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

Question

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 (A) 1008 K (B) 1200 K (C) 858 K (D) 1118 K

Tardigrade Admin · Accepted Answer

We know, ÎG = ÎHâ TÎS So, lets find the equilibrium temperature, i.e. at which ÎG = 0 ÎH = TÎS T=(179.1×1000/160.2) =1118 K So, at temperature above this, the reaction will become spontaneous. Hence, (4) is correct answer.

$1008 K$

$1200 K$

$858 K$

$1118 K$

We know, $Δ G = Δ H - T Δ S$
So, lets find the equilibrium temperature, i.e. at which $Δ G = 0$
$Δ H = T Δ S$
$T = \frac{179.1 \times 1000}{160.2}$
$= 1118 K$
So, at temperature above this, the reaction will become spontaneous.
Hence, (4) is correct answer.

Q. In conversion of lime-stone to lime, CaCO3​(s)→CaO(s)+CO2(g) the vales of ΔH° and ΔS° are +179.1kJmol−1 and 160.2J/K respectively at 298K and 1bar. Assuming that ΔH° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

1008K

1200K

858K

1118K

We know, ΔG=ΔH−TΔS So, lets find the equilibrium temperature, i.e. at which ΔG=0 ΔH=TΔS T=160.2179.1×1000​ =1118K So, at temperature above this, the reaction will become spontaneous. Hence, (4) is correct answer.

$1008 K$

$1200 K$

$858 K$

$1118 K$

We know, $Δ G = Δ H - T Δ S$
So, lets find the equilibrium temperature, i.e. at which $Δ G = 0$
$Δ H = T Δ S$
$T = \frac{179.1 \times 1000}{160.2}$
$= 1118 K$
So, at temperature above this, the reaction will become spontaneous.
Hence, (4) is correct answer.