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Answer :
Final answer:
The reaction will proceed in the direction that reduces the discrepancy between Qp and Kp; since Kp=0.30, it suggests reactants are initially favored, and if Qp > Kp, the reaction will proceed to the left.
Explanation:
In this scenario with a given Kp = 0.30, assess whether the reaction will proceed to the left, right, or is at equilibrium by comparing values of Qp and Kp. If Qp > Kp, the reaction will proceed in the reverse direction (to the left) to reach equilibrium. If Qp < Kp, the reaction will proceed in the forward direction (to the right) until equilibrium is reached. If Qp = Kp, the system is already at equilibrium.
An understanding of thermodynamics can be applied here. When products and reactants are in their standard states, the sign of the Gibbs free energy change (ΔG°) indicates the favorability of the reaction. If ΔG° is negative, Kp > 1 and products are favored; if ΔG° is positive, Kp < 1 and reactants are favored; and if ΔG° is zero, Kp = 1 and the reaction is at equilibrium.
Since we know that Kp is less than 1 (Kp = 0.30), it implies that the reactants are favored and that the reaction will tend to proceed towards the left to reach equilibrium if Qp > Kp. In the case where a mixture is prepared with all products and reactants in their standard states, typically Qp will be greater than Kp because initial concentrations (or pressures) of the reactants will be higher, indicating that the reaction will shift in the direction that reduces the concentrations of the reactants (to the left).
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