Answer:
The given reaction is:
2A(g) ⇌ B(g)
The equilibrium constant (Kp) is given as 7.17 × 10^−5 at 500 K.
We are given the initial pressure of A(g) as 3.10 atm.
Let's assume that the pressure of B(g) at equilibrium is x atm.
Using the equilibrium constant expression, we can write:
Kp = (P_B)^1 / (P_A)^2
Substituting the values, we get:
7.17 × 10^−5 = (x)^1 / (3.10)^2
Solving for x, we get:
x = 7.17 × 10^−5 × (3.10)^2
x = 6.81 × 10^−4 atm
Therefore, the pressure of B(g) at equilibrium is 6.81 × 10^−4 atm when a sample of A(g) at 3.10 atm is heated to 500 K.
Step-by-step explanation:
The given chemical reaction is in the form:
2A(g) ⇌ B(g)
This means that two molecules of A(g) can combine to form one molecule of B(g), and one molecule of B(g) can break down into two molecules of A(g). The double arrow indicates that the reaction can proceed in both directions, and the reaction is said to be in a state of equilibrium when the forward and reverse reactions occur at the same rate.
The equilibrium constant (Kp) is a measure of the relative concentrations of the reactants and products at equilibrium, and is given as 7.17 × 10^-5 at 500 K.
We are given the initial pressure of A(g) as 3.10 atm. Since we are asked to find the pressure of B(g) at equilibrium, we assume that the pressure of B(g) is x atm.
Using the equilibrium constant expression, we can write:
Kp = (P_B)^1 / (P_A)^2
where P_B is the pressure of B(g) at equilibrium, and P_A is the pressure of A(g) at equilibrium. Since we know the value of Kp and the initial pressure of A(g), we can solve for P_B.
Substituting the given values, we get:
7.17 × 10^-5 = (x)^1 / (3.10)^2
Simplifying this expression, we get:
x = 7.17 × 10^-5 × (3.10)^2
x = 6.81 × 10^-4 atm
Therefore, the pressure of B(g) at equilibrium is 6.81 × 10^-4 atm when a sample of A(g) at 3.10 atm is heated to 500 K.