Questions: For the reaction A(g) ⇌ 2 B(g), a reaction vessel initially contains only A at a pressure of PA=1.32 atm. At equilibrium, PA=0.25 atm. Calculate the value of Kp. (Assume no changes in volume or temperature.)

Solution Steps

Initially, the pressure of A is \( P_{\mathrm{A, initial}} = 1.32 \, \text{atm} \). At equilibrium, the pressure of A is \( P_{\mathrm{A, equilibrium}} = 0.25 \, \text{atm} \). The change in pressure for A, denoted as \( \Delta P_{\mathrm{A}} \), is:

\[ \Delta P_{\mathrm{A}} = P_{\mathrm{A, initial}} - P_{\mathrm{A, equilibrium}} = 1.32 - 0.25 = 1.07 \, \text{atm} \]

The reaction is \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\). For every mole of A that reacts, 2 moles of B are produced. Therefore, the change in pressure for B, \( \Delta P_{\mathrm{B}} \), is:

\[ \Delta P_{\mathrm{B}} = 2 \times \Delta P_{\mathrm{A}} = 2 \times 1.07 = 2.14 \, \text{atm} \]

Since the reaction vessel initially contains only A, the initial pressure of B is 0 atm. At equilibrium, the pressure of B, \( P_{\mathrm{B, equilibrium}} \), is equal to the change in pressure for B:

\[ P_{\mathrm{B, equilibrium}} = \Delta P_{\mathrm{B}} = 2.14 \, \text{atm} \]

The equilibrium constant \( K_{\mathrm{p}} \) for the reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\) is given by:

\[ K_{\mathrm{p}} = \frac{(P_{\mathrm{B, equilibrium}})^2}{P_{\mathrm{A, equilibrium}}} \]

Substituting the equilibrium pressures:

\[ K_{\mathrm{p}} = \frac{(2.14)^2}{0.25} = \frac{4.5796}{0.25} = 18.3184 \]

Final Answer