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Question
Following data is given for the reaction: \[\ce{CaCO3 (S) -> CaO (s) + CO2 (g)}\]
\[\ce{∆_fH^Θ [CaO(s)] = - 635.1 kJ mo^{-1}}\]
\[\ce{∆_fH^Θ [CO2(g)] = - 393.5 kJ mol^{-1}}\]
\[\ce{∆_fH^Θ [CaCO3(s)] = - 1206.9 kJ mol^{-1}}\]
Predict the effect of temperature on the equilibrium constant of the above reaction.
Solution
\[\ce{∆_rH^Θ = ∆_fH^Θ [CaO(s)] + ∆_fH^Θ [CO2(g)] – ∆_fH^Θ [CaCO3(s)]}\]
∴ \[\ce{∆_rH^Θ = 178.3 kJ mol^{-1}}\]
The reaction is endothermic. Hence, according to Le-Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.
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