淺談如何全面突破化學平衡圖像題.doc

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Antistatick opaten, jako jsou zazen, potrub, by mlo bt v souladu s technickmi pedpisy pro uzemnn design pro chemick podniky pslunch ustanoven. 7. nhl vbun petlaku nebo rychlho rozkladu vrobnch zazen nebo skladovacch zazen by mly bt vybaveny vbuchu (nevbun membrna), vvoz nonel trubice smrem k bezpenosti a aby prevence sekundrnch exploz, protiporn opaten. 8. vrobnthe principle of simplified EIA of construction projects in the region. In terms of land, linked to the implementation of urban and rural construction land increase and decrease, replacement indicators for priority areas project. Charges, into the projects of water, electricity, administrative charges and preferential policies. In the area of taxation, and settled in areas of industry and its production company, within 5 years after the completion of fiscal incentives to enterprises. In terms of financing, integration of land, tax, financial and other resources, and construct Government credit + business credit credit system, establishment of marketization, commercialization and modernization of the investment and financing platform; effective Bank-enterprise docking, encourages private capital into the Park, to raise industry development fund. 5, optimize the environment and service industries. To create policy lowlands, Highlands, integrity of service land, development land as the goal, to optimize the area under development environment. All administrative law enforcement departments to appoint a full-time personnel stationed in areas dedicated to coordinating and solving problems associated with businesses in this sector. When there are substantial issues, sector leaders arranged to personally intervene, in-person, in-person push tangible area building a green light, easy line. 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Conscientiously abide by the partys political discipline, abide by the Constitution and the rules and regulations of the party, in the political, ideological and maintain highly consistent with the CPC Central Committee on the action, there is no violation of the partys political discipline problems. 2, in the implementation of the central authorities of the eight provisions. Improving research, improving research methods, but there are less grass-roots units, primary first-hand an inadequate grasp of the problem, which is to be strengthened in the future. Second, construction, strictly in accordance with the provisions to streamline and improve the quality of meetings of the Conference. Third, streamlining file briefs, culture involves all aspects of propaganda and ideology, sometimes due to the practical needs of invention notifications, this area needs further淺談如何全面突破化學平衡圖像題呂 旭 平廣東省龍川縣教育局教研室 517300摘 要：化學平衡圖像題是近年高考化學的?？碱}型，它以圖像的形式將一些相關(guān)量之間的關(guān)系通過形象直觀的曲線表示出來，將化學原理抽象為數(shù)學問題，旨在考查學生的觀察能力、類比能力和推理能力。解答此類題目，一般要熟悉化學平衡移動的原理和化學平衡圖象的基本類型，掌握解答該題型的原則、方法和步驟等。關(guān)鍵詞：高考化學 化學平衡圖像 解題原則 方法與步驟化學平衡圖像題是近年高考化學試題中常見的一種題型。該類題型的特點是：圖像是題目的主要組成部分，把所要考查的知識寓于坐標曲線上，簡明、直觀、形象，旨在考查學生的觀察能力、類比能力和推理能力。綜合性強，思維難度大，許多學生感到困難，備考時應(yīng)給予重視。本文就如何解答化學平衡圖像題，談?wù)勛约旱目捶ǎＭ麑ν瑢W們有所啟發(fā)。一、化學平衡圖像題的解題原則1先拐先平，數(shù)值大。即在化學平衡圖像中，先出現(xiàn)拐點的曲線所對應(yīng)的反應(yīng)先達到平衡，同時，先出現(xiàn)拐點的曲線所對應(yīng)的溫度較高或壓強較大。2定一議二。即當圖像中有三個變量時，先確定一個量不變，再討論另外兩個量的關(guān)系（因為化學平衡移動原理只適用于外界“單因素”的改變而導致平衡移動的分析）。二、化學平衡圖像題的解題方法1明確坐標軸的意義，即橫坐標、縱坐標各表示什么。2弄清曲線上點的意義，特別是某些特殊的點(起點、拐點、交點等)所代表的含義。3弄清曲線斜率大?。辞€“平”與“陡”）的含義以及函數(shù)的增減性。4適當時可作“等溫線”、“等壓線”等輔助線。三、化學平衡圖象題的解題步驟1看圖像 一看面(即縱坐標與橫坐標的意義)；二看線(即線的走向和變化趨勢)；三看點(即起點、拐點、交點、終點、最高點與最低點等)；四看輔助線(如等溫線、等壓線、平衡線等) ；五看量的變化(如濃度變化、溫度變化、轉(zhuǎn)化率變化等) 。2想規(guī)律聯(lián)想外界條件對化學反應(yīng)速率和化學平衡的影響規(guī)律。3作判斷將圖像中所表現(xiàn)出的關(guān)系與所學規(guī)律對比，作出正確的判斷。四、化學平衡圖象的基本類型及例析1物質(zhì)的量（濃度）時間圖像解題關(guān)鍵：（1）分清反應(yīng)物和生成物。濃度減小的為反應(yīng)物，濃度增大的為生成物。（2）分清消耗濃度和增加濃度。反應(yīng)物的消耗濃度和生成物的增加濃度之比等于化學方程式中各物質(zhì)的化學計量數(shù)之比。（3）分清曲線斜率大小。斜率大（即曲線“陡”），速率大。典型例析：已知：2NO2(g) N2O4(g),H0。在恒溫恒容條件下，將一定量NO2和N2O4的混合氣體通入容積為2升的密閉容器中，反應(yīng)過程中各物質(zhì)的物質(zhì)的量濃度C隨時間t的變化關(guān)系如下圖所示:（1）a、b、c、d四個點中，化學反應(yīng)處于平衡狀態(tài)的是_點。（2）25min時，改變了什么條件會發(fā)生如圖所示的變化？_。（3）a、b、c、d四個點中所表示的反應(yīng)體系中，氣體顏色由深到淺的順序是_（填字母）解析：達到第一次平衡時，b點所處曲線代表的反應(yīng)濃度從0.2mol/L增加到0.6mol/L，代表NO2，c點所處曲線代表的反應(yīng)濃度從0.6mol/L減小到0.4mol/L，代表N2O4。水平線表示反應(yīng)平衡，因此，反應(yīng)處于平衡狀態(tài)的為b、d；25min時，NO2的濃度從0.6mol/L突變到1.0mol/L，而N2O4瞬間未變，顯然是加入了NO2，而且是加入了0.8mol的NO2；N2O4無色，NO2紅棕色，而c點代表的反應(yīng)還未達平衡，其所對應(yīng)的NO2的濃度值大于d點，因此，正確的順序應(yīng)是c、d、b、a。2速率（體積分數(shù)）溫度（壓強）圖像解題關(guān)鍵：（1）最高點（或最低點）為化學平衡點。（2）最高點（或最低點）的左端曲線，表示反應(yīng)未達平衡，隨溫度（T）升高，縱坐標所表示的量遞增(或遞減)。（3）最高點(或最低點)的右端曲線，表示反應(yīng)已達平衡，隨溫度升高平衡向吸熱反應(yīng)方向移動。典型例析：在5個容積相同的密閉容器中內(nèi)，分別充入同量的N2和H2，在不同溫度下，任其發(fā)生反應(yīng)：3H2 + N2 2NH3，并分別在t秒時測定其中NH3的體積分數(shù)x(NH3)，繪成如圖的曲線。 （1）A、B、C、D、E五點中，尚未達到化學平衡狀態(tài)的點是_，已達到化學平衡狀態(tài)的點是_。（2）此可逆反應(yīng)的正反應(yīng)是_反應(yīng)（填吸熱或放熱）。（3）AC段的曲線是增函數(shù)，CE段的曲線是減函數(shù)，試從反應(yīng)速率和平衡角度說明理由。_。解析：最高點為化學平衡點，最高點的右端曲線，表示反應(yīng)已達平衡，最高點的左端曲線，表示反應(yīng)未達平衡，因此，尚未達到化學平衡狀態(tài)的點是A、B，已達到化學平衡狀態(tài)的點是C、D、E；平衡后，隨溫度升高平衡向吸熱反應(yīng)方向移動，圖中NH3的體積分數(shù)遞減，因此，正反應(yīng)是放熱；AC段：反應(yīng)開始，V正V逆，反應(yīng)向右進行生成NH3，CE段：已達平衡，升溫使平衡左移，NH3%變小。3速率時間圖像解題關(guān)鍵：（1）分清正逆反應(yīng)。一般正反應(yīng)速率在上，平衡向正反應(yīng)方向移動；逆反應(yīng)速率在上，平衡向逆反應(yīng)方向移動。（2）分清瞬間變化點。一般改變反應(yīng)物或生成物濃度，圖像一點保持連續(xù)（無斷點）；改變反應(yīng)體系的壓強或改變反應(yīng)體系的溫度，兩點突變（一大一小，有斷點）。（3）若平衡無移動，則可能是由于以下原因所導致：反應(yīng)前后氣體分子個數(shù)不變；使用了催化劑。典型例析：1反應(yīng)A(g)+B(g)C(g)+Q已達平衡，升高溫度，平衡逆移，當反應(yīng)一段時間后反應(yīng)又達平衡，則速率對時間的曲線為（） 解析：首先要把握大方向，即升溫使正、逆反應(yīng)速率都變大，所以B不會是答案；再看起點和終點，由于原反應(yīng)是平衡的(V逆=V正)，因而起點相同，又由于再次達到平衡(V正V逆)，因而終點相同，故A不是答案；由于正反應(yīng)是放熱反應(yīng)，所以升溫使V正和V逆都增大，但逆反應(yīng)增大得更多，故V逆更大，即V逆的曲線斜率大，故答案為C。2向一定體積的密閉容器中加入2 mol A、0.6 mol C和一定量的B三種氣體。一定條件下發(fā)生反應(yīng)，各物質(zhì)的量濃度隨時間變化如圖（）所示，其中t0t1階段c（B）未畫出。圖（）為t2時刻后改變反應(yīng)條件，化學反應(yīng)速率隨時間變化的情況，四個階段改變的條件均不相同，每個階段只改變濃度、壓強、溫度、催化劑中的一個條件，其中t3t4階段為使用催化劑。請回答下列問題：（1）若t1=15 min，則t0t1階段以C物質(zhì)的濃度變化表示的反應(yīng)速率為 mol L1min1。（2）t4t5階段改變的條件為 ，B的起始物質(zhì)的量濃度為 mol L1。（3）t5t6階段保持容器內(nèi)溫度不變，若A的物質(zhì)的量共變化了0.01mol，而此過程中容器與外界的熱交換總量為a kJ，寫出此溫度下該反應(yīng)的熱化學方程式 。解析：本題是一道運用濃度-時間圖像與速率-時間圖像來確定化學方程式、條件改變和反應(yīng)物的起始濃度的綜合題。因圖（）為t2時刻后改變反應(yīng)條件，化學反應(yīng)速率隨時間變化的情況，四個階段改變的條件均不相同，每個階段只改變濃度、壓強、溫度、催化劑中的一個條件，其中t3t4階段為使用催化劑。根據(jù)正逆反應(yīng)速率變化與化學平衡的移動情況可推斷：t2t3階段和t5t6階段平衡發(fā)生了移動， t4t5階段平衡沒有移動。由于改變濃度、溫度化學平衡一定移動；而改變壓強，對于反應(yīng)前后氣體體積不變的可逆反應(yīng)，平衡不移動。由圖可知： t2t3階段為改變濃度；t5t6階段為升高溫度，且化學平衡正移，故該反應(yīng)的正反應(yīng)為吸熱反應(yīng)； t4t5階段為改變壓強，且改變壓強平衡不移動，故該反應(yīng)為反應(yīng)前后氣體體積不變的反應(yīng)。由圖（I）可知：t1時刻達到平衡，c（A）消耗0.2 molL1, c（C）增加了0.3 molL1，只有當c（B）消耗0.1molL1時，才滿足該反應(yīng)為反應(yīng)前后氣體體積不變的反應(yīng)這一條件，故B的起始物質(zhì)的量濃度為0.5mol L1，該反應(yīng)的化學方程式為：2A(g) + B(g) 3C(g) 。正確答案：0.02 mol L1min1；減小壓強，0.5 mol L1； 2A(g) + B(g) 3C(g) H= +200a kJmol14含量（轉(zhuǎn)化率）時間溫度（壓強）圖像解題關(guān)鍵：（1）由曲線的拐點作垂直于時間軸（t線）的垂線，其交點即為該條件下達到平衡的時間。（2）由達到平衡的時間長短，根據(jù)“先拐先平，數(shù)值大”的原則，推斷P1 與 P2、T1 與T2的相對大小。典型例析：已知某可逆反應(yīng)mA(g) + n B(g) pC(g) = Q在密閉容器中進行，圖中表示在不同反應(yīng)時間t時，溫度T和壓強P與反應(yīng)物B在混合氣體中的百分含量B%的關(guān)系曲線，由曲線分析下列判斷正確的（ ）A. T1 P2 m + n p Q T2P1 pQ 0 C. T1 P2 m + n 0 D. T1 T2P1 P2 m + n 0 解析：根據(jù)達到平衡的時間長短，可以斷定：壓強均為P2時，T2T1，溫度均為T1時，P2P1；根據(jù)“平臺”高低，可斷定低溫有利于B%增大，說明逆反應(yīng)是放熱反應(yīng)，即正反應(yīng)為吸熱，Q0；高壓有利于B%增大，說明增大壓強，平衡逆移，故mnp。故本題答案：D。5受兩個變量（P、T）影響的圖像解題關(guān)鍵：（1）根據(jù)“定一議二”的原則，固定溫度T（或壓強p），即作橫坐標軸的垂線，觀察分析圖中所示各物理量隨壓強p（或溫度T）的變化結(jié)果。（2）準確判斷所作垂線與原溫度（或壓強）曲線的交點的縱坐標。（3）縱坐標可以是某物質(zhì)的質(zhì)量分數(shù)、轉(zhuǎn)化率、濃度、濃度比值、體積分數(shù)、物質(zhì)的量分數(shù)等。典型例析：可逆反應(yīng)aA(g)+ bB(g)=dD(g)+eE(g)+Q在一定條件下達平衡狀態(tài)時，A的轉(zhuǎn)化率與溫度、壓強的關(guān)系如下圖所示。則在下列空格處填寫“大于”、“小于”或“等于”。 (1)a+b _ c+d； (2)Q _ 0。 解析：看任一根曲線，可知溫度升高，A的轉(zhuǎn)化率變大，送說明升溫使平衡正移，故正反應(yīng)為吸熱反應(yīng)，即Q0；垂直于橫坐標畫一直線，可知：溫度相同時，壓強越大，A的轉(zhuǎn)化率越小。這說明增大壓強使平衡逆移，即(a+b)(c+d) 。參考文獻1 王路全.解答化學平衡圖象題的方法和技巧.中學生數(shù)理化(高考版) .2005（13）2 邵國光.例談化學反應(yīng)速率與化學平衡圖像題的考查方式.人教網(wǎng).2011-3-24to clean and repair, it is strictly forbidden to head reaches into the mechanical stroke range. 17th article: in overhead transmission line following work should blackout, cannot blackout Shi, due isolation protection measures, crane shall not in overhead transmission line following work, through overhead transmission line should will lifting arm falls, in overhead line side work Shi, regardless of in any situation Xia, lifting arm, and rope or weights, and overhead transmission line of recently distance should not is less than Xia table provides: transmission line voltage 1000 v following 1-15 kV 20-40 kV 60-110 kV 220,000 v Allow minimum distances with the transmission line (m) vertical 1.5 level 1 1.5 2 3 4 5 6 4 6 18th: lamp voltage may not exceed 36 volts, when working in a damp place or metal containers, lamp voltage may not exceed 12 volts. 19th: pressure vessel should have a safety valve, pressure gauge, and avoid sun exposure, collisions, oxygen cylinders to prevent contaminated oil; of acetylene gas, LPG, must have a safety device to prevent tempering. 20th: x-ray and gamma-ray flaw detector operation area, non-operator, not allowed to enter. 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