question_answer 1)

  Following data is given for the reaction : \[CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)\] \[{{\Delta }_{f}}{{H}^{O-}}[CaO(s)]=-635.1kJ\,mo{{l}^{-1}}\] \[{{\Delta }_{f}}{{H}^{O-}}[C{{O}_{2}}(s)]=-393.5kJ\,mo{{l}^{-1}}\] \[{{\Delta }_{f}}{{H}^{O-}}[CaC{{O}_{3}}(s)]=-1206.9kJ\,mo{{l}^{-1}}\] Predict the effect of temperature on the equilibrium constant of the above reaction.

Answer:

  In the reaction: \[CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)\] \[{{\Delta }_{r}}{{H}^{o}}=[{{\Delta }_{f}}H_{CaO}^{o}+{{\Delta }_{f}}H_{C{{O}_{2}}}^{o}]-[{{\Delta }_{f}}H_{CaC{{O}_{3}}}^{o}]\]                 \[=-[-635.1-393.5]-[-1206.9]\] \[=+178.3kJmo{{l}^{-1}}\] The given reaction is endothermic, hence equilibrium constant will increase with temperature. We know, \[lo{{g}_{10}}\left( \frac{{{K}_{2}}}{{{K}_{1}}} \right)=\frac{\Delta H}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] Here, \[{{K}_{1}}\] = Equilibrium constant at temperature \[{{T}_{1}}\] \[{{K}_{2}}\] = Equilibrium constant at temperature \[{{T}_{2}}\] \[\Delta H\] = Heat of reaction