JEE Main 2025 Jan 29 Shift 2, Chemistry Q14: Hess’s Law and Bond Energy

JEE Main2025Jan 29, Shift 2Chemistry

If $C(diamond) \to C(graphite) + X$ kJ $m o l^{- 1}$

$C(diamond) + O_{2} (g) \to C O_{2} (g) + Y$ kJ $m o l^{- 1}$

$C(graphite) + O_{2} (g) \to C O_{2} (g) + Z$ kJ $m o l^{- 1}$

At constant temperature. Then

Solution

Each $“+ q kJ$ $m o l^{- 1} ”$ on the product side denotes heat released, so the enthalpy change of the reaction is $- q$ .

$Δ H_{1} = - X$ for diamond $\to$ graphite.

$Δ H_{2} = - Y$ for combustion of diamond.

$Δ H_{3} = - Z$ for combustion of graphite.

By Hess’s law, diamond $\to$ graphite is the same as: combust diamond, then reverse-combust graphite:

$Δ H_{1} = Δ H_{2} - Δ H_{3}$

$- X = - Y - (- Z) = - Y + Z$

$\Rightarrow X = Y - Z$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.