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

The Correct Option is C

The given chemical reaction is the decomposition of calcium bicarbonate:

\(Ca{{(HC{{O}_{3}})}_{2}}(s) \xrightarrow[{}]{{}} CaC{{O}_{3}}(s) + {{H}_{2}}O(g) + C{{O}_{2}}(g)\)

This reaction leads to the production of water vapor and carbon dioxide gas. The total pressure at equilibrium for these gases is given as 0.12 bar, which means:

\({{P}_{{{H_2}O}}} + {{P}_{{{CO_2}}}} = 0.12 \text{ bar}\)

Assuming ideal behavior, we have the same number of moles of each gas at equilibrium since they are produced from one mole of calcium bicarbonate:

• Let \(x\) be the partial pressure of each gas: \({{P}_{{{H_2}O}}} = x\) and \({{P}_{{{CO_2}}}} = x\).

Therefore, we can write:

\(x + x = 0.12\) bar

Simplifying, we find:

\(2x = 0.12\) bar

Solving for \(x\), we have:

\(x = 0.06\) bar

Now, we use the equilibrium constant expression for pressure, \({{K}_{p}}\), based on the partial pressures of the gaseous products:

\({{K}_{p}} = {{P}_{{{H_2}O}}} \cdot {{P}_{{{CO_2}}}}\)

Plugging in the values, we obtain:

\({{K}_{p}} = 0.06 \times 0.06 = 0.0036\)

Thus, the equilibrium constant, \({{K}_{p}}\), is 0.0036. This matches the correct option.