الأسبوع 11 · الباب الثالث (٢): مساحة السطح والتركيز وقانون فعل الكتلة وثابت الاتزان Kc وما تدل عليه قيمته (نص قصير: أضف أمثلة محلولة ومهام توقع) · 3 دروس
3
What Kc tells us: size, solids and water left out
Chemistry Sec 3 Term 1, Chapter 3: pp. 59–61
GoalRead the size of Kc, leave solids and pure liquids out of the expression, and find a missing concentration from a given Kc.
New words
extent of reaction · how far a reversible reaction goes before equilibrium: mostly products or mostly reactantspure solid or liquid · a substance whose concentration stays practically constant, so it is left out of Kcfixed temperature · the condition under which Kc has one value, whatever the starting concentrations
Explanation
The size of Kc says how far a reversible reaction goes before equilibrium. If Kc is much greater than 1, the products on top of the fraction are large: the reaction goes almost to completion. If Kc is much smaller than 1, the mixture holds mostly reactants, because the backward reaction wins. If Kc is near 1, reactants and products are both present in comparable amounts. Hydrogen with chlorine to give hydrogen chloride is an example of a huge Kc; we use it only as a direction, never to calculate. Kc says nothing about speed: a reaction with a large Kc may still be slow. Concentrations of pure solids and pure liquids do not appear, because they stay practically constant however much solid there is: CaCO₃(s) ⇌ CaO(s) + CO₂(g) gives Kc = [CO₂]. In dilute water solutions the solvent water is left out too. At one temperature Kc has one value, whatever the starting concentrations; a new temperature gives a new Kc (extra). Dissolving a sparingly soluble solid is treated in week 14 with its own constant, Ksp, because the solid is left out.
Size of Kc
What the equilibrium mixture holds
much less than 1 (such as 10⁻⁸)
mostly reactants
near 1
both, in comparable amounts
much greater than 1 (such as 10⁸)
mostly products, nearly complete
Worked examples
Reading the size. At one temperature three invented reactions have Kc = 2.5 × 10⁻⁸, Kc = 0.8 and Kc = 4.0 × 10⁹. The first holds almost only reactants. The second holds both, in comparable amounts. The third goes almost to completion, with almost no reactants left. Nothing here tells us which one is fastest.
ƒFind [H₂O]
[H₂O] = (0.30 × 0.30) ÷ (4.0 × 0.15) = 0.15 mol/L
A missing concentration. For CO + H₂O ⇌ CO₂ + H₂ the constant is Kc = 4.0 at the temperature used. A mixture at equilibrium has [CO] = 0.15, [CO₂] = 0.30 and [H₂] = 0.30 mol/L. Rearrange Kc = ([CO₂] × [H₂]) ÷ ([CO] × [H₂O]) to [H₂O] = ([CO₂] × [H₂]) ÷ (Kc × [CO]) = (0.30 × 0.30) ÷ (4.0 × 0.15) = 0.090 ÷ 0.60 = 0.15 mol/L. Check: (0.30 × 0.30) ÷ (0.15 × 0.15) = 0.090 ÷ 0.0225 = 4.0.
Equilibrium
Kc
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
[CO₂]
Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g)
[H₂O]⁴ ÷ [H₂]⁴
Solids left out. CaCO₃(s) ⇌ CaO(s) + CO₂(g): both solids are left out, so Kc = [CO₂]. At one temperature [CO₂] has one value whether the vessel holds 1 g or 100 g of CaCO₃, as long as some solid remains. For Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g), Kc = [H₂O]⁴ ÷ [H₂]⁴. With [H₂O] = 0.30 and [H₂] = 0.10 mol/L, Kc = (0.30 ÷ 0.10)⁴ = 3⁴ = 81. Both equations are paper only; the second one is the idea behind the iron oxide reductions of weeks 4 and 5.
Class activity
Equilibrium cards, with a grown-up. Make four cards. On the front of each write one of these: C(s) + CO₂(g) ⇌ 2CO(g); Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s); 2NO₂(g) ⇌ N₂O₄(g); H₂(g) + I₂(g) ⇌ 2HI(g). On the back write Kc, leaving the solids out. Swap the cards with a friend and check each other, then sort the four into two piles: those with a solid, and those without. Paper only.
Practice
1
Write Kc for C(s) + CO₂(g) ⇌ 2CO(g).
Kc = [CO]² ÷ [CO₂]; the solid carbon is left out.
2
Two invented reactions have Kc = 6.0 × 10⁻¹⁵ and Kc = 5.0 × 10¹⁰. Which one holds mostly products at equilibrium, and which mostly reactants?
More solid CaCO₃ is added to a closed vessel in which CaCO₃(s) ⇌ CaO(s) + CO₂(g) is at equilibrium at a fixed temperature. Does [CO₂] change? Give the reason.
No. Kc = [CO₂] has one value at a fixed temperature, and the solid does not appear in it.