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#enthalpy

35 public questions tagged with this topic.

CHEMISTRY

For the reaction CO(g) + 2H₂(g) -> CH₃OH(l), if Δ H = -128.1 kJ/mol at 298 K, what is Δ U (R = 8.314 J/mol · K)?

Given: For the reaction CO(g) + 2H₂(g) -> CH₃OH(l), if Δ H = -128.1 kJ/mol at 298 K, what is Δ U (R = 8.314 J/mol · K)? Formula: Δ n_g = 0 - (1 + 2) = -2, RT = 8.314 × 298 / 1000 = 2.48 kJ. Substitution & Calculation: Δ H = Δ U + Δ n_g RT. Δ U = -128.1 - (-2 × 2.48) = -123.14 kJ/mol. 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

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

The enthalpy change for the reaction Hâ‚‚O(l) -> Hâ‚‚O(g) at 373 K is 40.79 kJ/mol. What is the internal energy change i

Given: The enthalpy change for the reaction H₂O(l) -> H₂O(g) at 373 K is 40.79 kJ/mol. What is the internal energy change if Δ n_g = 1 and R = 8.314 J/mol · K? These values define the system as per NCERT data. Formula: Δ H = Δ U + Δ n_g RT. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given Δ H = 40.79 kJ/mol, Δ n_g = 1, T = 373 K, RT = 8.314 × 373 / 1000 = 3.1 kJ. So, Δ U = 40.79 - 3.1 = 37.69 kJ/mol. 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: Thermodynamics, Equilibrium and Chemical Kinetics, Topic: Energetics, equilibrium constants and reaction rates.

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CHEMISTRY

For an exothermic reaction at constant pressure, what is the relationship between Δ H and Δ U?

For an exothermic reaction, Δ H < 0. If Δ n_g < 0 (fewer gas moles in products), Δ H < Δ U because Δ H = Δ U + Δ n_g RT and Δ n_g RT < 0. This follows from latest NCERT 2026-27 principle explaining the concept clearly for NEET students in simple steps as per rationalized syllabus.

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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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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