Consider the following reaction: Xe(g) + 2F₂(g) ⇌ XeF₄(g)
A reaction mixture initially contains 2.24 bar Xe and 4.27 bar F₂. If the equilibrium pressure of Xe is 0.34 bar, find the equilibrium constant (K_p) for the reaction.

Question

Quizplus · Accepted Answer

The equilibrium constant (K_p) for the reaction is given by the expression:K_p = (P_XeF4) / (P_Xe * (P_F2)²)At equilibrium, the partial pressures of the reactants and products are:P_Xe = 0.34 barP_F2 = 4.27 bar - 2 * 0.34 bar = 3.59 barP_XeF4 = 2.24 bar - 0.34 bar = 1.90 barSubstituting these values into the expression for K_p, we get:K_p = (1.90 bar) / (0.34 bar * (3.59 bar)²) = 25 bar^-2

Consider the Following Reaction: Xe(g) + 2F2(g) ⇌ XeF4(g)

Consider the Following Reaction: Xe(g) + 2F₂(g) ⇌ XeF₄(g)