JEE Advance - Chemistry (2023 - Paper 2 Online - No. 15)
$$
\text { The value of enthalpy change, } \mathrm{H}_\beta-\mathrm{H}_\alpha \text { (in } \mathrm{J} \mathrm{mol}^{-1} \text { ), at } 300 \mathrm{~K} \text { is }
$$ ________.
Answer
300
Explanation
As the phase transition temperature is $600 \mathrm{~K}$
So at $600 \mathrm{~K} \quad \Delta \mathrm{G}_{\mathrm{rxn}}^{\circ}=0$ So $\Delta \mathrm{H}^{\circ}{ }_{\text {reaction (600) }}=\mathrm{T} \Delta \mathrm{S}^{\circ}{ }_{\text {reaction (600) }}$
$\Delta \mathrm{H}^{\circ}{ }_{(600)}=600 \times 1=600 \mathrm{Joule} / \mathrm{mole}$
So $\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=\Delta \mathrm{C}_{\mathrm{P}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
$\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=1 \times 300$
$\Delta \mathrm{H}_{300}=\Delta \mathrm{H}_{600}-300=600-300=300$ Joule $/$ mole.
So at $600 \mathrm{~K} \quad \Delta \mathrm{G}_{\mathrm{rxn}}^{\circ}=0$ So $\Delta \mathrm{H}^{\circ}{ }_{\text {reaction (600) }}=\mathrm{T} \Delta \mathrm{S}^{\circ}{ }_{\text {reaction (600) }}$
$\Delta \mathrm{H}^{\circ}{ }_{(600)}=600 \times 1=600 \mathrm{Joule} / \mathrm{mole}$
So $\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=\Delta \mathrm{C}_{\mathrm{P}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
$\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=1 \times 300$
$\Delta \mathrm{H}_{300}=\Delta \mathrm{H}_{600}-300=600-300=300$ Joule $/$ mole.
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