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#thermochemistry

26 public questions tagged with this topic.

Zeroth Law Thermal Equilibrium and Quasi-static

A solid has a molar specific heat capacity of 24.9 J mol⁻¹ K⁻¹ . Which element could it be?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Tungsten has C = 24.9 J mol⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields Tungsten, consistent with thermodynamic laws and energy conservation.

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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CHEMISTRY

For the combustion of 1 g of graphite in a bomb calorimeter, Δ U = -20.7 kJ. What is Δ U for 1 mole of graphite (molar

Given: For the combustion of 1 g of graphite in a bomb calorimeter, Δ U = -20.7 kJ. What is Δ U for 1 mole of graphite (molar mass = 12 g/mol)? These values define the system as per NCERT data. Formula: Δ U per mole = Δ U per gram × molar mass = -20.7 kJ/g × 12 g/mol = -248.4 kJ/mol.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

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

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CHEMISTRY

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.

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CHEMISTRY

For the reaction NO(g) + 1/2O₂(g) -> NO₂(g), if Δ_f H°: NO(g) = 90.25 kJ/mol, NO₂(g) = 33.18 kJ/mol, what is Δ_

Given: For the reaction NO(g) + 1/2O₂(g) -> NO₂(g), if Δ_f H°: NO(g) = 90.25 kJ/mol, NO₂(g) = 33.18 kJ/mol, what is Δ_r H? These values define the system as per NCERT data. Formula: Δ_r H = [33.18] - [90.25 + 0] = 33.18 - 90.25 = -57.07 kJ/mol.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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Heat Capacity and Calorimetry and Measurement of Enthalpy

Calculate the heat released when 22 g of CO₂ is formed from C(s) + O₂(g) → CO₂(g) . ( Δ H = -393.5 kJ/mol , molar mass o

Moles of CO₂ = 22 / 44 = 0.5 mol . For C(s) + O₂(g) → CO₂(g) , Δ H = -393.5 kJ/mol . Heat released = 0.5 × 393.5 = 196.75 kJ (positive, as heat is released in exothermic reaction).

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy

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