Question

At T(K), \(K_c\) for the reaction AO\(_2\)(g) + BO\(_2\)(g) \(\rightleftharpoons\) AO\(_3\)(g) + BO(g) is 16. One mole each of reactants and products are taken in a 1L flask and heated to T(K), and equilibrium is established. What is the equilibrium concentration of BO (in mol L\(^{-1}\))?

• A. 1.6
• B. 0.4
• C. 1.2
• D. 0.8

Hint:

Remember to set up an ICE table (Initial, Change, Equilibrium) to solve equilibrium problems.

Answer 1

The Correct Option is A

Let the reaction be: AO\(_2\)(g) + BO\(_2\)(g) \(\rightleftharpoons\) AO\(_3\)(g) + BO(g) Initial concentrations: [AO\(_2\)] = 1 mol/L [BO\(_2\)] = 1 mol/L [AO\(_3\)] = 1 mol/L [BO] = 1 mol/L Let x be the change in concentration at equilibrium. Equilibrium concentrations: [AO\(_2\)] = 1 - x [BO\(_2\)] = 1 - x [AO\(_3\)] = 1 + x [BO] = 1 + x The equilibrium constant \(K_c\) is given by: \[K_c = \frac{[\text{AO}_3][\text{BO}]}{[\text{AO}_2][\text{BO}_2]}\] Given \(K_c = 16\), we have: \[16 = \frac{(1+x)(1+x)}{(1-x)(1-x)} = \left(\frac{1+x}{1-x}\right)^2\] Taking the square root of both sides: \[4 = \frac{1+x}{1-x}\] \[4(1-x) = 1+x\] \[4 - 4x = 1 + x\] \[3 = 5x\] \[x = \frac{3}{5} = 0.6\] The equilibrium concentration of BO is: [BO] = 1 + x = 1 + 0.6 = 1.6 mol/L