To identify the correct equilibrium constant expression for the given reaction:
[tex]\[H_2(g) + F_2(g) \rightleftharpoons 2 HF(g)\][/tex]
we need to apply the law of mass action. This law states that the equilibrium constant expression [tex]\( K_{eq} \)[/tex] for a reaction is given by the ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their respective stoichiometric coefficients in the balanced chemical equation.
For the reaction:
[tex]\[H_2(g) + F_2(g) \rightleftharpoons 2 HF(g),\][/tex]
the stoichiometric coefficients are:
- 1 for [tex]\(H_2\)[/tex],
- 1 for [tex]\(F_2\)[/tex],
- 2 for [tex]\(HF\)[/tex].
Therefore, the equilibrium constant expression [tex]\( K_{eq} \)[/tex] would be:
[tex]\[
K_{eq} = \frac{\left[ HF \right]^2}{\left[ H_2 \right]\left[ F_2 \right]}
\][/tex]
Let's now match this expression with the given options:
1. [tex]\(K_{eq} = \frac{\left[ H_2 \right]\left[ F_2 \right]}{[ HF ]}\)[/tex]
2. [tex]\(K_{eq} = \frac{\left[ H_2 \right]\left[ F_2 \right]}{[ HF ]^2}\)[/tex]
3. [tex]\(K_{eq} = \frac{[ HF ]^2}{\left[ H_2 \right]\left[ F_2 \right]}\)[/tex]
4. [tex]\(K_{eq} = \frac{[2 HF ]}{\left[ H_2 \right]\left[ F_2 \right]}\)[/tex]
5. [tex]\(K_{eq} = \frac{[ HF ]}{\left[ H_2 \right]\left[ F_2 \right]}\)[/tex]
The correct option is:
[tex]\[K_{eq} = \frac{[ HF ]^2}{\left[ H_2 \right]\left[ F_2 \right]}\][/tex]
Thus, the answer is the third option:
[tex]\[
3
\][/tex]
