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

The correct option is B 165.6 kJFor given reaction, CaCO_3(s) → CaO(s)+CO_2(g) Δ n_g=1 Δ H=176 kJ/mol Δ E=Δ H Δ nRT Δ E=176000 (1 ...

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Introduction to Le Chatelier's Principle

For the react...

Question

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:

160

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165.6

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186.4

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180

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1240

1

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

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 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

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)

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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