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

Question

Quizplus · Accepted Answer

The change in pressure for Xe is 2.24 - 0.34 = 1.90 atm, so 1.90 atm of Xe is used. For F2, 2 * 1.90 atm = 3.80 atm is used, leaving 4.27 - 3.80 = 0.47 atm. The pressure of XeF4 formed is 1.90 atm. Kp = (P_XeF4) / (P_Xe * (P_F2)^2) = (1.90) / (0.34 * (0.47)^2) = 25.

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

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