Preparation, properties, uses of dihydrogen

Overview

Dihydrogen (H₂) is the most abundant element in the universe and a crucial industrial chemical. Understanding its preparation methods, physical and chemical properties, and applications is fundamental to industrial chemistry and energy systems.

Preparation of Dihydrogen

Laboratory Preparation

Method 1: Reaction of Metals with Acids

Principle: Active metals (Zn, Fe, Al) displace hydrogen from dilute acids.

Derivation from First Principles:

• Metals have low ionization energies → readily form cations
• Acids provide H⁺ ions in solution
• When metal loses electrons: M → M^n+^ + ne⁻
• H⁺ gains electrons: 2H⁺ + 2e⁻ → H₂
• Net reaction combines both half-reactions

WHY this works: The reduction potential of H⁺/H₂ (-0.00 V) is higher than active metals (Zn²⁺/Zn = -0.76 V), making the reaction spontaneous.

$\text{Zn(s) + 2HCl(aq) → ZnCl}_2\text{(aq) + H}_2\text{(g)}$

$\text{Zn(s) + H}_2\text{SO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq) + H}_2\text{(g)}$

HOW to perform:

1. Add granulated zinc to dilute H₂SO₄ or HCl in a conical flask
2. Collect gas over water (displaces water inverted tube)
3. Why granulated? → Larger surface area = faster reaction
4. Why dilute acid? → Concentrated H₂SO₄ is oxidizing and produces SO₂ instead

Solution:

• Step 1: Write balanced equation $\text{Zn + 2HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$ Why? Stoichiometry determines mole ratio

• Step 2: Calculate moles of Zn $n_{\text{Zn}} = \frac{13 \text{ g}}{65 \text{ g/mol}} = 0.2 \text{ mol}$ Why? Convert mass to moles using molar mass

• Step 3: Use stoichiometry (1:1 mole ratio) $n_{\text{H}_2} = 0.2 \text{ mol}$ Why? One mole Zn produces one mole H₂

• Step 4: Calculate volume at STP $V = n \times 22.4 \text{ L/mol} = 0.2 \times 22.4 = 4.48 \text{ L}$ Why? At STP, 1 mole gas = 22.4 L

Answer: 4.48 L of H₂

Method 2: Reaction of Alkali Metals/Hydrides with Water

$2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)$

$\text{CaH}_2\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow \text{Ca(OH)}_2\text{(aq)} + 2\text{H}_2\text{(g)}$

WHY CaH₂ works: In metal hydrides, H exists as H⁻ (hydride ion). Water provides H⁺. The reaction is acid-base: $\text{H}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{H}_2(\text{g})$

This method produces very pure H₂ because no acidic impurities are present.

Industrial Preparation

Method 1: Steam Reforming of Hydrocarbons (Bosch Process)

Most common industrial method (accounts for ~95% of H₂ production)

Step 1: Steam reforming at 1000-1200 K with Ni catalyst $\text{CH}_4\text{(g)} + \text{H}_2\text{O(g)} \xrightarrow{\text{Ni, 1000K}} \text{CO(g)} + 3\text{H}_2\text{(g)}$ $\Delta H = +206 \text{ kJ/mol (endothermic)}$

WHY Ni catalyst? Lowers activation energy for C-H bond breaking. High temperature needed because reaction is endothermic.

Step 2: Water-gas shift reaction at 673 K with Fe/Cr catalyst $CO(g) + H_2O(g) \xrightarrow{\text{Fe, 673K}} CO_2(g) + H_2(g)$ $\Delta H = -41 \text{ kJ/mol (exothermic)}$

WHY this step? Converts CO (poisonous) to CO₂ and produces additional H₂. Lower temperature favors exothermic forward reaction (Le Chatelier).

Derivation of Net Reaction: Adding both equations: $CH_4 + H_2O \rightarrow CO + 3H_2$ $\text{CO} + \text{H}_2\text{O} \rightarrow \text{CO}_2 + \text{H}_2$ $\text{CH}_4 + 2\text{H}_2\text{O} \rightarrow \text{CO}_2 + 4\text{H}_2$

Overall energy: (+206) + (-41) = +165 kJ/mol → Still endothermic, requires external heat

Solution:

• Step 1: Calculate theoretical moles of CH₄ $n_{\text{CH}_4} = \frac{16000 \text{ g}}{16 \text{ g/mol}} = 10000 \text{ mol}$ Why? Convert to moles for stoichiometry

• Step 2: Use stoichiometry (1:4 ratio) $n_{\text{H}_2} = 1000 \times 4 = 40000 \text{ mol (theoretical)}$ Why? 1 mol CH₄ → 4 mol H₂

• Step 3: Calculate theoretical mass $m_{\text{H}_2} = 4000 \times 2 = 80000 \text{ g} = 80 \text{ kg}$ Why? Molar mass of H₂ = 2 g/mol

• Step 4: Apply efficiency $\text{Actual} = 80 \times 0.80 = 64 \text{ kg}$ Why? Only 80% converts successfully

Answer: 64 kg H₂ per day

Method 2: Electrolysis of Water

$\text{2H}_2\text{O(l)} \xrightarrow{\text{electricity}} \text{2H}_2\text{(g)} + \text{O}_2\text{(g)}$

Derivation from electrochemical principles:

At cathode (reduction): $\text{2H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ $E° = -0.83 \text{ V}$

At anode (oxidation): $\ce{2OH^- -> \frac{1}{2}O2 + H2O + 2e^-}$ $E° = +0.40 \text{ V}$

Minimum voltage required: $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = -0.83 - 0.40 = -1.23 \text{ V}$

WHY negative? Non-spontaneous process, requires external energy input.

Practical voltage: 1.8-2.0 V (accounts for overpotential and resistance)

Advantages: Produces very pure H2 (>99.9%), useful for electronics and fuel cells Disadvantages: Energy-intensive (30-50 kWh per kg H2), only economic where electricity is cheap

[!formula] Faraday's Law for Electrolysis Amount of substance produced is proportional to charge passed: $n = \frac{Q}{zF} = \frac{It}{zF}$

Derivation:

• Q = total charge (coulombs) = I × t
• z = number of electrons per molecule (for H₂, z = 2)
• F = Faraday constant = 96485 C/mol
• WHY? Each electron transferred deposits 1/z of a molecule

$\text{Mass of } \ce{H2} = n \times M = \frac{It \times M}{zF}$

For H2: $m = \frac{It \times 2}{96485} = \frac{It}{96485} \text{ grams}$

[!example] Worked Example 3: Electrolysis Calculation Problem: A current of 5 A passes through water for 2 hours. Calculate (a) mass of H2 produced, (b) volume at STP.

Solution (a):

• Step 1: Calculate charge $Q = I \times t = 5 \times (2 \times 3600) = 36000 \text{ C}$ Why? Current × time = charge; convert hours to seconds

• Step 2: Calculate moles using Faraday's law $n_{\ce{H2}} = \frac{Q}{zF} = \frac{36000}{2 \times 96485} = 0.1865 \text{ mol}$ Why? 2 electrons needed per H₂ molecule

• Step 3: Calculate mass $m =0.1865 \times 2 = 0.373 \text{ g}$ Why? Multiply moles by molar mass

Solution (b): $V = n \times 22.4 = 0.1865 \times 22.4 = 4.18 \text{ L at STP}$ Why? 1 mole gas = 22.4 L at STP

Answer: (a) 0.373 g, (b) 4.18 L

Method 3: Coal Gasification

$\ce{C(s) + H2O(g) ->[\text{1270K}] CO(g) + H2(g)}$

Followed by water-gas shift reaction. Used where coal is abundant and cheap.

Properties of Dihydrogen

Physical Properties

Quantitative Properties:

• Molecular mass: 2.016 g/mol (lightest molecule)
• Density: 0.089 g/L at STP (compare: air = 1.29 g/L)
• Melting point: 14 K (-259°C)
• Boiling point: 20K (-253°C)
• Bond dissociation energy: 436 kJ/mol (very strong for a single bond)

WHY low boiling point?

• Only London dispersion forces between molecules
• Small molecular size → weak intermolecular forces
• Derivation: Boiling point ∝ intermolecular force strength ∝ molecular size/polarizability
• H₂ has minimal polarizability → lowest b.p. among all molecules

WHY low density? At STP: $\rho = \frac{M}{V_m} = \frac{2.016}{22.4} = 0.090$ g/L

Lightest molecule → lowest density

Diffusion: Diffuses rapidly (rate ∝ 1/√M by Graham's law) $\frac{r_{\ce{H2}}}{r_{\ce{O2}}} = \sqrt{\frac{M_{\ce{O2}}}{M_{\ce{H2}}}} = \sqrt{\frac{32}{2}} = 4$

H₂ diffuses 4× faster than O₂

Solubility: Poorly soluble in water (0.002 g/100 mL at 20°C) WHY? Nonpolar molecule, cannot form H-bonds with water

Chemical Properties

[!intuition] Reactivity Pattern H₂ is relatively unreactive at room temperature despite being thermodynamically favorable to react. WHY? High bond dissociation energy (436 kJ/mol) creates large activation barrier.

At elevated temperatures or with catalysts: H₂ becomes highly reactive.

1. Combustion in Air/Oxygen

Most exothermic reaction: $\ce{2H2(g) + O2(g) -> 2H2O(l)}$ $\Delta H = -286 \text{ kJ/mol}$ Derivation of enthalpy:

&\text{Breaking 2 H-H bonds: } 2(+436) = +872 \text{ kJ}\\ &\text{Breaking 1 O=O bond: } +498 \text{ kJ}\\ &\text{Forming 4 O-H bonds: } 4(-464) = -1856 \text{ kJ}\\ &\text{Net: } +872 + 498 - 1856 = -486 \text{ kJ/2 mol H}_2 = -243 \text{ kJ/mol H}_2 \end{aligned}$$ (Actual $\Delta H = -286 \text{ kJ/mol}$ includes condensation of water vapor) Explosive range: 4-75% H₂ in air WHY explosive? Highly exothermic + rapid chain reaction propagation Common Error: Confusing Combustion vs Explosion Wrong: "H₂ combustion always explodes." Why it feels right: H₂ + O₂ → loud pop in lab Reality: $$\begin{aligned} &\text{Combustion: Controlled, deflagration (subsonic flame propagation)}\\ &\text{Explosion: Uncontrolled, detonation (supersonic shock wave)}\\ &\text{Requires 4-75\% H}_2 \text{ in specific ratio} \end{aligned}$$ Fix: Pure H₂ burning in air with proper mixing is controlled combustion 2. Reaction with Halogens Reactivity order: F₂ > Cl₂ > Br₂ > I₂ With Fluorine (explosive, even in dark, at $-200°$C): $$\ce{H2 + F2 -> 2HF}$$

$\Delta H = -271 \text{ kJ/mol}$

WHY so reactive? F₂ has weakest X-X bond (158 kJ/mol) due to electron-electron repulsion in small F atom.

With Chlorine (requires sunlight or heating): $\ce{H2 + Cl2 ->[\text{hν or Δ}] 2HCl}$ $\Delta H = -185 \text{ kJ/mol}$

Mechanism (chain reaction):

1. Initiation: $\ce{Cl2 ->[\text{hν}] 2Cl·}$
2. Propagation: $\ce{Cl· + H2 -> HCl + H·}$
3. Propagation: $\ce{H· + Cl2 -> HCl + Cl·}$
4. Termination: $\ce{Cl· + Cl· -> Cl2}$

WHY photochemical? Light provides energy to break Cl-Cl bond (243 kJ/mol), initiating free radicals.

3. Reduction of Metal Oxides

H₂ acts as reducing agent, removing oxygen from metal oxides:

$\ce{CuO(s) + H2(g) ->[\Delta] Cu(s) + H2O(g)}$

$\ce{Fe2O3(s) + 3H2(g) ->[\Delta] 2Fe(s) + 3H2O(g)}$

Derivation of feasibility:

• H₂ has strong affinity for oxygen (forms stable H-O bonds)
• ΔG < 0 when: ΔH(formation of H₂O) < ΔH(formation of metal oxide)
• Works for: Cu, Pb, Fe, Zn oxides
• Does NOT work for: Alkali metal oxides, CaO, MgO, Al₂O₃ (too stable)

WHY temperature needed? Overcomes activation energy for breaking M-O bonds.

Solution:

• Step 1: Write balanced equation $\ce{Al2O3 + 3H2 -> 2Al + 3H2O}$ Why? Need stoichiometry for energy calculation

• Step 2: Calculate ΔH° for reaction $\Delta H° = [3(-286)] - [-1676] = -858 + 1676 = +818 \text{ kJ}$ Why? ΔH = Σ(products) - Σ(reactants)

• Step 3: Interpret sign Highly endothermic → Non-spontaneous Why? Al₂O₃ is more stable than H₂O

Answer: No, H₂ cannot reduce Al₂O₃ (Al₂O₃ too stable)

4. Reaction with Metals (Hydride Formation)

With reactive metals at high temperature: $\ce{2Na(s) + H2(g) ->[\Delta] 2NaH(s)}$ (ionic hydride) $\text{Ca(s) + H}_2\text{(g)} \xrightarrow{\Delta} \text{CaH}_2\text{(s)}$ (ionic hydride) $\text{Pd(s) + xH}_2\text{(g) → PdH}_x$ (interstitial hydride)

Classification by bonding:

• Ionic hydrides: s-block metals, contain H⁻ ion
• Covalent hydrides: p-block elements (H₂O, CH₄, NH₃)
• Metalic/interstitial hydrides: d-block metals, H atoms in metal lattice

5. Hydrogenation of Unsaturated Compounds

Addition across C=C double bonds with catalyst: $\text{C}_2\text{H}_4(g) + \text{H}_2(g) \xrightarrow{\text{Ni/Pt/Pd}} \text{C}_2\text{H}_6(g)$

$C_6H_5CH=CH_2 + H_2 \xrightarrow{Ni} C_6H_5CH_2CH_3$

Vegetable oil hardening: $\text{Liquid oil + H}_2 \xrightarrow{\text{Ni, 473K}} \text{Solid fat}$

WHY catalyst needed?

• H₂ bond is strong (436 kJ/mol)
• Catalyst adsorbs H₂ → weakens H-H bond
• Metal surface brings reactants close together
• Lowers activation energy from ~250 kJ/mol → ~50 kJ/mol

Uses of Dihydrogen

1. Ammonia Synthesis (Haber Process)

Largest use (~50% of global H₂ production)

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ $\Delta H = -92 \text{ kJ/mol}$

Conditions: 700K, 200-250 atm, Fe₃O₄/K₂O/Al₂O₃ catalyst

WHY these conditions?

• Exothermic: Low T favors products, but too slow
• ΔV negative (4 mol → 2 mol): High P favors products
• Compromise: Moderate T (fast enough) + High P (good yield)

Ammonia used for: Fertilizers (80%), plastics, explosives, cleaning agents

2. Methanol Production

$\text{CO(g)} + 2\,\text{H}_2\text{(g)} \xrightarrow[500\text{K},\ 250\text{atm}]{\text{ZnO/Cr}_2\text{O}_3} \text{CH}_3\text{OH(l)}$

WHY important? Methanol is feedstock for formaldehyde, acetic acid, and potential fuel

3. Petroleum Refining (Hydrocracking)

Breaks large hydrocarbons into smaller ones: $\text{C}_{20}\text{H}_{42} + \text{H}_2 \xrightarrow[\Delta, P]{\text{catalyst}} \text{C}_{10}\text{H}_{22} + \text{C}_{10}\text{H}_{20} + \ldots$

Hydrodesulfurization: Removes sulfur from crude oil $\text{RSH + H}_2 \rightarrow \text{RH + H}_2\text{S}$

WHY needed? Produces cleaner fuels, reduces SO₂ emissions

4. Margarine and Vanaspati Ghee Production

Hydrogenation of unsaturated fats: $\text{Vegetable oil (liquid) + H}_2 \xrightarrow[\text{473K}]{\text{Ni}} \text{Vanaspati (semi-solid)}$

Mechanism: Converts C=C to C-C, increasing melting point

5. Metal Extraction and Purification

Reduction of metal oxides: $\text{WO}_3 + 3\text{H}_2 \rightarrow \text{W} + 3\text{H}_2\text{O}$

Used for: Tungsten, molybdenum, copper extraction

6. Rocket Fuel (Liquid Hydrogen)

₂ + Liquid O₂ → Most efficient chemical rocket propellant

Specific impulse: 450s (gasoline: 250 s)

WHY best?

• Highest energy/mass ratio
• Clean exhaust (only H₂O)
• Calculation: $\text{Specific impulse} = \frac{\text{Thrust}}{\text{Weight flow rate}}$

Challenges: Cryogenic storage (-253°C), low density

7. Fuel Cells (Clean Energy)

$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O} + \text{Electrical energy}$

Efficiency: 40-60% (vs 2530% for combustion engines)

Advantages: Zero emissions, quiet, modular Disadvantages: H₂ storage, infrastructure, cost

8. Oxy-hydrogen Torch

$\text{2H}_2 + \text{O}_2 \rightarrow \text{2H}_2\text{O}$

Temperature achieved: 2800-3000°C

Used for: Metal cutting, welding (especially underwater)

9. Weather Balloons

WHY H₂? Lowest density → maximum lift Modern preference: Helium (non-flammable), but H₂ still used where cost matters

Making it: You can make it in your kitchen (don't try this!) by dropping a metal like zinc into acid - it fizes and releases hydrogen bubles. Factories make TONS of it by mixing natural gas with super-hot steam, kind of like breaking apart Lego blocks to get new pieces.

What's it like? It's invisible, has no smell, and if you tried to grab it, you'd feel nothing. It's 14 times lighter than air! But here's the cool part - it LOVES to burn. Mix it with oxygen and light it: BOOM! That explosion releases so much energy that rockets use liquid hydrogen as fuel to go to space.

What's it good for? We use hydrogen to make fertilizer (so plants grow better), to turn liquid vegetable oil into solid butter-like stuff (vanaspati), to clean gasoline so it doesn't pollute, and even to power special "fuel cell" cars that only "exhale" water!

The tricky part? Hydrogen is super reactive and can explode easily, plus it's hard to store because the molecules are so tiny they leak through almost anything. But scientists love it because burning it is super clean - it only makes water, no smoke or pollution!

For preparation methods: "MASH SCoRE"

• Metal + Acid → Small-scale Hydrogen
• Steam reforming → Commercial Really Effective

For properties: "Light Harry Burns Everything Rapidly"

• Light → Lowest density gas
• Harry → High diffusion rate (Graham's law)
• Burns → Combustible (exothermic combustion)
• Everything → Reduces metal oxides
• Rapidly → Reactive at high temperature

For reactivity with halogens: "Flowers Can't Bloom In Dark" F₂ (Flowers) → explosive even in cold/dark Cl₂ (Can't) → needs light (photochemical) Br₂ (Bloom) → needs heat + light I₂ (In Dark) → reversible, very slow

For uses: "AMP-MR FuelR"

• Ammonia synthesis
• Methanol production
• Petroleum refining (hydrocracking)
• Margarine production
• Reduction of metal oxides
• Fuel cells
• Rocket fuel

Connections

• Hydrogen - Introduction and Isotopes - Foundation for understanding H₂ molecular structure
• Water Gas Shift Reaction - Industrial CO removal process
• Haber Process - Major H₂ consumer for ammonia synthesis
• Hydrogenation Reactions - Addition of H₂ across multiple bonds
• Electrolysis and Faraday's Laws - Quantitative aspects of H₂ production
• Fuel Cells and Hydrogen Economy - Future energy applications
• Redox Reactions - H₂ as reducing agent
• Hydrides - Ionic, Covalent, Metallic - H₂ reactions with different elements
• Thermodynamics of Chemical Reactions - Spontaneity of H₂ reactions
• Graham's Law of Diffusion - Explains H₂ rapid diffusion

#flashcards/chemistry

What is the most common industrial method for producing dihydrogen and what are its two main steps? :: Steam reforming of hydrocarbons (Bosch Process). Step 1: CH₄ + H₂O → CO + 3H₂ (1000K, Ni catalyst). Step 2: CO + H₂O → CO₂ + H₂ (water-gas shift, 673K, Fe catalyst). Net: CH₄ + 2H₂O → CO₂ + 4H₂

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)