In conversion of lime-stone to lime, \(\mathrm{CaCO}_{3}(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) the vales of \(\Delta \mathrm{H}^{\circ}\) and \(\Delta \mathrm{S}^{\circ}\) are \(+179.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(160.2 \mathrm{~J} / \mathrm{K}\) respectively at 298 K and 1 bar. Assuming that \(\Delta \mathrm{H}^{\circ}\) do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

JEE Main · 2007 · 29 Apr 2007 · Offline · Chemical Thermodynamics

In conversion of lime-stone to lime,

CaCO_{3} (s) ⟶ CaO (s) + CO_{2} (g)

the vales of

Δ H^{\circ}

and

Δ S^{\circ}

are

+ 179.1 kJ mol^{- 1}

and

160.2 J / K

respectively at 298 K and 1 bar. Assuming that

Δ H^{\circ}

do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

Answer: (D)

The decomposition turns spontaneous once

Δ G = Δ H - T Δ S

becomes negative, and the threshold lies where

Δ G = 0

, that is

T = \frac{Δ H}{Δ S}

. With

Δ H = 179.1

kJ per mole and

Δ S = 160.2

J per kelvin,

T = \frac{179100}{160.2} \approx 1118

K. Above roughly

1118

K the conversion of limestone to lime proceeds spontaneously.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option