Ncert Exercises & Solutions

Following data is given for the reaction ${CaCO}_{3} (s) \to CaO (s) + {CO}_{2} (g)$

$∆_{f} H^{⊝} [CaO (s)] = - 635.1 kJ {mol}^{- 1}$
$∆_{f} H^{⊝} [{CO}_{2} (g)] = - 393.5 kJ {mol}^{- 1}$
$∆_{f} H^{⊝} [{CaCO}_{3} (s)] = - 1206.9 kJ {mol}^{- 1}$

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

Given that,

∆_{f} H^{⊝} [CaO (s)] = - 635.1 kJ {mol}^{- 1}

∆_{f} H^{⊝} [{CO}_{2} (g)] = - 393.5 kJ {mol}^{- 1}

∆_{f} H^{⊝} [{CaCO}_{3} (s)] = - 1206.9 kJ {mol}^{- 1}

In the reaction,

{CaCO}_{3} (s) \to CaO (s) + {CO}_{2} (g)

∆_{f} H^{⊝} = ∆_{f} H^{⊝} [CaO (s)] + ∆_{f} H^{⊝} [{CO}_{2} (g)] - ∆_{f} H^{⊝} [{CaCO}_{3} (s)]

∴ ∆_{f} H^{⊝} = - 635.1 + (393.5) - (- 1206.9) = 178.3 kJ {mol}^{- 1}

Because

∆ H

value is positive, so the reaction is endothermic. Hence, according to Le-Chateleir's principle, reaction will proceed in forward direction on increasing temperature. Thus, the value of equilibrium constant for the reaction increases.