Following data is given for the reaction: CaCO 3 (s) → CaO 3 (s) + CO 2 (g) Δ f H ⊖ [CaO (s)] = -635.1 kJ mol =-1 Δ f H ⊖ [CO 2 (g)] = -393.5 kJ mol -1 Δ f H ⊖ [CaCO 3 (s)] = -1206.9 kJ mol -1 Predict the effect of temperature on the Equilibrium constant of the above reaction.

Question

Philoid · Accepted Answer

Given; CaCO₃ (s) → CaO₃ (s) + CO₂ (g)

Δ_fH^⊖ [CaO (s)] = -635.1 kJ mol^-1

Δ_fH^⊖ [CO₂ (g)] = -393.5 kJ mol^-1

Δ_fH^⊖ [CaCO₃ (s)] = -1206.9 kJ mol^-1

We know that,

Where is the standard enthalpy change for the formation of one mole of compound from its elements in their most stable states of aggregation.

= Δ_fH^⊖ [CaO(s)] + Δ_fH^⊖ [CO₂(g)]- Δ_fH^⊖ [CaCO₃(s)]

= -635.1 -393.5+1206.9

=178.3 kJ mol^-1

=positive

i.e. the reaction is exothermic.

Hence according to Le Chatelier’s Principle on increasing the temperature the reaction will shift to forward direction.

Following data is given for the reaction: CaCO3 (s) → CaO3 (s) + CO2 (g)ΔfH⊖ [CaO (s)] = -635.1 kJ mol=-1ΔfH⊖ [CO2 (g)] = -393.5 kJ mol-1ΔfH⊖ [CaCO3 (s)] = -1206.9 kJ mol-1Predict the effect of temperature on the Equilibrium constant of the above reaction.

Following data is given for the reaction: CaCO₃ (s) → CaO₃ (s) + CO₂ (g)
Δ_fH^⊖ [CaO (s)] = -635.1 kJ mol^=-1

Δ_fH^⊖ [CO₂ (g)] = -393.5 kJ mol^-1

Δ_fH^⊖ [CaCO₃ (s)] = -1206.9 kJ mol^-1

Predict the effect of temperature on the Equilibrium constant of the above reaction.