Collisions, the law of mass action and Kc
Answers are for parents and teachers.
1
A cube of side 4 cm is cut into 64 cubes of side 1 cm. Find the total surface before and after, and how many times larger it becomes.
Before: 6 × 4² = 96 cm². After: 64 × 6 × 1² = 384 cm². 384 ÷ 96 = 4 times larger.
2
For N₂ + O₂ ⇌ 2NO the equilibrium concentrations are [N₂] = 0.40, [O₂] = 0.20 and [NO] = 0.060 mol/L. Find Kc.
Kc = [NO]² ÷ ([N₂] × [O₂]) = 0.060² ÷ (0.40 × 0.20) = 0.0036 ÷ 0.080 = 0.045.
3
For the invented reaction CO + 2H₂ ⇌ CH₃OH, Kc = 20 at the temperature used. At equilibrium [CO] = 0.10 and [H₂] = 0.20 mol/L. Find [CH₃OH].
[CH₃OH] = Kc × [CO] × [H₂]² = 20 × 0.10 × 0.040 = 0.080 mol/L.
4
Write Kc for NH₄Cl(s) ⇌ NH₃(g) + HCl(g), and say why NH₄Cl does not appear.
Kc = [NH₃] × [HCl]. NH₄Cl is a solid, and the concentration of a solid stays practically constant, so it is left out.
5
A student says: “A reaction with a large Kc is a fast reaction.” Write a two-sentence reply.
Kc tells how far the reaction goes before equilibrium, so a large Kc means mostly products. How fast the reaction gets there is a different question, decided by the rate.