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#chemical reactions

29 public questions tagged with this topic.

Why does nuclear fission release more energy than chemical reactions?

**Mass defect** Δm = Z m_p + N m_n - M_nucleus, binding energy BE = Δm c², 1 u =931.5 MeV/c², BE per nucleon = BE/A, measures stability, peak ~8.8 MeV at Fe-56, for A=36 BE=288 MeV BE/A=8 MeV, for A=12 BE=96 MeV BE/A=8 MeV, for A=16 BE=127.5 MeV BE/A≈7.97 MeV, higher BE/A more stable. Nuclear fission involves changes in nuclear binding energy (order of MeV), which is about a million times larger than the energy from chemical bond changes (order of eV). Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

What is the primary reason nuclear reactions release more energy than chemical reactions?

**Binding energy calculation** from mass defect, for nucleus mass number 28 BE 224 MeV BE/A=8 MeV, A=18 BE 144 MeV BE/A=8 MeV, BE/A indicates stability, fusion of light nuclei and fission of heavy release energy because product has higher BE/A, difference released. Nuclear reactions involve changes in nuclear binding energy, which operates on a scale of MeV, vastly exceeding the eV scale of chemical bond energy, due to the strong nuclear force. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields Higher nuclear density, consistent

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

What is Δ H for C₆H₆(l) + 15/2O₂(g) -> 6CO₂(g) + 3H₂O(l) if Δ_f H°: C₆H₆(l) = 49 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(l)

Given: What is Δ H for C₆H₆(l) + 15/2O₂(g) -> 6CO₂(g) + 3H₂O(l) if Δ_f H°: C₆H₆(l) = 49 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(l) = -285.8 kJ/mol? Formula: Δ_r H = [6(-393.5) + 3(-285.8)] - [49 + 0] = -2361 - 857.4 - 49 = -3267.4 kJ/mol.. Substitution: Substituting given values into formula as per NCERT 2026-27 method. Calculation: Simplifying step by step with proper SI units like m/s², J kg⁻¹ K⁻¹, 10⁻⁵, A m⁻¹ etc. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

Which compound does not undergo Friedel-Crafts acylation?

Pyridine does not undergo Friedel-Crafts acylation due to the presence of a lone pair on nitrogen, which deactivates the ring. This follows from NCERT principle where relation explains outcome clearly for students.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

Which of the following reactions represents the preparation of K 2 Cr 2 O 7 from chromite ore?

Chromite ore is oxidized with Na 2 CO 3 in the presence of air to form sodium chromate: 4FeCr 2 O 4 + 8Na 2 CO 3 + 7O 2 → 8Na 2 CrO 4 + 2Fe 2 O 3 + 8CO 2 . This is the key step in the preparation of K 2 Cr 2 O 7 .

Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Preparation and Properties of Potassium Dichromate and Permanganate

Which species is least likely to act as an electrophile in a reaction with CH₃CH=CH₂?

CH₃⁻ is electron-rich (carbanion), making it a nucleophile, not an electrophile, unlike the others which are electron-deficient.

Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Fundamental Concepts - Reaction Mechanism - Fission and Reaction Intermediates