Chemical Equilibrium

We are given these starting amounts:
[AO2]=1mol∕L[BO2]=1mol∕L[AO3]=1mol∕L[BO]=1mol∕LKc=16
We need to find the equilibrium concentration of AO3.
Step 1: Write the reaction equation.
AO2(g)+BO2(g)⇌AO3(g)+BO(g)
Step 2: Set up an ICE (Initial, Change, Equilibrium) table.
At the start (initial):
[AO2]=1,[BO2]=1,[AO3]=1,[BO]=1
Let x be the amount that reacts.
‌ Change in concentrations: ‌begin‌array‌cccc‌A‌O2‌BO2‌AO3‌BO\−x‌−x‌+x‌+x
At equilibrium:
‌[AO2]=1−x
‌[BO2]=1−x
‌[AO3]=1+x
‌[BO]=1+x
Step 3: Write the equilibrium expression and solve for x.
Kc=‌

[AO3][BO]

[AO2][BO2]

=‌

(1+x)(1+x)

(1−x)(1−x)

=16
Take the square root of both sides: ‌

1+x

1−x

=4
Solve for x :
‌1+x=4(1−x)
‌1+x=4−4x
‌1+x+4x=4
‌1+5x=4
‌5x=3
‌x=0.6
Step 4: Find the equilibrium concentration of AO3.
[AO3]=1+x=1+0.6=1.6mol∕L