Questions: Write the pressure equilibrium constant expression for this reaction. 2 H2(g)+O2(g) → 2 H2O(l)

Solution Steps

The given reaction is a chemical reaction involving gases and a liquid. The reactants are gaseous hydrogen (\(\mathrm{H}_2\)) and oxygen (\(\mathrm{O}_2\)), and the product is liquid water (\(\mathrm{H}_2\mathrm{O}\)).

The pressure equilibrium constant, \(K_p\), is used for reactions involving gases. It is expressed in terms of the partial pressures of the gaseous reactants and products. For a general reaction:

\[ aA(g) + bB(g) \rightarrow cC(g) + dD(g) \]

The expression for \(K_p\) is:

\[ K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \]

where \(P_X\) represents the partial pressure of species \(X\).

For the given reaction:

\[ 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \]

Since the product \(\mathrm{H}_2\mathrm{O}\) is a liquid, it does not appear in the \(K_p\) expression. Therefore, the pressure equilibrium constant expression is based only on the reactants:

\[ K_p = \frac{1}{(P_{\mathrm{H}_2})^2 (P_{\mathrm{O}_2})^1} \]

Final Answer