We know, ΔG=ΔH−TΔS So, lets find the equilibrium temperature, i.e. at which $\begin{aligned}
& \Delta \mathrm{G}=0 \
& \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \
& \mathrm{T}=\frac{179.1 \times 1000}{160.2} \
& =1118 \mathrm{~K}
\end{aligned}$ So, at temperature above this, the reaction will become spontaneous. Hence, (D) is correct answer.
In conversion of lime-stone to lime, CaCO_3( ~s) longrightarrow CaO( s)+ CO_2( ~g) the vales of Δ H° and Δ S° are +179.1 ~kJ ~mol^-1 and 160.2 ~J / K…
JEE Main 2007 — Chemistry Physical Chemistry
2007mcqmedium
In conversion of lime-stone to lime, CaCO3( s)⟶CaO(s)+CO2( g) the vales of ΔH∘ and ΔS∘ are +179.1 kJ mol−1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ΔH∘ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is
Official previous-year question
Held on 30 Apr 2007 · Verified 6 Jul 2026.
Options
- A
1008 K
- B
1200
- C
845 K
- D
1118 K
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