,在Cu2+/Cu半电池中使c(Cu2+)=1.0 mol·L-1,在Pb2+/Pb半电池中加入SO42-,产生PbSO4沉淀,并调至c(SO42-)=1.0 mol·L-1,实验测得:电动势E=0.62V(已知铜为正极),计算PbSO4的。(Eθ(Pb2+/Pb)=-0.1263V,Eθ(Cu2+/Cu)=0.34V)Р解: 由题意知:Eθ(PbSO4/Pb) = 0.34 - 0.62 = -0.28VР而Eθ(PbSO4/Pb) = E(Pb2+/Pb) = Eθ(Pb2+/Pb) + 0.0592/2 lgc(Pb2+)Р所以:-0.28 = -0.1263 + 0.0592/2 lgc(Pb2+)Р可求得:c(Pb2+) = 6.42 ×10-6 mol·L-1Р则:(PbSO4) = c(SO42-)/cθ· c(Pb2+)/cθ= 6.42×10-6Р3. 450℃时HgO的分解反应为:2HgO (s) 2Hg (g) + O2 (g),若将0.05mol HgO固体放在1L密闭容器中加热到450℃,平衡时测得总压力为108.0KPa,求该反应在450℃时的平衡常数Kθ,及HgO的转化率。Р解:Hg蒸汽的分压为:2/3×108.0 kPa = 72.0 kPaРO2的分压为:1/3×108.0 kPa = 36.0 kPaР所以:Kθ= (p(Hg)/pθ)2 ·(p(O2)/pθ) = 0.722 × 0.36 = 0.187Р = -2.303RTlg KθР = -2.303 × 8.314 × (450 + 273.15)lg0.187 = 10.1 kJ·mol-1РpV = nRTР72 × 1 = n × 8.314 ×(450 + 273.15)Р求得:n = 0.012 molР所以转化率为:0.012/0.05 × 100% = 24.0%
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