Question

$Ca{{(HC{{O}_{3}})}_{2}}(s)$ decomposes as $Ca{{(HC{{O}_{3}})}_{2}}(s)\xrightarrow[{}]{{}}CaC{{O}_{3}}(s)+{{H}_{2}}O(g)$ $+C{{O}_{2}}(g)$ Total pressure at equilibrium is found to be 0.12 bar. Thus, ${{K}_{p}}$ is

• A. 0.24
• B. 0.06
• C. 0.0036
• D. 0.0144

Answer 1

Answer: (C)

0.0036

Answer 2

$Ca(HC0_3)_2(s) \rightarrow CaO(s) + C0_2(g) + H_20 (g)$
Total pressure is due to $C{{O}_{2}}(g)$ one ${{H}_{2}}O(g)$ molar ratio being equal.
Thus, ${{p}_{C{{O}_{2}}}}={{p}_{{{H}_{2}}O}}$
${{p}_{C{{O}_{2}}}}={{p}_{{{H}_{2}}O}}=0.06\,bar$
${{k}_{p}}={{p}_{C{{O}_{2}}}}\times {{p}_{{{H}_{2}}O}}$
$=0.06\times 0.06$ $=0.0036$