|
BiCl3(s)、PCl5(g)是两种重要的无机氯化物,采用热力学方法计算其水解常数Kθh,对了解BiCl3(s)、PCl5(g)的水解特性及获取有关物质的热力学数据有一定意义.
1. BiCl3(s)的水解
BiCl3(s)水解反应如下:BiCl3(s)+H2O(l)=BiOCl(s)+2HCl(aq) (1)
式(1)中有关物质的热力学数据参见表1[1].
Tbl. 1 ΔfGθm values of related substances
Substance | BiCl3(aq) | H2O(l) | BiOCl(s) | HCl(aq) |
ΔfGθm/(kJ▪mol-1) | -315.1 | -237.129 | -322.2 | -131.25 |
依题: ΔrGθm=Σνi▪ΔfGθm,i=ΔfGθm[BiOCl,s]+2×ΔfGθm[HCl,aq]-ΔfGθm[BiCl3,aq]-ΔfGθm[H2O,l]
=-322.2+2×(-131.25)-(-315.1)-(-237.129)=-32.471(kJ▪mol-1)<0
又因为:ΔrGθm=-RT▪lnKθh, 1
将已知数据代入可得:lnKθh, 1=13.0994
解之得:Kθh, 1=4.886×105
2. PCl5(g)的水解
2.1 H3PO4(aq) 的 ΔfGθm
25℃,H3PO4(aq)的解离方程式如下:
H3PO4(aq)=3H+(aq) + PO43-(aq) (2)
查表可得:ΔfGθm( PO43-,aq)=-1018.8 kJ▪mol-1
又可查的H3PO4(aq)的三级解离常数依次为:Ka,1=7.1×10-3; Ka,2=6.3×10-8; Ka,3=4.8×10-13.
所以H3PO4(aq)的总解离常数为:Ka=Ka,1×Ka,2×Ka,3=7.1×10-3×6.3×10-8×4.8×10-13=2.147×10-22
式(2)ΔrGθm=-RT▪lnKa=-8.314×298.15×ln(2.147×10-22)=123.6752 kJ▪mol-1
又因为:ΔrGθm=ΔfGθm( PO43-,aq)-ΔfGθm(H3PO4,aq)
将已知数据代入可得:ΔfGθm(H3PO4,aq)=-1018.8-123.6752=-1142.5 kJ▪mol-1
同理可计算得到:ΔfGθm(H2PO4-,aq)=-1130.2 kJ▪mol-1
ΔfGθm(HPO42-,aq)=-1089.1 kJ▪mol-1
2.2 PCl5(g)的水解常数
PCl5(g)的水解反应如下: PCl5(g)+4H2O(l)=H3PO4(aq)+5HCl(aq) (3)
式(3)中有关物质的热力学数据参见表2.
Tbl. 2 ΔfGθm values of related substances
Substance | PCl5(g) | H2O(l) | H3PO4(aq) | HCl(aq) |
ΔfGθm/(kJ▪mol-1) | -305.0 | -237.129 | -1142.5 | -131.25 |
依题: ΔrGθm=Σνi▪ΔfGθm,i=ΔfGθm[H3PO4,aq]+5×ΔfGθm[HCl,aq]-ΔfGθm[PCl5,g]-4×ΔfGθm[H2O,l]
=-1142.5+5×(-131.25)-(-305.0)-4×(-237.129)=-545.234(kJ▪mol-1)<0
又因为:ΔrGθm=-RT▪lnKθh, 2
将已知数据代入可得:lnKθh, 2=219.957
解之得:Kθh, 2=3.3582×1095
3. 结论
(1)298.15K 时,BiCl3(s)及PCl5(g)的水解常数分别为4.886×105和3.3582×1095;
(2)298.15K时,H3PO4(aq)、H2PO4-(aq)、HPO42-(aq)的ΔfGθm分别为-1142.5、-1130.2、-1089.1kJ▪mol-1.
参考文献
[1] Lide D R. CRC Handbook of Chemistry and Physics. 89th ed, Chemical Co, 2008,17:2688
备注:尽管PCl5常温下为固态,但由于缺少PCl5(s)的热力学数据,博文选择计算PCl5(g).
Archiver|手机版|科学网 ( 京ICP备07017567号-12 )
GMT+8, 2024-7-17 05:55
Powered by ScienceNet.cn
Copyright © 2007- 中国科学报社