1. Tardigrade
  2. Question
  3. Chemistry
  4. Given (I) 2 Fe 2 O 3( s ) → 4 Fe ( s )+3 O 2( g ) ; Δr G°=+1487.0 kJ mol-1 (II) 2 CO ( g )+ O 2( g ) arrow 2 CO 2( g ) ; Δr, G°=-514.4 kJ mol -1 Free energy change, Δr G° for the reaction 2 Fe 2 O 3( s )+ 6 CO ( g ) arrow 4 Fe ( s )+6 CO 2( g ) will be

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Q. Given
(I) $ 2 Fe _{2} O _{3}( s ) \to 4 Fe ( s )+3 O _{2}( g ) ; \Delta_{r} G^{\circ}=+1487.0 \,kJ\,mol^{-1}$
(II) $2 CO ( g )+ O _{2}( g ) \rightarrow 2 CO _{2}( g ) ; \Delta_{r}, G^{\circ}=-514.4 kJ\, mol ^{-1}$
Free energy change, $\Delta_{r} G^{\circ} $ for the reaction $ 2 Fe _{2} O _{3}( s )+ 6 CO ( g ) \rightarrow 4 Fe ( s )+6 CO _{2}( g )$ will be

JEE MainJEE Main 2018Thermodynamics

Solution:

We have $2Fe _{2} O _{3}( s ) \rightarrow 4 Fe ( s )+3 O _{2}( g ) ; \Delta_{r} G^{\circ}=+1487.0 \,kJ \,mol ^{-1}$

$2 CO ( g )+ O _{2}( g ) \rightarrow 2 CO _{2}( g ) ; \Delta_{r} G^{\circ}=-514.4 \,kJ \,mol ^{-1}$

$\Delta_{r} G^{\circ}$ for the reaction $2 Fe _{2} O _{3}( s )+6 CO ( g ) \rightarrow 4 Fe ( s )+6 CO _{2}( g )$ can be obtained by Eq. (1) $+3 \times$ Eq .(2)

$\Delta G^{\circ} =1487.0+3 \times(-514.4) $

$=1487-1543.2 $

$=-56.2 \,kJ \,mol ^{-1}$