At 298 K , the enthalpy of fusion of a solid $(\mathrm{X})$ is $2.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the enthalpy of vaporization of the liquid $(X)$ is $98.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The enthalpy of sublimation of the substance $(X)$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is (in nearest integer)

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Chemical Thermodynamics

At 298 K , the enthalpy of fusion of a solid

(X)

2.8 kJ mol^{- 1}

and the enthalpy of vaporization of the liquid

(X)

98.2 kJ mol^{- 1}

. The enthalpy of sublimation of the substance

(X)

kJ mol^{- 1}

is (in nearest integer)

Answer: 101

Enthalpy is a state function, so taking the solid to the vapour directly costs the same as melting it and then boiling the liquid:

Δ H_{sub} = Δ H_{fus} + Δ H_{vap} = 2.8 + 98.2 = 101.0 kJ mo l^{- 1}

The enthalpy of sublimation is

101 kJ mo l^{- 1}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer