Thermodynamics and ThermochemistrymediumMCQ SINGLE

The thermochemical equation for the dissociation of hydrogen gas into atoms may be written as: . WhaThermodynamics and Thermochemistry Chemistry Question

Question

The thermochemical equation for the dissociation of hydrogen gas into atoms may be written as: $\text{H}_2 \to 2\text{H}; \Delta H = 432.0\text{ kJ}$. What is the ratio of the energy yield on combustion of hydrogen atoms to steam to the yield on combustion of an equal mass of hydrogen molecules to steam? Heat of formation of steam is –240.0 kJ/mol.

Answer: A

💡 Solution & Explanation

Combustion of $\text{H}_2$ molecules: $\text{H}_2 + 0.5\text{O}_2 \to \text{H}_2\text{O}$ yields 240.0 kJ per mole. Combustion of equal mass of H atoms (2 moles of H): Since forming $2\text{H}$ from $\text{H}_2$ takes 432.0 kJ, the combustion $2\text{H} + 0.5\text{O}_2 \to \text{H}_2\text{O}$ yields the molecule's heat plus the atomization energy $= 240.0 + 432.0 = 672.0\text{ kJ}$. Ratio $= 672.0 / 240.0 = 2.80$.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Advanced Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry