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NEET 2010
Exam
NEET
Year
2010
Subject
Chemistry
Chapter
Thermodynamics (C)

NEET 2010 · Chemistry · Thermodynamics (C)

The following two reactions are known aligned & Fe_ 2 O_ 3 (~s)+3...

The following two reactions are known Fe2O3( s)+3CO(g)⟶2Fe(s)+3CO2( g),ΔH=−26.8 kJFeO(s)+CO(g)⟶Fe(s)+CO2( g)ΔH=−16.5 kJ\begin{aligned}& \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}( \mathrm{~g}),\\ & \Delta \mathrm{H}=-26.8 \mathrm{~kJ}\\ & \mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\\ & \Delta \mathrm{H}=-16.5 \mathrm{~kJ} \end{aligned}The value of ΔH\Delta \mathrm{H} for the following reaction Fe2O3( s)+CO(g)⟶2FeO(s)+CO2( g)\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) is

Options

  1. A+10.3 kJ+10.3 \mathrm{~kJ}
  2. B−43.3 kJ-43.3 \mathrm{~kJ}
  3. C−10.3 kJ-10.3 \mathrm{~kJ}
  4. D+6.2 kJ+6.2 \mathrm{~kJ}

Correct answer

D. +6.2 kJ+6.2 \mathrm{~kJ}

Step-by-step solution

Given, (I) Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g)\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{ s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{ g}); ΔH=−26.8kJ\Delta \mathrm{H}=-26.8 \mathrm{ kJ} (II) FeO\mathrm{FeO} (s) +CO(g)⟶Fe(s)+CO2(g)+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_2(\mathrm{ g}); ΔH=−16.5kJ\Delta \mathrm{H}=-16.5 \mathrm{ kJ} On multiplying Eq (II) with 2, we get (III) 2FeO(s)+2CO(g)⟶2Fe(s)+2CO2(g)2 \mathrm{FeO}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+2 \mathrm{CO}_2(\mathrm{ g}); ΔH=−33kJ\Delta \mathrm{H}=-33 \mathrm{ kJ} On subtracting Eq (III) from I, we get \begin{array}{r}\mathrm{Fe}_{2} \mathrm{O}}_{3}(\mathrm{ s})+\mathrm{CO}(\mathrm{g}) \longrightarrow2 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}}_{2}(\mathrm{ g}); \\ \Delta \mathrm{H}=-26.8-(-33) \\ =+6.2 \mathrm{ kJ} \end{array}
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