问题 填空题

硫酸用途十分广泛,工业上合成硫酸时,将SO2转化为催化氧化是一个关键步骤。请回答下列问题:

(1)该反应在恒温恒容密闭容器中进行,判断其达到平衡状态的标志是  。(填字母)

a.SO2和SO3浓度相等

b.SO2百分含量保持不变

c.容器中气体的压强不变

d.SO3的生成速率与SO2的消耗速率相等

e.容器中混合气体的密度保持不变

(2)某温度下,2SO2(g)+O2(g)2SO3(g) △H=-196 kJ•mol-1。在一个固定容积为5 L的密闭容器中充入0.20 mol SO2和0.10 mol O2,半分钟后达到平衡,测得容器中含SO3为0.18 mol,则v(O2)=    mol•L-1•min-1,放出的热量为            kJ。

(3)一定温度时,SO2的平衡转化率(α)与体系总压强(p)的关系如图所示。试分析工业生产中采用常压的原因是                 

(4)将一定量的SO2和0.7 molO2放入一定体积的密闭容器中,在550℃和催化剂作用下发生反应。反应达到平衡后,将容器中的混合气体通过过量NaOH溶液,气体体积减少了21.28 L;再将剩余气体通过焦性没食子酸的碱性溶液吸收O2,气体的体积又减少了5.6 L(以上气体体积均为标准状况下的体积)。则该反应达到平衡时SO2的转化率是多少?(要写出计算过程,计算结果保留一位小数。)

答案

(1)bc   

(2)0.036  17.64

(3)在常压下该反应转化率已很高,若再加压,对设备要求高,同时要消耗大量的能源,不经济。

(4)94.7%

题目分析:(1)“变量不变达平衡”,a。反应物和生成物的浓度不变时达平衡,而不是相等,错误;b。SO2百分含量是变量,不变时达到平衡,正确;c.反应是在恒容的容器中进行,反应前后气体的物质的量会发生变化,故容器中气体的压强是变量,正确; d.SO3的生成速率与SO2的消耗速率是同一个方向,一定相等,错误;e.容器体积固定,气体的质量固定,气体的密度是定量,错误;

(2)v(O2)=v (SO3)/2=0.18÷5÷0.5÷2=0.036mol•L-1•min-1,放出的热量为196÷2×0.18=17.64kJ;

(3)在常压下该反应转化率已很高,若再加压,对设备要求高,同时要消耗大量的能源,不经济。

(4)通过过量NaOH溶液,被吸收的为SO2和SO3,物质的量为21.28÷22.4=0.95mol;剩余氧气的量为5.6÷22.4=0.25mol,反应的氧气为0.7-0.25=0.45mol,生成SO3的量为0.45×2=0.9mol;故剩余SO2的量为0.95-0.9=0.05mol,其始量为0.95+0.05=1.00mol,转化率为0.95÷1.00=0.95

阅读理解

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单项选择题