/nCO/r/n2/r/n～CH/r/n4/r/n～8e/r/n-/r/n，故/r/n1molCH/r/n4/r/n生成轉(zhuǎn)移電子為/r/n8mol/r/n，/r/nD/r/n錯誤；/r/n故答案選/r/nD/r/n。/r/n16．（天津高三一模）/r/n正戊烷異構(gòu)化為異戊烷是油品升級的一項重要技術(shù)。在合適催化劑和一定壓強下，正戊烷的平衡轉(zhuǎn)化率/r/n(a)/r/n隨溫度變化如圖所示。/r/n名稱/r/n熔點/r/n/℃/r/n沸點/r/n/℃/r/n燃燒熱/r/nΔ/r/nH/r/n/kJ/r/n·/r/nmol/r/n-1/r/n正戊烷/r/n-130/r/n36/r/n-3506.1/r/n異戊烷/r/n-159.4/r/n27.8/r/n-3504.1/r/n下列說法不正確的是/r/nA．/r/n正戊烷異構(gòu)化為異戊烷反應(yīng)的/r/nΔ/r/nH/r/n<0/r/nB．28～36/r/n℃時，隨溫度升高，正戊烷的平衡轉(zhuǎn)化率增大，原因是異戊烷氣化離開體系，產(chǎn)物濃度降低，平衡正向移動/r/nC．/r/n尋找更好的催化劑可使正戊烷異構(gòu)化為異戊烷的轉(zhuǎn)化率提升/r/nD．150℃/r/n時，體系壓強從/r/n100kPa/r/n升高到/r/n500kPa/r/n，正戊烷的平衡轉(zhuǎn)化率基本不變/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．根據(jù)正戊烷和異戊烷的燃燒熱，①/r/n/r/n，/r/n/r/n，根據(jù)蓋斯定律①/r/n?/r/n②得正戊烷異構(gòu)化為異戊烷反應(yīng)/r/n，故/r/nA/r/n正確；/r/nB/r/n．根據(jù)表中沸點數(shù)值，在/r/n28/r/n～/r/n36℃/r/n時，隨溫度升高，異戊烷氣化離開體系，產(chǎn)物濃度降低，導(dǎo)致平衡正向移動，正戊烷的平衡轉(zhuǎn)化率增大，故/r/nB/r/n正確；/r/nC/r/n．催化劑不影響平衡，不能提高平衡轉(zhuǎn)化率，故/r/nC/r/n錯誤；/r/nD/r/n．/r/n150℃/r/n時，正戊烷和異戊烷都是氣態(tài)，此時反應(yīng)前后氣體體積不變，增大壓強，平衡不移動，正戊烷的平衡轉(zhuǎn)化率不變，故/r/nD/r/n正確；/r/n答案選/r/nC/r/n。/r/n17．（天津高三一模）/r/n如圖是/r/nCH/r/n4/r/n與/r/nCl/r/n2/r/n生成/r/nCH/r/n3/r/nCl/r/n的部分反應(yīng)過程中各物質(zhì)物質(zhì)的能量變化關(guān)系圖/r/n(E/r/na/r/n表示活化能/r/n)/r/n，下列說法錯誤的是/r/nA．/r/n增大/r/nCl/r/n2/r/n的濃度，可提高反應(yīng)速率，但不影響/r/nΔH/r/n的大小/r/nB．/r/n第一步反應(yīng)的速率小于第二步反應(yīng)/r/nC．/r/n總反應(yīng)為放熱反應(yīng)/r/nD．/r/n升高溫度，/r/nE/r/na1/r/n、/r/nE/r/na2/r/n均增大，反應(yīng)速率加快/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．/r/nCl/r/n2/r/n是該反應(yīng)的反應(yīng)物，增大反應(yīng)物的濃度，反應(yīng)速率增大，但增大氯氣的濃度不影響/r/nΔH/r/n的大小，故/r/nA/r/n正確；/r/nB/r/n．第一步反應(yīng)所需活化能/r/nE/r/na1/r/n大于第二步反應(yīng)所需活化能/r/nE/r/na2/r/n，第一步反應(yīng)單位體積內(nèi)活化分子百分數(shù)低于第二步反應(yīng)，故第二步反應(yīng)速率更大，故/r/nB/r/n正確；/r/nC/r/n．反應(yīng)物總能量大于生成物總能量，為放熱反應(yīng)，故/r/nC/r/n正確；/r/nD/r/n．/r/nE/r/na1/r/n、/r/nE/r/na2/r/n分別為第一步反應(yīng)、第二步反應(yīng)所需活化能，升高溫度，反應(yīng)所需活化能不變，即/r/nE/r/na1/r/n、/r/nE/r/na2/r/n不變，故/r/nD/r/n錯誤；/r/n故選/r/nD/r/n。/r/n二、原理綜合題/r/n18．（安徽高三一模）/r/n甲醇是目前人類認知最為安全、高效、清潔的替代燃料，可用如下方法制備：碳的氧化物和氫氣合成法，其主要反應(yīng)包括：/r/nⅠ/r/n.CO(g)+2H/r/n2/r/n(g)/r/nCH/r/n3/r/nOH(g)/r/n△/r/nH/r/n1/r/n=/r/n—/r/n90.8kJ/r/n?/r/nmol/r/n—/r/n1/r/nⅡ/r/n.CO/r/n2/r/n(g)+3H/r/n2/r/n(g)/r/nCH/r/n3/r/nOH(g)+H/r/n2/r/nO(g)/r/n△/r/nH/r/n2/r/n=/r/n—/r/n49.0kJ/r/n?/r/nmol/r/n—/r/n1/r/nⅢ/r/n.CO/r/n2/r/n(g)+H/r/n2/r/n(g)/r/nCO(g)+H/r/n2/r/nO(g)/r/n△/r/nH/r/n3/r/n(1)/r/n在恒溫的合成塔中發(fā)生上述反應(yīng)，達到平衡時合成塔出口處總壓強為/r/n5.0MPa/r/n，各組分的體積分數(shù)如表所示：/r/nCO/r/nCO/r/n2/r/nH/r/n2/r/nCH/r/n3/r/nOH/r/n其他/r/n10.0%/r/n5.0%/r/n50.0%/r/n2.5%/r/n32.5%/r/n①反應(yīng)Ⅲ的焓變/r/n△/r/nH/r/n3/r/n=___/r/n。/r/n②此溫度下反應(yīng)Ⅰ的平衡常數(shù)/r/nK/r/np/r/n=___(MPa)/r/n—/r/n2/r/n。/r/n(2)/r/n反應(yīng)Ⅰ是合成甲醇的主要反應(yīng)，在二氧化碳存在時，單位體積催化劑上生成甲醇的速率如圖：/r/n①加入二氧化碳對甲醇反應(yīng)速率的影響是/r/n___/r/n。/r/n②在無二氧化碳的體系中加入適量氧氣也可以起到和二氧化碳相似的作用，原因是/r/n___/r/n。/r/n(3)/r/n一氧化碳與氫氣的反應(yīng)歷程如圖，其中吸附在催化劑表面上的物種用/r/n*/r/n標注。/r/n①該反應(yīng)歷程中控速步驟的能壘為/r/n___eV/r/n。/r/n②寫出虛線框內(nèi)發(fā)生的化學(xué)反應(yīng)方程式/r/n___/r/n。/r/n【KS5U答案】/r/n+41.8kJ/mol0.04/r/n先增大后減小/r/n/r/n氧氣能與一氧化碳反應(yīng)生成二氧化碳/r/n0.8H/r/n3/r/nCO/r/n*/r/n+/r/nH/r/n2/r/n=CH/r/n3/r/nOH/r/n*/r/n/r/n【KS5U解析】/r/n(1)①/r/n由蓋斯定律可知，反應(yīng)/r/nⅡ/r/n—反應(yīng)/r/nⅠ/r/n得反應(yīng)/r/nⅢ/r/n，則焓變/r/n△/r/nH/r/n3/r/n=△/r/nH/r/n2/r/n—/r/n△/r/nH/r/n1/r/n=(/r/n—/r/n49.0kJ/mol)/r/n—/r/n(/r/n—/r/n90.8kJ/mol)=+41.8kJ/mol/r/n，故/r/n+41.8kJ/mol/r/n；/r/n②/r/n由題給數(shù)據(jù)可知，一氧化碳、氫氣和甲醇的平衡分壓分別為/r/n10.0%/r/n×/r/n5.0MPa=0.5MPa/r/n、/r/n50%/r/n×/r/n5.0MPa=2.5MPa/r/n、/r/n2.5%/r/n×/r/n5.0MPa=0.125MPa/r/n，則反應(yīng)/r/nⅠ/r/n的平衡常數(shù)/r/nK/r/np/r/n=/r/n=/r/n=0.04(MPa)/r/n—/r/n2/r/n，故/r/n0.04/r/n；/r/n(2)①/r/n由圖可知，隨著二氧化碳百分含量的增加，單位體積催化劑上，甲醇反應(yīng)速率先增大后減小，故先增大后減?。?r/n②/r/n若在無二氧化碳的體系中加入適量氧氣，氧氣能與一氧化碳反應(yīng)生成二氧化碳，從而起到和二氧化碳相似的作用，故氧氣能與一氧化碳反應(yīng)生成二氧化碳；/r/n(3)①/r/n化學(xué)反應(yīng)速率取決于化學(xué)反應(yīng)速率最慢的一步，反應(yīng)的能壘越大，反應(yīng)速率越慢，由圖可知，反應(yīng)歷程中控速步驟的能壘為/r/n(/r/n—/r/n0.1eV)/r/n—/r/n(/r/n—/r/n0.9eV)=0.8eV/r/n，故/r/n0.8/r/n；/r/n②/r/n由圖可知，虛線框內(nèi)發(fā)生的反應(yīng)為/r/nH/r/n3/r/nCO/r/n*/r/n與/r/nH/r/n2/r/n反應(yīng)生成/r/nCH/r/n3/r/nOH/r/n*/r/n，反應(yīng)的化學(xué)反應(yīng)方程式為/r/nH/r/n3/r/nCO/r/n*/r/n+/r/nH/r/n2/r/n=CH/r/n3/r/nOH/r/n*/r/n，故/r/nH/r/n3/r/nCO/r/n*/r/n+/r/nH/r/n2/r/n=CH/r/n3/r/nOH/r/n*/r/n。/r/n19．（四川成都市·成都七中）/r/n近年來，隨著聚酯工業(yè)的快速發(fā)展，氯氣的需求量和氯化氫的產(chǎn)出量也隨之迅速增長。因此，將氯化氫轉(zhuǎn)化為氯氣的技術(shù)成為科學(xué)研究的熱點?；卮鹣铝袉栴}：/r/n(1)Deacon/r/n發(fā)明的直接氧化法為：/r/n。下圖為剛性容器中，進料濃度比/r/nc(HCl)/r/n∶/r/nc(O/r/n2/r/n)/r/n分別等于/r/n1/r/n∶/r/n1/r/n、/r/n4/r/n∶/r/n1/r/n、/r/n7/r/n∶/r/n1/r/n時/r/nHCl/r/n平衡轉(zhuǎn)化率隨溫度變化的關(guān)系：可知該反應(yīng)在/r/n___________(/r/n高溫或低溫/r/n)/r/n條件下能自發(fā)進行。設(shè)/r/nHCl/r/n初始濃度為/r/nc/r/n0/r/n，根據(jù)進料濃度比/r/nc(HCl)/r/n∶/r/nc(O/r/n2/r/n)=1/r/n∶/r/n1/r/n的數(shù)據(jù)計算/r/nK(400/r/n℃/r/n)=___________(/r/n列出計算式/r/n)/r/n。/r/n按化學(xué)計量比進料可以保持反應(yīng)物高轉(zhuǎn)化率，同時降低產(chǎn)物分離的能耗。進料濃度比/r/nc(HCl)/r/n∶/r/nc(O/r/n2)/r/n過低、過高的不利影響分別是/r/n___________/r/n。/r/n(2)Deacon/r/n直接氧化法可按下列催化過程進行：/r/n/r/n△/r/nH/r/n1/r/n=83kJ·mol/r/n-1/r/n則/r/n的/r/n______/r/n(3)/r/n在一定溫度的條件下，進一步提高/r/nHCl/r/n的轉(zhuǎn)化率的方法是/r/n___________(/r/n寫出/r/n2/r/n種/r/n)/r/n(4)/r/n在傳統(tǒng)的電解氯化氫回收氯氣技術(shù)的基礎(chǔ)上，科學(xué)家最近采用碳基電極材料設(shè)計了一種新的工藝方案，主要包括電化學(xué)過程和化學(xué)過程，如下圖所示：/r/n負極區(qū)發(fā)生的反應(yīng)有/r/n___________(/r/n寫反應(yīng)方程式/r/n)/r/n。電路中轉(zhuǎn)移/r/n1mol/r/n電子，需消耗氧氣/r/n___________L(/r/n標準狀況/r/n)/r/n。/r/n【KS5U答案】/r/n高溫/r/n/r/nO/r/n2/r/n和/r/nCl/r/n2/r/n分離能耗較高、/r/nHCl/r/n轉(zhuǎn)化率較低/r/n-58kJ/r/n?/r/nmol/r/n-1/r/n/r/n增加反應(yīng)體系壓強；及時分離出產(chǎn)物/r/nFe/r/n3+/r/n+e/r/n-/r/n═/r/nFe/r/n2+/r/n、/r/n4Fe/r/n2+/r/n+O/r/n2/r/n+4H/r/n+/r/n═/r/n4Fe/r/n3+/r/n+2H/r/n2/r/nO5.6/r/n【KS5U解析】/r/n(1)/r/n根據(jù)圖知，進料濃度比/r/nc(HCl)/r/n：/r/nc(O/r/n2/r/n)/r/n一定時，升高溫度/r/nHCl/r/n的轉(zhuǎn)化率降低，說明平衡逆向移動，升高溫度平衡向吸熱方向移動，則正反應(yīng)為放熱反應(yīng)，/r/n?H<0/r/n，反應(yīng)為氣體分子數(shù)減小的反應(yīng)，/r/n?S<0/r/n，由/r/n?G=?H-T?S>0/r/n，則高溫下自發(fā)進行；溫度一定時進料濃度比/r/nc(HCl)/r/n：/r/nc(O/r/n2/r/n)/r/n越大，/r/nHCl/r/n的轉(zhuǎn)化率越小，所以/r/n400/r/n℃進料濃度比/r/nc(HCl)/r/n：/r/nc(O/r/n2/r/n)=1/r/n：/r/n1/r/n時，/r/nHCl/r/n的轉(zhuǎn)化率為最上邊曲線，/r/nHCl/r/n轉(zhuǎn)化率為/r/n84%/r/n，列三段式：/r/n，/r/n400/r/n℃時化學(xué)平衡常數(shù)/r/n/r/n；進料濃度比/r/nc(HCl)/r/n：/r/nc(O/r/n2/r/n)/r/n過低/r/nO/r/n2/r/n和/r/nCl/r/n2/r/n分離能耗較高、過高時/r/nHCl/r/n轉(zhuǎn)化率較低，所以投料濃度比過高或過低都不好；/r/n(2)/r/n根據(jù)蓋斯定律：/r/n?H=?H/r/n1/r/n+?H/r/n2/r/n+?H/r/n3/r/n=(83-20-121)kJ?mol/r/n-1/r/n=-58kJ?mol/r/n-1/r/n；/r/n(3)/r/n提高/r/nHCl/r/n的轉(zhuǎn)化率，即要反應(yīng)正向進行，/r/n4HCl(g)+O/r/n2/r/n(g)═2Cl/r/n2/r/n(g)+2H/r/n2/r/nO(g)/r/n，正向是氣體分子數(shù)減少的反應(yīng)，可以增大壓強，使反應(yīng)正向；亦可分離出生成物，使反應(yīng)正向；/r/n(4)/r/n根據(jù)圖示，電解池左側(cè)發(fā)生反應(yīng)/r/nFe/r/n3+/r/n+e/r/n-/r/n═Fe/r/n2+/r/n，該反應(yīng)為還原反應(yīng)，屬于電解池的陰極，負極通入氧氣后/r/nFe/r/n2+/r/n被/r/nO/r/n2/r/n氧化而再生成/r/nFe/r/n3+/r/n，該反應(yīng)為/r/n4Fe/r/n2+/r/n+O/r/n2/r/n+4H/r/n+/r/n═4Fe/r/n3+/r/n+2H/r/n2/r/nO/r/n；根據(jù)電子守恒及/r/n4Fe/r/n2+/r/n+O/r/n2/r/n+4H/r/n+/r/n═4Fe/r/n3+/r/n+2H/r/n2/r/nO/r/n可知，電路中轉(zhuǎn)移/r/n1mol/r/n電子，消耗氧氣的物質(zhì)的量為：/r/n1mol×1/4=0.25mol/r/n，標況下/r/n0.25mol/r/n氧氣的體積為：/r/n22.4L/mol×0.25mol=5.6L/r/n。/r/n20．（河南新鄉(xiāng)市·新鄉(xiāng)縣一中高三其他模擬）/r/n在一定條件下，由/r/nCO/r/n2/r/n和/r/nH/r/n2/r/n合成甲醇已成為現(xiàn)實，該合成對解決能源問題具有重大意義。該過程中有兩個競爭反應(yīng)，反應(yīng)過程能量關(guān)系如圖。/r/n(1)/r/n請寫出/r/nCO(g)/r/n與/r/nH/r/n2/r/n(g)/r/n生成/r/nCH/r/n3/r/nOH(g)/r/n的熱化學(xué)方程式為/r/n___________/r/n。/r/n(2)/r/n對于/r/nCO/r/n2/r/n(g)/r/n＋/r/n3H/r/n2/r/n(g)/r/nCH/r/n3/r/nOH(g)/r/n＋/r/nH/r/n2/r/nO(g)/r/n反應(yīng)，已知/r/nv/r/n正/r/n=k/r/n正/r/nc(CO/r/n2/r/n)c/r/n3/r/n(H/r/n2/r/n)/r/n，/r/nv/r/n逆/r/n=k/r/n逆/r/nc(CH/r/n3/r/nOH)c(H/r/n2/r/nO)/r/n，升高溫度/r/nk/r/n正/r/n增大的倍數(shù)/r/n___________k/r/n逆/r/n增大的倍數(shù)/r/n(/r/n填/r/n“/r/n＜/r/n”/r/n、/r/n“/r/n＞/r/n”/r/n或/r/n“=”)/r/n，為了提高/r/nH/r/n2/r/n的轉(zhuǎn)化率，可采取的措施有/r/n___________(/r/n填選項/r/n)/r/n。/r/nA/r/n．加壓/r/nB/r/n．升溫/r/nC/r/n．加催化劑/r/nD/r/n．增加/r/nCO/r/n2/r/n的濃度/r/n(3)/r/n在一容積可變的密閉容器中，充入/r/n1molCO/r/n2/r/n與/r/n3molH/r/n2/r/n發(fā)生反應(yīng)：/r/nCO/r/n2/r/n(g)/r/n＋/r/n3H/r/n2/r/n(g)/r/nCH/r/n3/r/nOH(g)/r/n＋/r/nH/r/n2/r/nO(g)/r/n，/r/nCO/r/n2/r/n在不同溫度下的平衡轉(zhuǎn)化率與總壓強的關(guān)系如下圖所示，/r/n圖中/r/nM/r/n點時/r/nCH/r/n3/r/nOH/r/n的物質(zhì)的量分數(shù)為/r/n___________/r/n，該反應(yīng)的壓強平衡常數(shù)為/r/nK/r/np/r/n=___________atm/r/n-2/r/n(/r/n用平衡分壓代替平衡濃度計算，分壓/r/n=/r/n總壓/r/n×/r/n物質(zhì)的量分數(shù)/r/n)/r/n(4)/r/n由/r/nCO/r/n2/r/n和/r/nH/r/n2/r/n合成甲醇有兩個競爭反應(yīng)，為提高/r/nCH/r/n3/r/nOH/r/n的選擇性，在原料氣中摻入一定量/r/nCO/r/n，原因是/r/n___________/r/n。另外，可以通過控制雙組份催化劑/r/n(CuO-ZnO)/r/n中/r/nCuO/r/n的含量，可提高甲醇產(chǎn)率，根據(jù)下圖判斷，催化劑選擇性最好的/r/nCuO/r/n的含量為/r/n___________/r/n。/r/n【KS5U答案】/r/nCO(g)/r/n＋/r/n2H/r/n2/r/n(g)/r/nCH/r/n3/r/nOH(g)/r/n△/r/nH=/r/n－/r/n90.56kJ·mol/r/n－/r/n1/r/n<AD16.7%5.33×10/r/n－/r/n4/r/n/r/n加入一定濃度/r/nCO/r/n，使反應(yīng)/r/nCO/r/n2/r/n(g)/r/n＋/r/nH/r/n2/r/n(g)/r/nCO(g)/r/n＋/r/nH/r/n2/r/nO(g)/r/n向逆向移動，提高生產(chǎn)/r/nCH/r/n3/r/nOH/r/n的選擇性/r/n50%/r/n【KS5U解析】/r/n(1)/r/n根據(jù)圖像可知兩個競爭反應(yīng)為：/r/ni/r/n：/r/nCO/r/n2/r/n(g)/r/n＋/r/n3H/r/n2/r/n(g)/r/nCH/r/n3/r/nOH(g)/r/n＋/r/nH/r/n2/r/nO(g)Δ/r/nH/r/n=/r/n－/r/n53.66kJ/mol/r/nii/r/n：/r/nCO/r/n2/r/n(g)/r/n＋/r/nH/r/n2/r/n(g)/r/nCO(g)/r/n＋/r/nH/r/n2/r/nO(g)Δ/r/nH/r/n=36.90kJ/mol/r/n根據(jù)蓋斯定律/r/ni-ii/r/n可得/r/nCO(g)/r/n＋/r/n2H/r/n2/r/n(g)/r/nCH/r/n3/r/nOH(g)Δ/r/nH/r/n=/r/n－/r/n90.56kJ·mol/r/n－/r/n1/r/n；/r/n(2)/r/n該反應(yīng)焓變小于/r/n0/r/n，升高溫度平衡逆向移動，說明逆反應(yīng)速率比正反應(yīng)速率增大的多，所以/r/nk/r/n正/r/n＜/r/nk/r/n逆/r/n；升溫平衡逆向移動，/r/nH/r/n2/r/n的轉(zhuǎn)化率減小，加催化劑不影響平衡轉(zhuǎn)化率，增大壓強，/r/n增加/r/nCO/r/n2/r/n的濃度都能使平衡正向移動，/r/nH/r/n2/r/n轉(zhuǎn)化率增大，故選/r/nAD/r/n；/r/n(3)/r/n根據(jù)圖中/r/nM/r/n點是/r/nCO/r/n2/r/n的平衡轉(zhuǎn)化率為/r/n50%/r/n，得：/r/n所以平時/r/nCH/r/n3/r/nOH/r/n的物質(zhì)的量分數(shù)為/r/n×100%≈16.7%/r/n；/r/nK/r/np/r/n=/r/n≈5.33×10/r/n－/r/n4/r/n；/r/n(4)/r/n加入一定量的/r/nCO/r/n可以使反應(yīng)/r/nCO/r/n2/r/n(g)/r/n＋/r/nH/r/n2/r/n(g