The principle of multiple equilibria K=(K)2=0.67×0.051=0.023 Conclusion: if a reaction equation is obtained by a linear combination of a couple of reaction equations,the reaction equilibrium constant of the resulting combined reaction is equal to the product /or quotient of the participating reactions'reaction equilibrium constant. ------a principle of multi-equilibria 结论：如果多个反应的计量式经过线性组合得到一个总的化学计量式，则总反应 的标准平衡常数等于各反应的平衡常数之积或商 多重平衡原理
4.1.3 Experimental measurement of ke Example:GeO(g)and W2O6(g)can react to give GeWO4(g)at high temperature 2GeO(g)+W2O(g)-2 GeWO4(g) The reaction starts with a mixture of GeO and W2O6, both at pressures 100.0kPa in a constant volume vessel at a certain temperature.Analysis shows that the partial pressure of GewO4(g)is 98.0 kPa at equilibrium. Calculate the partial pressures of Geo and W2Os at equilibrium and Ke for the reaction
Answer: 2GeO (g)+W2O(g)-2 GeWO4(g) Initial(pB/kPa)100.0 100.0 0 98.0 Change(pg/kPa)-98.0 98.0 2 Bqui0 P:/kPa)100.0-98.0100.0.98.0 98.0 2 p(GeO)=100.0kPa -98.0kPa=2.0kPa 20W0a0-10.0Pa- 98.0 kPa-51.0kPa
K= [p(Gewo )pe 12 [p(Ge0)/pe]2[p(W206)/ps] (98.0/1002 =4.7×103 (2.0/100)(51.0/100)
Equilibrium conversion: &(B)▣(B)-2(B） n。(B) Example: xGe0）)=o0020Pa×100%=98% 100.0kPa x(w,0.)=00.0-51.0kPa 100.0kPa ×100%=49% Recall::专= △ne_n(5)-s(0) =3 VB VB Question:If po(W2O)=Po(Ge)=1000KPa,a(W2O)=?