Chemical Equilibrium

We are given these starting amounts:$[ \mathrm{AO}_2 ] = 1 \text{ mol/L} \quad [ \mathrm{BO}_2 ] = 1 \text{ mol/L} \quad [ \mathrm{AO}_3 ] = 1 \text{ mol/L} \quad [ \mathrm{BO} ] = 1 \text{ mol/L} \quad K_c = 16$We need to find the equilibrium concentration of $A\,O_3$.Step 1: Write the reaction equation.$\mathrm{AO}_2(\mathrm{g})+\mathrm{BO}_2(\mathrm{g}) \rightleftharpoons \mathrm{AO}_3(\mathrm{g})+\mathrm{BO}(\mathrm{g})$Step 2: Set up an ICE (Initial, Change, Equilibrium) table. At the start (initial):$[ \mathrm{AO}_2 ] = 1, [ \mathrm{BO}_2 ] = 1, [ \mathrm{AO}_3 ] = 1, [ \mathrm{BO} ] = 1$Let $x$ be the amount that reacts.$\;\text{ Change in concentrations: }\; \begin{array}{cccc} \mathrm{AO}_2 & \mathrm{BO}_2 & \mathrm{AO}_3 & \mathrm{BO} \\ -x & -x & +x & +x \end{array}$At equilibrium:$\; [ \mathrm{AO}_2 ] = 1 - x$$\; [ \mathrm{BO}_2 ] = 1 - x$$\; [ \mathrm{AO}_3 ] = 1 + x$$\; [ \mathrm{BO} ] = 1 + x$Step 3: Write the equilibrium expression and solve for $x$.$K_c = \frac{ [ \mathrm{AO}_3 ] [ \mathrm{BO} ] }{ [ \mathrm{AO}_2 ] [ \mathrm{BO}_2 ] } = \frac{ (1+x)(1+x) }{ (1-x)(1-x) } = 16$Take the square root of both sides: $\frac{1+x}{1-x} = 4$Solve for $x$ :$1 + x = 4(1 - x)$$1 + x = 4 - 4x$$1 + x + 4x = 4$$1 + 5x = 4$ $5x = 3$$x = 0.6$Step 4: Find the equilibrium concentration of $\mathrm{AO}_3$.$[ \mathrm{AO}_3 ] = 1 + x = 1 + 0.6 = 1.6 \text{ mol/L}$