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JEE CHEMISTRY MCQ: SOME BASIC CONCEPTS OF CHEMISTRY

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The following sequence of reaction occurs in commercial production of aqueous nitric acid.

4NH3( g)+5O2( g)⟶4NO(g)+6H2O(l)ΔH=−904 kJ.........(i)2NO(g)+O2( g)⟶2NO2( g)ΔH=−112 kJ.........(ii)3NO2( g)+H2O(l)⟶2HNO3(aq)+NO(g)ΔH=−140 kJ.........(iii)\begin{array}{cc} 4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) & \Delta \mathrm{H}=-904 \mathrm{~kJ}.........(i) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g}) & \Delta \mathrm{H}=-112 \mathrm{~kJ}.........(ii) \\ 3 \mathrm{NO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_3(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) & \Delta \mathrm{H}=-140 \mathrm{~kJ}.........(iii) \end{array}

Determine the total heat (in kJ/mole\mathrm{kJ} / \mathrm{mole} ) liberated at constant pressure for the production of exactly 1 mole of aqueous nitric acid by this process.
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Correct option: A
493
Option A (493) is correct because the total heat liberated is calculated by summing the enthalpy changes for the reactions required to produce 1 mole of aqueous nitric acid.

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