Q15. The standard enthalpy of reaction (in kJ mol^-1) for obtaining three moles of H₂(g) from atomic hydrogen in gas phase is ______.

The standard enthalpy of reaction for producing three moles of H₂(g) from atomic hydrogen is -1308 kJ/mol.

This value comes from the Hess’s law application using the given standard enthalpy of formation for H(g).

Reaction Breakdown

The reaction is 6H(g) → 3H₂(g).

Standard enthalpy of formation Δ_fH° for H(g) is +218 kJ/mol, meaning ½H₂(g) → H(g), ΔH = +218 kJ/mol.

The reverse, H(g) → ½H₂(g), has ΔH = -218 kJ/mol per mole of H(g).

Calculation Steps

For one mole H₂: 2H(g) → H₂(g), ΔH = 2 × (-218) = -436 kJ/mol.

Scale to three moles H₂: 3 × (-436) = -1308 kJ/mol.

Using formation enthalpies directly: Δ_rH° = 3Δ_fH°(H₂(g)) – 6Δ_fH°(H(g)) = 0 – 6 × 218 = -1308 kJ/mol.

No Options Provided

The query mentions “explain every option,” but no options are listed, typical for numerical fill-in CSIR NET questions.

Common errors include +1308 (wrong sign), -654 (for 1.5 moles H₂), or -436 (single mole).

The standard enthalpy of reaction for obtaining three moles of H₂(g) from atomic hydrogen in the gas phase is a key thermodynamics problem for CSIR NET aspirants in Life Sciences and Chemical Sciences. Given the standard enthalpy of formation of atomic hydrogen (H(g)) = 218 kJ mol⁻¹, this numerical requires applying Hess’s law precisely.

Understanding the Reaction

The balanced equation is 6H(g) → 3H₂(g). This recombination is exothermic as bonds form. Δ_fH°(H₂(g)) = 0 kJ/mol (elemental standard state).

Step-by-Step Solution Using Formation Enthalpies

Δ_rH° = ΣΔ_fH°(products) – ΣΔ_fH°(reactants).

Products: 3 × 0 = 0 kJ/mol.

Reactants: 6 × 218 = 1308 kJ/mol.

Δ_rH° = 0 – 1308 = -1308 kJ/mol.

Bond Energy Perspective

H-H bond dissociation: H₂(g) → 2H(g), ΔH ≈ +436 kJ/mol (twice 218). Reverse: -436 kJ/mol per H₂. For three moles: 3 × -436 = -1308 kJ/mol.

CSIR NET Exam Tips

• Sign convention: Formation of atoms endothermic (+), recombination exothermic (-).
• Verify with NIST value: Δ_fH°(H(g)) = 218 kJ/mol confirms.
• Practice similar: Enthalpy for 4H(g) → 2H₂(g) = -872 kJ.

This yields -1308 (fill-in answer). Master for competitive exams!