问题 问答题

工业上用CO生产燃料甲醇.一定条件下发生反应:CO(g)+2H2(g)

 CH3OH(g).图1表示反应中能量的变化;图2表示一定温度下,在体积为2L的密闭容器中加入4mol H2和一定量的CO后,CO和CH3OH(g)的浓度随时间变化.

请回答下列问题:

(1)在“图1”中,曲线______(填:a或b)表示使用了催化剂;该反应属于______(填:吸热、放热)反应.

(2)下列说法正确的是______

a.该反应的反应热为:△H=91kJ•mol-1

b.起始充入的CO为2mol

c.容器中压强恒定时,反应已达平衡状态    

d.增加CO浓度,CO的转化率增大

e.保持温度和密闭容器容积不变,再充入1molCO和2molH2,再次达到平衡时n(CH3OH)/n(CO)会减小

(3)从反应开始到建成平衡,v(H2)=______;该温度下CO(g)+2H2(g) 

CH3OH(g)的平衡常数为______.

(4)请在“图3”中画出平衡时甲醇百分含量(纵坐标)随温度(横坐标)变化的曲线,要求画压强不同的2条曲线(在曲线上标出P1、P2,且P1<P2).

答案

(1)由图可知,曲线b降低了反应所需的活化能,则b使用了催化剂,又该反应中反应物的总能量大于生成物的总能量,则该反应为放热反应,故答案为:b;放热;  

(2)a、由图可知CO(g)+2H2(g)

 CH3OH(g)△H=419KJ/mol-510KJ/mol═-91kJ•mol-1,故a错误;

b、由图2可知生成0.75mol/LCH3OH,则反应了0.75mol/LCO,平衡时有0.25mol/LCO,即CO的起始物质的量为

(0.75mol/l+0.25mol/L)×2L=2mol,故b正确;

c、该反应为反应前后压强不等的反应,则压强不变时,该反应达到平衡状态,故c正确;

d、增加CO浓度,会促进氢气的转化,氢气的转化率增大,但CO的转化率减小,故d错误;

e、再充入1molCO和2molH2,体积不变,则压强增大,平衡正向移动,再次达到平衡时n(CH3OH)/n(CO)会增大,故e错误;故答案为:bc;

(3)由图2可知,反应中减小的CO的浓度为1mol/L-0.25mol/L=0.75mol/L,10min时达到平衡,

则用CO表示的化学反应速率为

0.75mol/L
10min
=0.075mol•L-1•min-1

因反应速率之比等于化学计量数之比,则v(H2)=0.075mol•L-1•min-1×2=0.15mol•L-1•min-1

      CO(g)+2H2(g)

 CH3OH(g)

开始1mol/L    2mol/L               0

转化0.75mol/L 1.5mol/L             0.75mol/L

平衡0.25mol/L 0.5mol/L             0.75mol/L

则化学平衡常数K=

0.75mol/L
0.25mol/L×(0.5mol/L )2
=12 L2•mol-2

故答案为:0.15mol•L-1•min-1;12 L2•mol-2

(4)CO(g)+2H2(g)

 CH3OH(g)△H<0,

则温度越高,逆向反应进行的程度越大,甲醇的含量就越低,

压强增大,反应正向进行的程度大,则甲醇的含量高,则图象为

,故答案为:

阅读理解

阅读理解。

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or weekend classes.

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单项选择题