The enthalpy change for the conversion of $\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})$ to $\mathrm{Cl}^{-}(\mathrm{aq})$ is $(-)$ ______ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer) Given: $\Delta_{\text{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}}^{\mathrm{o}}=240 \mathrm{kJmol}^{-1}$. $\Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl}_{(\mathrm{g})}}^{\mathrm{o}}=-350 \mathrm{kJmol}^{-1}$, $\Delta_{\text{hyd}} \mathrm{H}_{\mathrm{Cl}_{(\mathrm{g})}^{-}}^{\mathrm{o}}=-380 \mathrm{kJmol}^{-1}$

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Chemical Thermodynamics

The enthalpy change for the conversion of

\frac{1}{2} Cl_{2} (g)

Cl^{-} (aq)

(-)

______

kJ mol^{- 1}

(Nearest integer)

Given:

Δ_{dis} H_{Cl_{2 (g)}}^{o} = 240 kJmol^{- 1}

Δ_{eg} H_{Cl_{(g)}}^{o} = - 350 kJmol^{- 1}

Δ_{hyd} H_{Cl_{(g)}^{-}}^{o} = - 380 kJmol^{- 1}

Answer: 610

The conversion is assembled from three steps: dissociation of half a mole of chlorine,

\frac{1}{2} \times 240 = + 120 kJ

; electron gain by the chlorine atom,

- 350 kJ

; and hydration of the chloride ion,

- 380 kJ

. Their sum is

Δ H = 120 - 350 - 380 = - 610 kJ mo l^{- 1}

, so the magnitude of the enthalpy change is

610 kJ mo l^{- 1}

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer