Highly selective Backlog Qs for JEE MainChemistryThermodynamics (C)
Data given for the following reaction is as follows : FeO ( s ) + C ( graphite ) ⟶ Fe ( s ) + CO ( g ) Substance Δ f H ° kJmol - 1 Δ S ° Jmol - 1 K - 1 FeO ( s ) - 266 . 3 57 . 49 C ( graphite ) 0 5 . 74 Fe ( s ) 0 27 . 28 CO ( g ) - 110 . 5 197 . 6 The minimum temperature in K at which the reaction becomes spontaneous is_____.(Integer answer)
Correct answer
964
Step-by-step solution
T min = 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 . 8 kJ / 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 . 65 J / mol . K T min = 155 . 8 × 10 3 J / mol 161 . 65 J / mol - K = 963 . 8 K ≈ 964 K (nearest integer)