C0(g)+H20(g)====C02(g)+H2(g)
Given that at the same temperature, two reaction equations and their standard equilibrium constant are as below:
?C(graphite)+H20(g) ====C0(g)+H2(g) K1?T? ? C(graphite)+2H20(g)====C02 (g)+2H2(g) K2?T?
Calculate K??T? of the following reaction: C0(g)+H20(g)====C02(g)+H2(g)
??答案：K??K2 /K14．某理想气体反应如下：
A(g) + 2B(g) = Y(g) + 4Z(g)
已知有关数据如下表:
物质 A(g) B(g) Y(g) Z(g) ??B,298K??fHmkJ?mol?1 ??B,298K?Sm ?1?1kJ?mol?KCp,m?B?kJ?mol?K3 14 11 5 ?1?1 -74.84 -241.84 -393.42 0 186.0 188.0 214.0 130.0 (1)经计算说明：当A、B、Y和Z的摩尔分数分别为0.3，0.2，0.3和0.2，T=800K,p=0.1MPa时反应进行的方向；
(2)其它条件与(1)相同，如何改变温度使反应向着与(1)的方向相反的方向进行？ Relative data of pg reaction A(g) + 2B(g)==Y(g) + 4Z(g) are： substance A(g) B(g) Y(g) Z(g) ?fHm?(298.15K)/ kJ·mol-1 －74.84 －241.84 －393.42 0 Sm?(298.15K)/ J·K-1·mol-1 186.0 188.0 214.0 130.0 Cp,m?(B) / J·K-1·mol-1 3 14 11 5 Please calculate and determine:(1) the direction of chemical reaction while T=810K，p=0.1MPa and the mole fraction of A, B,Y and Z are 0.3，0.2，0.3 and 0.2 respectively ; (2) if the condition is same as (1), how to change temperature to make the direction backward. 答案：（1）反应向反方向进行；（2）T>827K
5．在真空的容器中放人固态的NH4HS，于25℃下分解为NH3(g)和H2S(g)，平衡时容器内的压力为66.66kPa。
（1） 当放入NH4HS (s)时容器中已有39.99 kPa的H2S(g)，求平衡时容器中的压力； （2） 容器中原有6.666kPa的NH3(g),问需加多大压力的H2S(g)，才能形成NH4HS
固体?
The solid NH4HS was put into vacuum container , and was decomposed to NH3(g)和H2S(g) at 25℃, When the reaction reached in equilibrium the pressure was 66.66kPa.
(1) When NH4HS (s) was put into the container, the pressure of H2S(g) was 39.99 kPa. Please calculate the pressure in container when the reaction reaches equilibrium.
(2)Beginning with 6.666kPa的NH3(g) in the container, how much pressure of H2S(g) is needed to form solid NH4HS. 答案：（1）p=77.7kPa;(2) p（H2S）>166.65 kPa才可能有NH4HS (s)生成 6．现有理想气体间反应 A(g)+ B(g) ==== C(g) +D(g) 开始时，A与B均为lmol，在25℃时反应达到平衡，此时A与B物质的量各为(1／3)mol。 (1) 求此反应的K?；
(2) 开始时，A为lmol，B为2mol；
(3) 开始时，A为lmol，B为lmol，C为0．5mol；
(4) 开始时，C为1mol ，D为2mol；分别求反应达平衡时C的物质的量。
There is a reaction of perfect gas A(g)+ B(g) ==== C(g) +D(g). At the beginning, both A and B are 1mol, when the reaction reach equilibrium at 25℃, the substance amount of A and B are (1／3)mol respectively. (1) Calculate K? of reaction
(2) At the beginning, A is 1mol,B is 2mol;
(3) At the beginning, A is 1mol,B is 2mol,C is 0.5mol; (4) At the beginning, C is 1mol,D is 2mol;
Please Calculate the substance amout of C respectively when the reaction reacheas equilibrium.
答案：(1) K?＝4;(2) nC=0.845mol;(3) nC=1.096mol;(4) nC=0.543mol 7．在高温下水蒸气通过灼热的煤层，按下式生成水煤气
C(石墨)+H20(g)====C0(g)+H2(g)
若在1000K及1200K时，K?分别为2.472及37.58，试计算此温度范围内的平均摩尔
?
反应焓?rHm及在1100K时反应的标准平衡常数K
Under high temperature, the water vapor go through coal layer which is scorching hot ,
produce water gas as the reaction below:
C(graphite)+H20(g)====C0(g)+H2(g)
?
If K? is2.472 and 37.58 under 1000K and 1200K respectively, Calculate ?r H m and K of
the reaction at 1100K. 答案：K?（1100K）=11.0
?8．在100℃下，下列反应的K=8.1×10-9，?rSm =125.6J·mol-1·K-1。计算:
?COCl2 (g)====CO(g)+C12 (g) (1)100℃且总压为200kPa时COCl2的解离度； (2)100℃时上述反应的 ?rHm?；
(3)总压为200kPa、COCl2解离度为0.1％时的温度(设?Cp,m =0)。
?At 100℃, K=8.1×10-9，?rSm =125.6J·mol-1·K-1 of the following reaction. Please
?calculate:
COCl2 (g)====CO(g)+C12 (g) (1) dissociated degree of COCl2 at 100℃ and 200kPa; (2) ?r H m? of the above reaction at 100℃；
(3) the temperature when dissociated degree of COCl2 is 0.1％ and total pressure is 200kPa (?Cp,m =0)。
答案：（1）??6.37?10?5；（2）ΔrHm=105 kJ·mol-1；（3）T 2=446K 9．某理想气体反应2A(g) =Y(g)有关数据如下： 物质 ?fHm?(298.15K)/ kJ·mol-1 A(g) Y(g) 35 10 Sm? (298.15K)/ J·K-1·mol-1 250 300 Cp,m? (B)/ J·K-1·mol-1 38.0 76.0 求：（1）在310K、100KPa下，A、Y各为y=0.5的气体混合物反应向哪个方向进行？
（2）：欲使反应向与上述（1）相反的方向进行，在其他条件不变时：（a）改变压力，P应控制在什么范围？(b)改变温度，T应控制在什么范围？(c)改变组成，yA应控制在什么范围？
pg reaction: 2A(g)==Y(g) substance ?fHm?(298.15K)/ Sm?(298.15K)/ Cp,m?(B) kJ·mol-1 A(g) Y(g) 35 10 J·K-1·mol-1 250 300 / J·K-1·mol-1 38.0 76.0 Please calculate and determine:(1) the direction of chemical reaction while T=310K，p=100KPa and the mole fraction of A and Y are 0. 5 respectively ; (2) if the other conditions are not change, how to change ① pressure p ② composition yA to make the direction backward.
1p>434.8kPa; ○2T<291.6K; ○3yA>0.745 答案：(1) 向左进行;(2)○
?10．反应3CuCl(g) = Cu3Cl3(g) 的?rGm与温度T的关系如下：
??rGm/?J?mol?1???528858?52.34?T/K?lg?T/K??438.2?T/K?
??求：(1) 2000K时，此反应的?rHm； ,?rSm(2) 此反应在2000K，100kPa平衡混合物中Cu3Cl3的摩尔分数为0.5时系统的总压
?The relationship between ?rGm and T of the reaction 3CuCl(g) = Cu3Cl3(g) as follows:
??rGm/?J?mol?1???528858?52.34?T/K?lg?T/K??438.2?T/K?
??Calculate :(1) ?rHmof the reaction at 2000K ; ,?rSm(2) the mole fraction of Cu3Cl3 in the equilibrium mixture at 2000K and 100kPa.
??答案：(1)?rSm?2000K???242.6J?K?1?mol?1;?rHm?2000K???483.4kJ?mol?1
(2)p=211.2kPa
第六章 相平衡
一、选择题
1.NaHCO3(s)在真空容器中部分分解为Na2CO3(s)、H2O(g)和CO2(g)处于如下的化学平衡时： NaHCO3(s) =Na2CO3(s)+H2O(g)+CO2 (g) 该系统的独立组分数，相数及自由度符合( ). A. 3，2 ; B.3，1 ; C. 2，0； D. 2，1
2.将过量的NaHCO3 (s)放入一真空密闭容器中,在50℃下，NaHCO3(s)按下式分解： 2NaHCO3 (s) =Na2CO3 (s)+H2O(g)+CO2 (g)
该系统达平衡后，则其独立组分数C=( ), 自由度数F=( ).