In the conversion of limestone to lime- CaCO3s- CaOs-CO2gthe values of -Ho and -So are -179-1kJ mol -1 and 160-2kJ mol -1 respectively at 298K and 1 bar- Assuming that- -Ho and -So do not change with temperature- temperature above which conversion of limestone to lime will be spontaneous is-

The correct option is A 1118K CaCO _ 3 ( s ) → CaO( s ) + CO _ 2 ( g ) Δ H ^ 0 =+179.1KJ/mol Δ S ^ 0 =160.2KJ/mol Δ G ^ 0 =Δ H ^ 0 TΔ ...

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Byju's Answer

Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

In the conver...

Question

In the conversion of limestone to lime,
$C a {C O}_{3} (s) \to C a O (s) + {C O}_{2} (g)$
the values of $Δ H^{o}$ and $Δ S^{o}$ are $+ 179.1 k J {m o l}^{- 1}$ and $160.2 k J {m o l}^{- 1}$ respectively at $298 K$ and $1$ bar. Assuming that, $Δ H^{o}$ and $Δ S^{o}$ do not change with temperature; temperature above which conversion of limestone to lime will be spontaneous is:

1118 K

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1008 K

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1200 K

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845 K

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Solution

The correct option is A $1118 K$
${C a C O}_{3} (s) \to C a O (s) + {C O}_{2} (g)$
${Δ H}^{0} = + 179.1 K J / m o l$
${Δ S}^{0} = 160.2 K J / m o l$
${Δ G}^{0} = {Δ H}^{0} - T Δ S^{0}$
For spontaneous process, ${Δ G}^{0} < 0$
$∴ {Δ H}^{0} - T {Δ S}^{0} < 0$
$∴ 10^{3} \times 179.1 < 160.2 \times T$
$∴ T > \frac{179.1 \times 10^{3}}{160.2}$
$∴ T > 1118 K$
Therefore, above $1118 K$ , the reaction will be spontaneous.

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The correct option is A 1118KCaCO3(s)→CaO(s)+CO2(g)ΔH0=+179.1KJ/molΔS0=160.2KJ/molΔG0=ΔH0−TΔS0For spontaneous process, ΔG0<0∴ΔH0−TΔS0<0∴103×179.1<160.2×T∴T>179.1×103160.2∴T>1118KTherefore, above 1118K, the reaction will be spontaneous.