The principle of multiple equilibria Example:Equilibrium constants for the following reactions have been determined: ①BrCI(g)=1/2Cl2(g+1/2Br2(g)K,o=0.67 ②L2(g+Br2(g)=21IBr(g)K9=0.051 Using the above information,calculate the standard equilibrium constant for the following reaction: 32BrCl(g)+12(g)--2IBr(g)+Cl2(g) Answer:①x2+②=③ 2BrCl (g)+12(g)=2IBr(g)+Cl(g) K月=()29=0.672×0.051=0.023 Example：Equilibrium constants for the following reactions have been determined: The principle of multiple equilibria Answer：①x2 + ② = ③ ②I2 (g)+Br2 (g) 2IBr(g) K 2 = 0.051 Using the above information, calculate the standard equilibrium constant for the following reaction: ①BrCl(g) 1/2Cl2 (g)+1/2Br2 (g) K1 = 0.67 2BrCl (g)+ I2 (g) 2IBr(g)+ Cl2 (g) ③2BrCl (g)+ I2 (g) 2IBr(g)+ Cl2 (g) K3 = · = 0.67 K3 K1 K2 2×0.051=0.023 2 ( )