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Byju's Answer

Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

In conversion...

Question

In conversion of limestone to lime, $C a C O_{3 (s)} \to C a O_{(s)} + C O_{2 (g)}$ ; the values of $Δ H^{0}$ and $Δ S^{0}$ are + 179.1 kJ/mol and 160.2J/K respectively, at 298K and 1 bar. Above what temperature, conversion of limestone to lime will be spontaneous (assuming that $Δ H^{0}$ and $Δ S^{0}$ do not change with temperature)?

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 D $1118 K$
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$

For spontaneous process, $Δ G < O$

$∴ Δ H - T Δ S < O$

$∴ \frac{Δ H^{\circ}}{Δ S^{\circ}} < T$

$Δ H^{\circ} = 179.1$ $K J m o l$ $^{- 1}$ and $Δ S^{\circ} = 160.2 J / K$

$∴ T > \frac{179.1 \times 1000}{160.2}$ $K m o l$ $^{- 1}$

$≃$ $1117.9$

$≃$ $1118 k$

So, the correct option is $A$

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$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
The values of $△ H^{\circ}$ and $△ S^{\circ}$ are 179.1 kJ $m o l^{- 1}$ and 160.2 J/K respectively at 298 K and 1 bar. Assuming $△ H^{\circ}$ and $△ S^{\circ}$ remains constant with temperature, at what temperature will the conversion of limestone to lime be spontaneous?

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In conversion of limestone to lime, CaCO3(s)→CaO(s)+CO2(g); the values of ΔH0 and ΔS0 are + 179.1 kJ/mol and 160.2J/K respectively, at 298K and 1 bar. Above what temperature, conversion of limestone to lime will be spontaneous (assuming that ΔH0and ΔS0 do not change with temperature)?

The correct option is D 1118KΔG∘=ΔH∘−TΔS∘For spontaneous process, ΔG<O∴ΔH−TΔS<O∴ΔH∘ΔS∘<TΔH∘=179.1 KJmol −1 and ΔS∘=160.2J/K∴T>179.1×1000160.2 Kmol −1≃ 1117.9≃ 1118kSo, the correct option is A