Answer: From largest to smallest
K(Cl) > K(Br) > K(I2)
Step-by-step explanation:
According to the first figure, two Cl2 molecules, two C2H4molecules, and eight C2H4Cl2 molecules are left. Using the law of mass action, you can find the equilibrium constant:
K= [C2H4Cl2]/ [C2H4][Cl2]=8/(2×2)=
K= 2
According to the second figure, four Br2molecules, four C2H4molecules, and six C2H4Br2 molecules are left. Using the law of mass action you can find the equilibrium constant:
K=[C2H4Br2]/[C2H4][Br2] =6/(4×4)=
K= 0.4
According to the third figure, seven I2 molecules, seven C2H4molecules, and three C2H4I2 molecules are left. Using the law of mass action you can find the equilibrium constant:
K=[C2H4I2]/ [C2H4][I2]= 3/(7×7)=
K= 0.06
Therefore, the equilibrium constant for Cl2 is the largest, and the equilibrium constant for I2 is the smallest, with Br2 falling in the middle of the two.