Given that2Cs +2Fe2O3s → 4Fes + 3CO2g; ∆H0=-93657 kcal at 250C 3Cs +3Og → 3CO2g ∆H0=-94050 kcal at 250C The value of ∆Hf0 Fe2O3 is

2Fes +3/2O3g → Fe2O3s; ∆H=? Given, 3Cs +Fe2O3s → 4Fes+3CO2g; ∆H=-93657 kcal And 3Cs +3O2g → 3CO2g; ∆H=-94050 kcal Equation ii-Equation i, we get 4Fes +3O2 → 2Fe2O3s; ∆H=-393 kcal ∴ 2Fe +3/2O2 → Fe2O3s; ∆H=-196.5 kcal at 250C

Given that2Cs +2Fe2O3s → 4Fes + 3CO2g; ∆H0=-93657 kcal at 250C 3Cs +3Og → 3CO2g ∆H0=-94050 kcal at 250C The value of ∆Hf0 Fe2O3 is

2Fes +3/2O3g → Fe2O3s; ∆H=? Given, 3Cs +Fe2O3s → 4Fes+3CO2g; ∆H=-93657 kcal And 3Cs +3O2g → 3CO2g; ∆H=-94050 kcal Equation ii-Equation i, we get 4Fes +3O2 → 2Fe2O3s; ∆H=-393 kcal ∴ 2Fe +3/2O2 → Fe2O3s; ∆H=-196.5 kcal at 250C

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Given that
$2 C (s) + 2 {Fe}_{2} O_{3} (s) \overset{}{\to} 4 Fe (s) + 3 {CO}_{2} (g); ∆ H^{0} = - 93657 kcal at 25^{0} C 3 C (s) + 3 O (g) \overset{}{\to} 3 {CO}_{2} (g) ∆ H^{0} = - 94050 kcal at 25^{0} C The value of ∆ H_{f}^{0} ({Fe}_{2} O_{3}) is$

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16.750 kcal

$- 16.750 kcal$

$- 196.5 kcal$

$- 393 kcal$

answer is C.

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Detailed Solution

$2 Fe (s) + 3 / 2 O_{3} (g) \overset{}{\to} {Fe}_{2} O_{3} (s); ∆ H = ? Given, 3 C (s) + {Fe}_{2} O_{3} (s) \overset{}{\to} 4 Fe (s) + 3 {CO}_{2} (g); ∆ H = - 93657 kcal And 3 C (s) + 3 O_{2} (g) \overset{}{\to} 3 {CO}_{2} (g); ∆ H = - 94050 kcal Equation (ii) - Equation (i), we get 4 Fe (s) + 3 O_{2} \overset{}{\to} 2 {Fe}_{2} O_{3} (s); ∆ H = - 393 kcal ∴ 2 Fe + 3 / 2 O_{2} \overset{}{\to} {Fe}_{2} O_{3} (s); ∆ H = - 196.5 kcal at 25^{0} C$