If $\quad \mathrm{C}$ (diamond) $\rightarrow \mathrm{C}$ (graphite) $+\mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}$ \[ \begin{aligned} & \mathrm{C}(\text { diamond })+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{Y} \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{Z} \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \] At constant temperature. Then

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Chemical Thermodynamics

C

(diamond)

\to C

(graphite)

+ X kJ mol^{- 1}

C (diamond) + O_{2} (g) \to CO_{2} (g) + Y kJ mol^{- 1} C (graphite) + O_{2} (g) \to CO_{2} (g) + Z kJ mol^{- 1}

At constant temperature. Then

Answer: (D)

Subtracting the graphite-combustion equation from the diamond-combustion equation cancels

O_{2}

and

C O_{2}

[C (diamond) + O_{2} \to C O_{2} + Y] - [C (graphite) + O_{2} \to C O_{2} + Z]

gives

C (diamond) - C (graphite) \to (Y - Z)

, i.e.

C (diamond) \to C (graphite) + (Y - Z)

. Comparing with the first equation,

X = Y - Z

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option