Respuesta :
Answer:
A) is true
Explanation:
For the reaction:
O₂(g) + 2F₂(g) ⇄ 2OF₂(g); Kp = 2,3x10⁻¹⁵
kp is defined as:
Kp = 2,3x10⁻¹⁵ = [OF₂]²/[O₂] [F₂]²
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!