the equilibrium system shown below was analyzed and the concentrations of HI(g, and I2(g) were found, in mol/L, to be 4.4, 3.2, 1.5 respectively. the equilibrium constant must be which f the following?
H2(g) + I2(g)<> 2HI(g) + 65KJ
To determine the equilibrium constant (Kc) for the given reaction, we need to use the concentrations of the reactants and products at equilibrium. We are given the concentrations of HI(g), H2(g), and I2(g) as 4.4 mol/L, 3.2 mol/L, and 1.5 mol/L, respectively.
The balanced equation for the reaction is:
H2(g) + I2(g) ⇌ 2HI(g)
The equilibrium constant expression (Kc) for this reaction is:
Kc = [HI]^2 / ([H2] * [I2])
Substituting the given concentrations into the equilibrium constant expression:
Kc = (4.4^2) / (3.2 * 1.5)
Calculating this expression:
Kc = 19.36 / 4.8
Simplifying further:
Kc ≈ 4.033
Therefore, the equilibrium constant (Kc) for the reaction is approximately 4.033.
To find the equilibrium constant in this case, you need to use the concentrations of the reactants and products at equilibrium. The equilibrium constant (K) can be calculated using the following formula:
K = [C]^c * [D]^d / [A]^a * [B]^b
In this equation, A and B represent the molar concentrations of the reactants (H2 and I2), while C and D represent the molar concentrations of the products (HI). The exponents a, b, c, and d represent the coefficients of the respective species in the balanced chemical equation.
Given that the concentrations of HI, H2, and I2 at equilibrium are 4.4, 3.2, and 1.5 mol/L, respectively, we can substitute these values into the equation to calculate the equilibrium constant.
K = (4.4)^2 / (3.2)^1 * (1.5)^1
Simplifying the calculation:
K ≈ 19.36 / 4.8 ≈ 4.033
So, the equilibrium constant for the given reaction is approximately 4.033.
Please note that the value of the equilibrium constant is dimensionless and does not have any units.
............H2 + I2 ==> 2HI
Equil......4.4..3.2....1.5
Substitute into Kc expression and solve for kc.