发布时间 : 星期三 文章双语物理化学各章习题更新完毕开始阅读
(4) NH4HS(s) is in equilibrium with arbitrary NH3(g) and H2S(g);
(5)Put I2 as solute in immiscible liquid H2O and CCl4, reaching distribution equilibrium(condensed system).
答案: (1)1,2,1;(2) 2,3,1;(3) 1,2,1; (4)2,2,2; (5)3,2,2
2.已知液体甲苯(A)和液体苯(B)在90℃时的饱和蒸气压分别为p?A?54.22kPa和
?pB?136.12kPa.两者可形成理想液态混合物.今有系统组成为xB,0?0.3的甲苯-苯
混合物5mol,在90℃下成气-液两相平衡,若气相组成为yB?0.4556,求: (1)平衡时液相组成xB及系统的压力p;
(2)平衡时气-液两相的物质的量n(l),n(g). At 90 ℃, toluene(A) and benzene(B) can form ideal liquid mixture. The saturated vapor pressure of them is 54.22kPa and 136.13kPa respectively, Now 5 mol mixture of them withxB,0?0.3,when g-l reaches equilibrium at 90℃,the the composition of the gas phase is yB?0.4556,please Calculate :
(1) how the composition of the liquid phasexB and total pressure of system p. (2) The amout of substance n(l),n(g) when g-l reach equilibrium
答案: (1)xB?0.2500,p?74.70kPa;(2)n?l??3.784mol,n?g??1.216mol 3. .已知液体甲苯(A)和液体苯(B)在90℃时的饱和蒸气压分别为p?A?54.22kPa和
?pB?136.12kPa.两者可形成理想液态混合物.取200.0g甲苯和200.0g苯置于带活塞
的导热容器中,始态为一定压力下90℃的液态混合物.在恒温90℃下逐渐降低压力,问: (1)压力降到多少时,开始产生气相,此气相的组成如何?
(2)压力降到多少时,液相开始消失,最后一滴液相的组成如何?
(3)压力为92.00kPa时,系统内气-液两相平衡,两相的组成如何?两相物质的量各为多少? At 90 ℃, toluene(A) and benzene(B) can form ideal liquid mixture. The saturated vapor pressure of them is 54.22kPa and 136.13kPa respectively, If 200.0g C7H8 and 200.0g C6H6 was put into a thermal conducting container with a piston, keep the temperature of 90 ℃and decrease the pressure gradually, Calculate :
(1)How much the pressure is,when the gas begins to product?And how the composition of
the gas phase is?
(2)How much the pressure is, when the liquid begins to disappear?And how the composition of the last drop of liquid is?
(3)If the pressure is 92kPa, how the gas-liquid composition in the system is?And how much the amount of substance both gas and liquid phase? 答案:
(1)p?98.54kPa,yB?0.7476;?2?p?80.40kPa,xB?0.3197;(3)n?l??3.022mol,n?g??1.709mol
4.已知水-苯酚系统在30℃液-液平衡时共轭溶液的组成w(苯酚)为:L1(苯酚溶于水),8.75%;L2(水溶于苯酚),69.9%. (1)在30℃,100g苯酚和200g水形成的系统达液-液平衡时,两液相的质量各为多少? (2)在上述系统中若再加入100g苯酚,又达到相平衡时,两液相的质量各为多少? Given that C6H5OH and H2O reaches liquid-liquid equilibrium at 30℃, the composition of the two conjugate solutions of this system are w(C6H5OH，L1)= 8.75%，w(C6H5OH，L2)= 69.9% respectively.
(1) At 30℃,t he system formed by 100g C6H5OH and 200g H2O reaches liquid-liquid equilibrium ,please calculate the mass of the two liquid phase.
(2) If 100g C6H5OH was added into the above system ,when the system reaches
liquid-liquid equilibrium again, please calculate the mass of the two liquid phase. 答案: (1)m?L1??179.6g,m?L2??120.4g;?2?m?L1??130.2g,m?L2??269.8g 6.利用下列数据,粗略地描绘出Mg-Cu二组分凝聚系统相图,并标出各区的稳定相.已知Mg和Cu的熔点分别为648℃,1085℃.两者可形成两种稳定化合物,Mg2Cu,MgCu2,其熔点依次为580℃,800℃.两种金属与两种化合物四者之间形成三种低共熔混合物.低共熔化合物的组成及熔点对应为:35%,380℃;66%,560℃;90.6%,680℃.
Using the data below, a rough describe Mg-Cu two component condensing system phase diagram and point out the stable phase of every region. Given Melting point of Mg and Cu is 648℃ and 1085℃ respectively.They can form of two stable compounds Mg2Cu and MgCu2 whose melting point are 580℃,800℃ respectively. Two metal and two compounds can form three kinds of eutectic mixture.Composition of eutectic mixture w(Cu) and eutectic point corresponding are: 35%,380℃;66%,560℃;90.6%,680℃. 7.A和B固态时完全不互溶，102325Pa时A(s)的熔点为30℃，B(s)的熔点为50℃，A和B在10℃具有最低共熔点，其组成为xB,E=0.4，设A和B相互溶解度均为直线。
(1)画出该系统的熔点-组成图(t-xB图)；
(2)今由2molA和8molB组成一系统，根据画出的t-xB图，列表回答系统在5℃， 30℃，50℃时的相数、相的聚集态及成分、各相的物质的量、系统所在相区的条件自由度。 A and B are immiscible in solid phase, at p=101325Pa, the melting point of A and B are30?C and 50?C respectively. A and B has the lowest cocrystalline point at 10℃, its component is xB,E=0.4,supposing the dissolving line is straight line.
(1)Draw melting point-composition diagram (t-xB)of this system.
(2)Now the system formed by 2mol A and 8mol B , according to (t-xB)Phase diagram, list table and write P , state of aggregation(g, l or s) and ingredient of phase, amount of the phases and f ’ of regions when the system temperature is 5℃， 30℃，50℃ respectivly . 答案: (2) 系统的温度/℃ 5 30 50 相数 2 2 1 相的聚集态及成分 S(A)+ S(B) l(A+B)+ S(B) l(A+B) 各相的量 nA=2mol nB=8mol nl=6.67mol nB=3.33mol nl=10mol 系统所在相区的F’ 1 1 2 8. A和B在液态部分互溶，A和B在100kPa下的沸点分别为100? C和120?C，该二组分的气、液平衡相图如图所示，且知C，E，D三个相点的组成分别为xB,C = 0.05, yB,E = 0.60， xB,D = 0.97
(1)试将图中各相区及CED线所代表的相区的相数、聚集态及成分（聚集态用g，l及s表示气、液及固；成分用A，B或A+B表示）、条件自由度f ’列成表格；
(2)试计算3mol B与7mol A的混合物，在100kPa，80?C达成平衡时气、液两相各相的物质的量各为多少摩尔？
A and B are partially miscible in liquid phase,
at p=100kPa, the boiling point are100?C and 120?C respectively. In the gas-liquid
equilibrium phase diagram, the composition of phase point C, E and D are xB,C = 0.05, yB,E = 0.60 and xB,D = 0.97 respectively.
(1) list table and write P , state of aggregation(g, l or s) and ingredient(A, B or A+B) of phase and f ’ of all regions and CED line.
(2) mixture formed by 3mol B and 7mol A reaches gas-liquid equilibrium at 100kPa，80?C . Please calculate n(l) and n(g). 答案: n (g) = 5.7 mol n ( l ) = 4.3 mol
9.如图所示，在101.325 kPa 下，A，B 二组分液态完全互溶，固态完全不互溶，其低共熔混合物wB= 0.60 今有180 g，wB= 0.40 的溶液，试回答：
（1）冷却时，最多可得多少克纯A(s)？
（2）在三相平衡时，若低共熔混合液的质量为60 g ,与其平衡的固体A及B各为多少克？
As shown in the diagram, A and B are miscible in liquid phase and immiscible in solid phase at p=101.325kPa, the lowest cocrystalline component is wB=
0.60, Now 180g solution with wB= 0.40.Calculate:
(1)How much pure A(s) we can get when the solution was cooled.
(2) the amount of A and B when the system reaches triple phase equilibrium and the mass of the lowest cocrystalline solution is 60g.
答案: (1) 得到60g纯A(s);(2) 若低共熔混合物的质量为60g时，与其平衡的固
体A为84g 固体B为36g
10. 图为A，B二组分凝聚系统平衡相图。tA*，tB* 分别为A，B 的熔点。
（1）请根据所给相图列表填写I 至 VI 各相区的相数、相的聚集态及成分、条件自由度数；
（2）系统点a0 降温经过a1，a2，a3，a4，写出在a1，a2，a3 和a4点系统相态发生的变化并画出不冷曲线。
It’s the phase diagram of A and B. tA*，tB* is the melting point of A and B respectively.
(1) According to the phase diagram, list table and write P , state of aggregation(g, l or s) and ingredient(A, B or A+B) of phase and f ’ of regions from I to VI.
(2) The system point pass through a1，a2，a3，a4 when cooling, write out the system phase state changes at point a1，a2，a3,a4 and draw the cooling curve.