at constant temperature. Then [2025] JEE Main & Advanced Thermodynamics PYQ | Topic-Wise Practice

Question

If $C (diamond) ⟶ C (graphite) + X {kJ mol}^{- 1}$

$C (diamond) + O_{2} (g) ⟶ {CO}_{2} (g) + Y {kJ mol}^{- 1}$

$C (graphite) + O_{2} (g) ⟶ {CO}_{2} (g) + Z {kJ mol}^{- 1}$

at constant temperature. Then [2025]

Smart Achievers · Accepted Answer

(1)

$C(diamond) + O_{2} (g) \to {CO}_{2} (g) + Y \dots (i)$

$C(graphite) + O_{2} (g) \to {CO}_{2} (g) + Z \dots (i i)$

(i) - (ii) gives

$C(diamond) \to C(graphite) + X$

By Hess’s law, $X = Y - Z$

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