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
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