Question

In the reaction H₂ (g) + Cl₂ (g) → 2HCl (g), Kg = 50.0. If 1.00 mol H₂ (g) and 1.00 mol Cl, (g) are placed in a 0.50 L container and allowed to reach equilibrium, what is the concentration of HCI at equilibrium?

the answer to this question is (HCL)=2x2(1.56)=3.12mol but i need the explanation ​

Answer 1

In the given reaction, hydrogen gas (H₂) and chlorine gas (Cl₂) combine to form hydrogen chloride gas (HCl). The equilibrium constant expression, which helps us determine the concentration of HCl at equilibrium, is written as Kg = [HCl]² / ([H₂][Cl₂]).

In this case, the equilibrium constant (Kg) is given as 50.0. Initially, we have 1.00 mole of H₂ gas and 1.00 mole of Cl₂ gas in a 0.50-liter container. To find the concentration of HCl at equilibrium, we can set up the following equation using the equilibrium constant expression:

50.0 = [HCl]² / ([H₂][Cl₂])

We know that the initial concentrations of H₂ and Cl₂ are both 1.00 mole divided by the volume of the container, which is 0.50 liters, giving us a concentration of 2.00 M.

Substituting these values into the equation, we have:

50.0 = [HCl]² / (2.00 - x)(2.00 - x)

To solve this equation, we can rearrange it as a quadratic equation:

[HCl]² = 50.0 * (2.00 - x)(2.00 - x)

Simplifying further:

[HCl]² = 100.0 * (2.00 - x)(2.00 - x)

To find the value of x, we solve this quadratic equation. The solutions to the equation are x = -1.56 and x = 1.56. However, since a negative value for x does not make physical sense in this context, we can conclude that x = 1.56.

Thus, the concentration of HCl at equilibrium, [HCl], is equal to 2x, which is 2 times 1.56, resulting in [HCl] = 3.12 M.