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1lianqian

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[求助] 一道物理化学题 已有1人参与

1. 某物质有和两种晶型。25℃时和型的标准摩尔生成热分别为-200.0和-1980 kJmol-1。标准摩尔熵分别为70.0和71.5 JK-1mol-1。它们都能溶于CS2中,在CS2中溶解度为10.0 molkg-1。假定, 溶解后活度系数皆为1。
     (1)  求25℃,由型转化为型的ΔG ;
     (2)  求25℃,型在CS2中的溶解度(molkg-1)。
第二问怎么解?
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maxue115223

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【答案】应助回帖

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sddtc888: 金币+1, 感谢应助,欢迎常来! 2014-01-13 22:24:17
ΔGm=μ(a)-μ(b)=RT*ln(c(a)/c(b))推得c(a)=c(b)*exp(ΔGm/RT)=10*EXP(-2227/8.314/298)=4.07(mol/kg)
2楼2014-01-02 22:31:46
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1lianqian

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引用回帖:
2楼: Originally posted by maxue115223 at 2014-01-02 22:31:46
ΔGm=μ(a)-μ(b)=RT*ln(c(a)/c(b))推得c(a)=c(b)*exp(ΔGm/RT)=10*EXP(-2227/8.314/298)=4.07(mol/kg)

这个ΔG的单位是KJ啊,数值很大啊
3楼2014-01-10 17:37:28
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