The equilibrium for the following reaction at 700 K is, H₂(g) + I₂(g) ⇄ 2 HI(g), K₁ = 55.17: What is the value of the equilibrium constant for the following reaction ( = K₂ )? HI(g) ⇄ 1/2 H₂(g) + 1/2 I₂(g), K₂ = ?
k2 = sqrt (1/k1)
To find the equilibrium constant, K₂, for the reaction HI(g) ⇄ 1/2 H₂(g) + 1/2 I₂(g), we can use the relationship between equilibrium constants and chemical equations.
The given reaction, H₂(g) + I₂(g) ⇄ 2 HI(g), has an equilibrium constant, K₁ = 55.17.
Now let's write the equation for the reaction HI(g) ⇄ 1/2 H₂(g) + 1/2 I₂(g):
HI(g) ⇄ 1/2 H₂(g) + 1/2 I₂(g)
We can find the equilibrium constant, K₂, for this reaction by using the relationship between equilibrium constants and stoichiometric coefficients:
K₂ = (K₁)^(coefficients of products) / (K₁)^(coefficients of reactants)
In this case, we have 1 mol of HI on the left side and 1/2 mol of H₂ and 1/2 mol of I₂ on the right side. Therefore, we have:
K₂ = (K₁)^(1/2 + 1/2 - 1) = (K₁)^0 = 1
So, the value of the equilibrium constant, K₂, for the reaction HI(g) ⇄ 1/2 H₂(g) + 1/2 I₂(g), is 1.
To find the value of the equilibrium constant, K₂, for the given reaction HI(g) ⇄ 1/2 H₂(g) + 1/2 I₂(g), we can use the relationship between equilibrium constants for reactions that are the reverse of each other.
First, let's write the equation for the reverse reaction of the given equation: 1/2 H₂(g) + 1/2 I₂(g) ⇄ HI(g)
Now, we know that for a reaction and its reverse, the equilibrium constants are related as follows:
K₁ * K₂ = 1
Therefore, we can rearrange the equation to solve for K₂:
K₂ = 1/K₁
Plugging in the given value for K₁:
K₂ = 1/55.17
So, the value of the equilibrium constant K₂ for the given reaction HI(g) ⇄ 1/2 H₂(g) + 1/2 I₂(g) is approximately 0.0181.