For 4/3 Al + O2 -> 2/3 Al2O3 take Delta G deg = -1115 + 0.210 T kJ. For 2Mg(g) + O2 -> 2MgO(s) (T above 1363 K, the boiling point of Mg) take Delta G deg = -1450 + 0.400 T kJ. Aluminium can reduce MgO to magnesium vapour only above about

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JEE Advanced levelSingle correct optionDifficulty 3 of 3mt220

For $\frac{4}{3} Al + O X_{2} \frac{2}{3} Al X_{2} O X_{3}$ take $Δ G^{\circ} = - 1115 + 0.210 T$ kJ. For $2 Mg (g) + O X_{2} 2 MgO (s)$ (T above 1363 K, the boiling point of Mg) take $Δ G^{\circ} = - 1450 + 0.400 T$ kJ. Aluminium can reduce MgO to magnesium vapour only above about

A3625 K
B1763 K
C549 K
D5310 K

Answer: B

Al reduces MgO when its line lies below the Mg line:

- 1115 + 0.210 T < - 1450 + 0.400 T

, i.e.

335 < 0.190 T

and

T > 1763

K. This is above 1363 K, so the Mg-vapour line is the valid one. The Mg line rises more steeply because of the gas formation, which reverses the usual order at high T.

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For 34​Al+OX2​​32​AlX2​OX3​ take ΔG∘=−1115+0.210T kJ. For 2Mg(g)+OX2​​2MgO(s) (T above 1363 K, the boiling point of Mg) take ΔG∘=−1450+0.400T kJ. Aluminium can reduce MgO to magnesium vapour only above about

Answer: B

More questions on Ellingham diagrams and thermodynamics of reduction

For $\frac{4}{3} Al + O X_{2} \frac{2}{3} Al X_{2} O X_{3}$ take $Δ G^{\circ} = - 1115 + 0.210 T$ kJ. For $2 Mg (g) + O X_{2} 2 MgO (s)$ (T above 1363 K, the boiling point of Mg) take $Δ G^{\circ} = - 1450 + 0.400 T$ kJ. Aluminium can reduce MgO to magnesium vapour only above about