Consider the following equilibrium:

O2(g) + 2F2(g)--> 2OF2(g); Kp = 2.3 x 10-15

Which of the following statements is true?

A) If the reaction mixture initially contains only OF2(g), then at equilibrium, the
reaction mixture will consist of essentially only O2(g) and F2(g).

B) For this equilibrium, Kc = Kp.

C) If the reaction mixture initially contains only OF2(g), then the total pressure at
equilibrium will be less than the total initial pressure.

D) If the reaction mixture initially contains only O2(g) and F2(g), then at equilibrium,
the reaction mixture will consist of essentially only OF2(g).

E) If the reaction mixture initially contains only O2(g) and F2(g), then the total
pressure at equilibrium will be greater than the total initial pressure.

A) If the reaction mixture initially contains only OF₂(g), then at equilibrium, the reaction mixture will consist of essentially only O₂(g) and F₂(g). TRUE. As the kp is 2,3x10⁻¹⁵ means per 1 of [O₂] [F₂]² you will have just 2,3x10⁻¹⁵ of [OF₂]²

B) For this equilibrium, Kc = Kp. FALSE. That is true just when moles of reactants are the same than moles of products. Here there are 3 moles of reactants vs 2 moles of products.

C) If the reaction mixture initially contains only OF₂(g), then the total pressure at equilibrium will be less than the total initial pressure. FALSE. Because per 2 moles of OF₂(g) you will produce 3 moles of gas increasing pressure.

D) If the reaction mixture initially contains only O₂(g) and F₂(g), then at equilibrium, the reaction mixture will consist of essentially only OF₂(g). FALSE. For the same reason of A), the mixture will contains essentially only O₂(g) and F₂(g)

E) If the reaction mixture initially contains only O₂(g) and F₂(g), then the total pressure at equilibrium will be greater than the total initial pressure. FALSE. If mixture initially contains only O₂(g) and F₂(g), 3 moles will of gas will react to produce 2 moles of gas doing pressure decreases.

I hope it helps!