For the reaction at 1240 K and 1 atm CaCO3s- CaOs-CO2g - H-176 kJ-mol- - E will be-

The correct option is A 165.6 kJ Δ E = Δ H Δ n RT = 176000 J 1 mol × 8.314 J/mol.K × 1240 K = 165691 J = 165.6 kJ ...

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Derivation of Kp and Kc

For the react...

Question

For the reaction (at $1240$ K and $1$ atm)
$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
$Δ H = 176 k J / m o l; Δ E$ will be:

160

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165.6

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186.4

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180

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For the reaction $C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$ (at $1240$ K and $1$ atm.) $Δ H = 176 k J / m o l$ , then $Δ E$ will be:

Solids $C a C O_{3}$ and CaO and gaseous $C O_{2}$ are placed in a vessel and allowed to reach equilibrium

$C a O (s) + C O_{2} (g) ⇌ C a C O_{3} (s) Δ H = - 180 k J m o l^{- 1}$

The quantity of CaO in the vessel could be increased by:

C O_{2}

- 94.0

- 152

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

C a C O_{3} (s)

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

977^{o} C

Δ H = 174 k J / m o l

Δ E

I

E_{a} = 15

k J

{m o l}^{- 1};

Δ H = - 70

k J

{m o l}^{- 1}

I I

E_{a} = 30

k J

{m o l}^{- 1};

Δ H = - 15

k J

{m o l}^{- 1}

I I I

E_{a} = 60

k J

{m o l}^{- 1};

Δ H = + 20

k J

{m o l}^{- 1}

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