Chemistry - Metallurgy Question with Solution | TestHub

$$\Delta G^{\circ }=\Delta H-T\Delta S$$

slope =$-\Delta S$.

For P line ⇒ slope = 0, i.e. $\Delta S=0$.

C(s) + O₂(g) → CO₂(g) ($DS=0$).

Hence, P line represents C ⟶ CO₂

For Q line ⇒ slope = negative,

i.e. $\Delta S$ = positive.

2C(s) + O₂(g) → 2CO(g) has a positive $\Delta S$.

Hence, line Q represents the conversion of C ⟶ CO.

R ⟶ 2 (line which represents a metal oxide that is reduced by carbon above the boiling point of the metal)

S ⟶ 1 (line of metal oxide that can be reduced by carbon, producing the metal in the liquid phase. )

Therefore, the correct matching is: P - 3, Q - 4, R - 2, S - 1.