A) \[{{K}_{P}}={{P}_{(CaC{{O}_{3}})}}\]
B) \[{{K}_{P}}={{P}_{(C{{O}_{2}})}}\]
C) \[{{K}_{P}}=\frac{1}{{{P}_{(CaC{{O}_{3}})}}}\]
D) \[{{K}_{P}}=\frac{1}{{{P}_{(C{{O}_{2}})}}}\]
Correct Answer: B
Solution :
\[CaC{{O}_{3}}(s)CaO(s)+C{{O}_{2}}(g)\] Applying law of mass reaction, \[{{K}_{P}}=\frac{{{P}_{CaO(s)}}.\,{{P}_{C{{O}_{2(g)}}}}}{{{P}_{Ca{{O}_{3(s)}}}}}\] In the above reaction partial pressures of \[CaC{{O}_{3}}\] and \[CaO\] are. their vapour pressures. Every solid has a vapour pressure, although it is too small to be measured. The V.P. of solid at constant temperature has a constant value irrespective of the solid present. Thus, \[{{K}_{p}}={{P}_{C{{O}_{2}}}}\]
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