Calculate the standard enthalpy for the given reaction- CaCO3 - CaO-CO2 Hf CaCO3-1206-9 Hf CaO-635-1 kJmol Hf CO2-393-5

CaCO_3 → CaO+CO_2 H_f CaCO_3=1206.9 H_f CaO= 635.1 kJmol H_f CO_2= 393.5 H_reaction= ∑ H_product ^f ∑ H_reactants^f H_reaction= ...

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Question

Calculate the standard enthalpy for the given reaction.
$C a C O_{3} \to C a O + C O_{2}$
$△ H_{f} C a C O_{3} = 1206.9$
$△ H_{f} C a O = - 635.1 k J m o l$
$△ H_{f} C O_{2} = - 393.5$

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

Δ_{f} H^{o} [C a O (s)] = - 635.1 {kJ mol}^{- 1}

Δ_{f} H^{o} [C O_{2} (g)] = - 393.5 {kJ mol}^{- 1}

Δ_{f} H^{o} [C a C O_{3} (s)] = - 1206.9 {kJ mol}^{- 1}

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} ⇌ C a O (s) + C O_{2} (g)

Δ_{f} H^{⊖} [C a O (s)] = - 635.1 k J m o l^{- 1}

Δ_{f} H^{⊖} [C O_{2} (g)] = - 393.5 k J m o l^{- 1}

Δ_{f} H^{⊖} [C a C O_{3} (s)] = - 1206.9 k J m o l^{- 1}

The equilibrium constant for the reaction,

CaCO₃(s) ⇌ CaO(s) + CO₂(g), is

K_{C}

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

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