Final answer:
The equilibrium constant for the given reaction is 0.210. To find the equilibrium constant for the new equation, ½ H₂ (g) ½ F₂ (g) ⇌ HF (g), we use the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
Step-by-step explanation:
The equilibrium constant, K, is a measure of the relative concentrations of reactants and products at equilibrium for a particular chemical reaction. It is determined by the stoichiometry of the reaction and the temperature at which it occurs. In the given equation, 2 HF (g) ↔ H₂ (g) + F₂ (g), the equilibrium constant is 0.210.
To find the equilibrium constant for the equation ½ H₂ (g) ½ F₂ (g) ↔ HF (g), we can use the fact that the equilibrium constant is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. In this case, the stoichiometric coefficients are 1/2 for H₂ and F₂, and 1 for HF.
So, the equilibrium constant for the equation ½ H₂ (g) ½ F₂ (g) ↔ HF (g) can be calculated as: K = ([HF] / ([H₂]^1/2 [F₂]^1/2)). Given that the equilibrium constant for the reaction 2 HF (g) ↔ H₂ (g) + F₂ (g) is 0.210, we can substitute these values into the equation and solve for the equilibrium constant for the new equation.
K = (0.210 / ([H₂]^1/2 [F₂]^1/2))