☰ خطة العام · الكيمياء (بالإنجليزية)

Properties, magnetism and catalysts of the first series

الأسبوع 3 · الباب الأول (٣): الخواص العامة لعناصر السلسلة الأولى (الكتلة الذرية ونصف القطر والكثافة والخاصية الفلزية) والخاصيتان البارامغناطيسية والديامغناطيسية وعدّ الإلكترونات المفردة والنشاط الحفزي للفلزات الانتقالية · 3 دروس
3

Transition metals as catalysts

Chemistry Sec 3 Term 1, Chapter 1: pp. 12–15
GoalExplain what a catalyst does, name four catalysts of the series, and calculate masses and volumes for the reactions they speed up.
New words
catalyst · a substance that speeds up a reaction and is recovered unchanged at the endactivation energy · the minimum energy that colliding particles need for a reaction to happensurface · the outer layer of a solid catalyst, where reactant molecules attach
Explanation

A reaction happens only when colliding particles have enough energy, the activation energy. A catalyst offers the reaction an easier route with a lower activation energy, so more collisions succeed and the reaction is faster. The catalyst is not used up and the amount of product from a given amount of reactant stays the same. Transition metals and their compounds make good catalysts because their 4s and 3d electrons can form temporary bonds with reactant molecules lying on the metal surface. The molecules gather there and their own bonds are weakened, which lowers the activation energy; then the products leave and the surface is free again. This is why a catalyst is used as a fine powder or gauze with a large surface. Four examples: finely divided nickel in hardening oils with hydrogen; iron in N₂ + 3H₂ ⇌ 2NH₃; vanadium(V) oxide in 2SO₂ + O₂ ⇌ 2SO₃, followed by SO₃ + H₂O → H₂SO₄; manganese(IV) oxide in 2H₂O₂ → 2H₂O + O₂. The catalyst is written above the arrow and is not part of the balanced equation. These industrial processes are paper reactions here.

020406080reactantspeakproductsenergy, invented unitswithout catalyst (sketch)with catalyst (sketch)
Worked examples
ƒEquation
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
catalyst: MnO₂
Hydrogen peroxide on paper. 2H₂O₂ → 2H₂O + O₂, catalyst MnO₂. A task has 6.8 g of H₂O₂ (H = 1, O = 16). M = 2 × 1 + 2 × 16 = 34 g/mol, so n = 6.8 ÷ 34 = 0.2 mol. The ratio H₂O₂ : O₂ is 2 : 1, so n(O₂) = 0.1 mol, m = 0.1 × 32 = 3.2 g, and V = 0.1 × 22.4 = 2.24 L at STP. The MnO₂ is unchanged at the end.
ƒEquation
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
catalyst: V₂O₅
The contact process on paper. 2SO₂ + O₂ ⇌ 2SO₃, catalyst V₂O₅ (S = 32, O = 16). From 6.4 g of SO₂: M = 64 g/mol, so n = 0.1 mol. The ratio SO₂ : O₂ : SO₃ is 2 : 1 : 2, so 0.05 mol of O₂ (1.6 g) is needed, and the most SO₃ possible is 0.1 mol × 80 g/mol = 8.0 g. “Most possible” because the reaction is reversible.
ƒEquation
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
catalyst: Fe
Ammonia on paper. N₂ + 3H₂ ⇌ 2NH₃, catalyst Fe (N = 14, H = 1). From 5.6 g of N₂: n = 5.6 ÷ 28 = 0.2 mol. It needs 3 × 0.2 = 0.6 mol of H₂ = 1.2 g and can give at most 2 × 0.2 = 0.4 mol of NH₃ = 0.4 × 17 = 6.8 g. Check the mass: 5.6 g + 1.2 g = 6.8 g.
Class activity

A hill on paper. Draw two valleys side by side, one for the reactants and one for the products, with a hill between them. A ball has to be given enough energy to roll over the top: mark that height as the activation energy. Draw a second, lower hill with another colour for the route with a catalyst, and mark the smaller energy needed. Write under it what stays the same (the two valleys) and what changes (the hill). No chemicals are used.

Practice
1
A task has 17 g of hydrogen peroxide H₂O₂ (H = 1, O = 16), decomposing as 2H₂O₂ → 2H₂O + O₂. Find the mass of oxygen and its volume at STP (22.4 L/mol).
2
In N₂ + 3H₂ ⇌ 2NH₃ (N = 14, H = 1), what is the largest mass of ammonia that 14 g of nitrogen can give with plenty of hydrogen?
3
Name the catalyst of the series for each: (a) hardening oils, (b) making ammonia, (c) making sulphuric acid, (d) decomposing hydrogen peroxide.
4
A pupil says: “A catalyst is a reactant, so it is used up.” Correct this in one or two sentences.