Chemistry - Metallurgy Question with Solution | TestHub

The options A, C, and D are correct based on the Ellingham diagram principles:

Carbon Reduction (A): Carbon can reduce $M_a$ at a lower temperature than it can reduce $M_b$ because the carbon line ($C$) intersects the $M_a$ line earlier.

Melting Points (B): The first "kink" (slope change) on the $M_b$ line occurs at a lower temperature than on the $M_a$ line, indicating that the M.P. of $M_b<$ M.P. of $M_a$.

Displacement (C): Since the $M_a$ line is above the $M_b$ line at 500°C, $M_b$ is more stable as an oxide; thus, $M_a$ cannot spontaneously reduce $M_{b}O$ ($\Delta G^{\circ }$ is positive).

Boiling Points (D): The second sharp slope increase (boiling point) for $M_b$ happens at a higher temperature than for $M_a$, making B.P. of $M_b>$ B.P. of $M_a$.