At 1400 K the standard Gibbs energies of formation of oxides, per mole of O2 consumed, are (in kJ): 2C + O2 -> 2CO, -440; 2Fe + O2 -> 2FeO, -380; 4/3 Cr + O2 -> 2/3 Cr2O3, -520; 2Mn + O2 -> 2MnO, -570; 4/3 Al + O2 -> 2/3 Al2O3, -890. Select the correct statement(s) at 1400 K.

Aluminium can reduce MnO to manganese

FeO + C -> Fe + CO has Delta G deg = -30 kJ per mole of FeO

Cr2O3 can be reduced to chromium by carbon with CO as the product

Aluminium can reduce MnO to manganese

FeO + C -> Fe + CO has Delta G deg = -30 kJ per mole of FeO

Iron can reduce MnO to manganese

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3mt034

At $1400 K$ the standard Gibbs energies of formation of oxides, per mole of $O X_{2}$ consumed, are (in $kJ$ ): $2 C + O X_{2} 2 CO$ , $- 440$ ; $2 Fe + O X_{2} 2 FeO$ , $- 380$ ; $\frac{4}{3} Cr + O X_{2} \frac{2}{3} Cr X_{2} O X_{3}$ , $- 520$ ; $2 Mn + O X_{2} 2 MnO$ , $- 570$ ; $\frac{4}{3} Al + O X_{2} \frac{2}{3} Al X_{2} O X_{3}$ , $- 890$ . Select the correct statement(s) at $1400 K$ .

A $Cr X_{2} O X_{3}$ can be reduced to chromium by carbon with $CO$ as the product
BAluminium can reduce $MnO$ to manganese
C $FeO + C Fe + CO$ has $Δ G^{\circ} = - 30 kJ$ per mole of $FeO$
DIron can reduce $MnO$ to manganese

A reduction is feasible when

Δ G^{\circ} = Δ G^{\circ} (reductant oxide) - Δ G^{\circ} (metal oxide)

(per mole

O X_{2}

) is negative. For FeO with carbon:

- 440 + 380 = - 60

per mol

O X_{2}

, i.e.

- 30

per mol

FeO

(statement 1 true). For

Cr X_{2} O X_{3}

with C:

- 440 + 520 = + 80

(false). Al with MnO:

- 890 + 570 = - 320

(true). Fe with MnO:

- 380 + 570 = + 190

(false).