If Kc = 6.38 × 105 for the reaction
A(g) ⇀ ↽ 2B(g),
what is Kc for the reaction written as
2B(g) ⇀ ↽ A(g)?
How would I solve this?
A--> 2B Kc
2B --> A K'c = 1/Kc
To solve this problem, we can use the concept of the equilibrium constant expression. The equilibrium constant (Kc) is a ratio of the concentrations of products to the concentrations of reactants, with each concentration raised to the power of its coefficient in the balanced equation.
Given the equilibrium constant expression for the reaction A(g) ⇀ ↔ 2B(g) as Kc = 6.38 × 10^5, we can establish the following equilibrium constant expression:
Kc = [B]^2 / [A]
To find the equilibrium constant expression for the reaction 2B(g) ⇀ ↔ A(g), we need to reverse the reaction. By doing so, the concentration of A(g) will be in the numerator, and the concentration of 2B(g) will be in the denominator. However, since we have twice the amount of B in the reverse reaction, the equilibrium constant expression will be squared.
So, the equilibrium constant expression for 2B(g) ⇀ ↔ A(g) is:
Kc' = [A] / [B]^2
To find the value of Kc' for this reaction, we can use the value of Kc from the given reaction.
Kc' = 1 / Kc
Kc' = 1 / (6.38 × 10^5)
Therefore, Kc' for the reaction 2B(g) ⇀ ↔ A(g) is approximately 1.57 × 10^-6.
To determine the value of Kc for the reaction 2B(g) ⇀ ↽ A(g), you can use the relationship between the forward and reverse reaction constants.
First, recall the formula relating the equilibrium constants of the forward and reverse reactions:
Kc = (Products)^(coefficients) / (Reactants)^(coefficients)
In the given reaction, Kc = 6.38 × 10^5 for the reaction A(g) ⇀ ↽ 2B(g). This means that:
Kc = ([2B]^2) / [A]
Now, we want to find Kc for the reverse reaction 2B(g) ⇀ ↽ A(g), which involves swapping the products and reactants. Thus, we can write:
Kc(reverse) = ([A]) / ([2B]^2)
Since we know the value of Kc for the forward reaction, we can substitute it into the equation:
6.38 × 10^5 = ([2B]^2) / [A]
Next, isolate Kc(reverse) to solve for it. Multiply both sides of the equation by [A] and take the square root:
Kc(reverse) = √(6.38 × 10^5 × [A])
Therefore, the value of Kc for the reaction 2B(g) ⇀ ↽ A(g) is √(6.38 × 10^5 × [A]).