To determine the equilibrium concentration of BO(g), we need to consider the reaction and the given equilibrium constant: AO2(g)+BO2(g)⇌AO3(g)+BO(g) The reaction quotient, Qc, is given by: Qc=
[AO3][BO]
[AO2][BO2]
Given Kc=16 at equilibrium, the system satisfies: Kc=
[AO3][BO]
[AO2][BO2]
=16 Let x be the change in concentration of products and reactants at equilibrium. The initial concentration for all species is 1mol∕L, thus: Initial concentrations: [AO2]=[BO2]=[AO3]=[BO]=1mol∕L Change in concentrations: [AO2]=[BO2]=1−x [AO3]=[BO]=1+x At equilibrium: Kc=
(1+x)(1+x)
(1−x)(1−x)
=16
(1+x)2
(1−x)2
=16 Taking the square root of both sides:
1+x
1−x
=4 Solving for x : 1+x=4−4x 1+5x=4 5x=3 x=0.6 Thus, the equilibrium concentration of BO is: [BO]=1+x=1+0.6=1.6mol∕L
