Data given for the following reaction is as follows : FeO _ ( s ) + C _ ( graphite ) ⟶ Fe _ ( s ) + CO _ ( g ) Substance Δ _ f H ° ( kJmol ^ - 1 ) ° ( Jmol ^ - 1 K ^ - 1 ) FeO _ ( s ) - 266 . 3 57 . 49 C _ ( graphite )

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Correct answer: 0

T _ = ((Δ ° H)/(Δ ° S)) Δ ^ o H _ rxn = [ Δ _ f ^ ° H ( Fe _ (s) ) + Δ _ f ^ ° H ( CO _ (g) )] - [ Δ _ f ^ ° H ( FeO _ (s) ) + Δ _ f ^ ° H ( C _ (graphite) )] = [ 0 - 110 . 5 ] - [ - 266 . 3 + 0 ] = 155 . 8kJ / mol Δ ° S _ rxn = [Δ ° S ( Fe _ (s) ) + Δ ° S ( CO _ (g) )] - [Δ ° S ( FeO _ (s) ) + Δ ° S( C _ ( graphite ) )] = [ 27 . 28 + 197 . 6 ] - [ 57 . 49 + 5 . 74 ] = 161 . 65J / mol . K T _ = 155 . 8 × 10 ^ 3 J / mol 161 . 65J / mol - K = 963 . 8K 964K (nearest integer)

Substance	${\Delta}_{f}H ° \left({kJmol}^{- 1}\right)$	$\DeltaS ° \left({Jmol}^{- 1}{K}^{- 1}\right)$
${FeO}_{( s )}$	$- 266 . 3$	$57 . 49$
${C}_{( graphite )}$	$0$	$5 . 74$
${Fe}_{( s )}$	$0$	$27 . 28$
${CO}_{( g )}$	$- 110 . 5$	$197 . 6$

Data given for the following reaction is as follows : FeO _ ( s ) +...

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