1. Tardigrade
  2. Question
  3. Chemistry
  4. In conversion of lime-stone to lime, CaCO3(s) arrow 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

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Q. In conversion of lime-stone to lime,
$CaCO_3(s) \rightarrow CaO(s) +CO2(g)$
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

AIEEEAIEEE 2007Thermodynamics

Solution:

We know, $ΔG = ΔH− TΔS$
So, lets find the equilibrium temperature, i.e. at which $ΔG = 0$
$ΔH = TΔS$
$T=\frac{179.1\times1000}{160.2}$
$=1118\,K$
So, at temperature above this, the reaction will become spontaneous.
Hence, (4) is correct answer.