Question

For the reaction A (g) → 3 B (g), Kp = 80100 at 298 K. When ∆G = -14.2 kJ/mol, what is the partial pressure of A when the partial pressure of B is 2.00 atm for this reaction at 298 K.

Answer 1

The partial pressure of A (g) at equilibrium, given the partial pressure of B is 2.00 atm and Kp = 80100, is found to be 1.00 × 10^-4 atm by using the equilibrium constant equation and the reaction quotient.

Step-by-step explanation:

To find the partial pressure of gas A for the given reaction A (g) → 3 B (g) with a known Kp and ΔG, we use the relation between Gibb's free energy (ΔG) and the equilibrium constant (Kp), which is ΔG = -RTlnKp. From this equation and given that ΔG = -14.2 kJ/mol and Kp = 80100, we calculate the partial pressure of A when the partial pressure of B is 2.00 atm. First, we need to consider the relationship between the reaction quotient Q and Kp when the system is at equilibrium (Q = Kp).

The reaction quotient Q for the reaction A (g) → 3 B (g) can be expressed as:

Q = (P_B)^3 / (P_A),

where P_A and P_B are the partial pressures of A and B respectively. As the system is at equilibrium, Q equals Kp, and therefore we can write:

80100 = (2.00)^3 / (P_A).

By solving for P_A, we can find its value:

P_A = (2.00)^3 / 80100.

P_A = 8.00 atm / 80100,

P_A = 1.00 × 10^-4 atm.

The partial pressure of A at equilibrium when B has a partial pressure of 2.00 atm is therefore 1.00 × 10^-4 atm.

Answer 2

To find the partial pressure of A when the partial pressure of B is 2.00 atm, divide the partial pressure of B cubed by the equilibrium constant, Kp. Plugging in the values, the partial pressure of A is 0.099 atm.

Step-by-step explanation:

To find the partial pressure of A when the partial pressure of B is 2.00 atm, we can use the expression for the equilibrium constant, Kp, which relates the partial pressures of the reactants and products. For the reaction A (g) → 3 B (g), the equilibrium constant expression is Kp = (Pb)^3 / Pa. Rearranging the equation to solve for Pa, we get Pa = (Pb)^3 / Kp. Plugging in the given values, Pa = (2.00 atm)^3 / 80100 = 0.099 atm.