In the conversion of lime stone to lime- CaCO 3 s - CaO s - CaO 2 g the values of - H - and - S - are -179-1 kJ mol -1 ad160-2 JK -1 mol -1 respectively at 298 K and 1 bar- Assuming- - H - and - S - do not change with temperature- temperature above which conversion of lime stone to lime will be spontaneous isA- 1118 KB- 1008 KC- 1200 KD- 845 K

The correct option is A 1118 K Δ G^∘ = Δ H^∘ T Δ S^∘ At equilibrium Δ G^∘ = 0 Δ H^∘ = TΔ S^∘ T = 179.1 × 1000/160.2 = 1118K So, at the temperature above 118K ...

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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

In the conver...

Question

In the conversion of lime stone to lime, $C a C O_{3} (s) \to C a O (s) + C a O_{2} (g)$ the values of $Δ H^{\circ}$ and $Δ S^{\circ}$ are $+ 179.1 k J m o l^{- 1}$ ad $160.2 J K^{- 1} m o l^{- 1}$ respectively at $298 K$ and $1 b a r$ . Assuming, $Δ H^{\circ}$ and $Δ S^{\circ}$ do not change with temperature; temperature above which conversion of lime stone 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

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

At equilibrium $Δ G^{\circ} = 0$

$Δ H^{\circ} = T Δ S^{\circ}$

$T = \frac{179.1 \times 1000}{160.2} = 1118 K$

So, at the temperature above 118K, the reaction will becomes spontaneous.

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$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$ the values of $Δ H^{0}$ and $Δ S^{0}$ are $+ 179.1 k J m o l^{- 1}$ and $160.2 J / K$ respectively at 298 K and 1 bar. Assuming that $Δ H^{0}$ and $Δ S^{0}$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

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C a C O_{3 (s)} \to C a O_{(s)} + C O_{2 (g)}

; the values of

Δ H^{0}

and

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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

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do not change with temperature)?

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 $△ H^{\circ}$ and $△ S^{\circ}$ are 179.1 kJ $m o l^{- 1}$ and 160.2 J $K^{- 1}$ respectively at 298 K and 1 bar. Assuming $△ H^{\circ}$ and $△ S^{\circ}$ remains constant with temperature, at which temperature conversion of limestone to lime will be spontaneous:

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The correct option is A 1118 K ΔG∘ = ΔH∘ − TΔS∘ At equilibrium ΔG∘ = 0 ΔH∘ = TΔS∘ T = 179.1 × 1000160.2 = 1118K So, at the temperature above 118K, the reaction will becomes spontaneous.

The correct option is A
1118 K

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

At equilibrium $Δ G^{\circ} = 0$

$Δ H^{\circ} = T Δ S^{\circ}$

$T = \frac{179.1 \times 1000}{160.2} = 1118 K$

So, at the temperature above 118K, the reaction will becomes spontaneous.