Question

2HF(g)<-----> H2(g)+F2(g)

Answer 1

Answer : The value of $K_{eq}$ for the following reaction will be, $2.1\times 10^{-2}$

Explanation :

$K_{eq}$ is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The given balanced equilibrium reaction is,

$2HF(g)\rightleftharpoons H_2(g)+F_2(g)$

As we know that the concentrations of pure solids are constant that means they do not change. Thus, they are not included in the equilibrium expression.

The expression for equilibrium constant for this reaction will be,

$K_{eq}=\frac{[H_2][F_2]}{[HF]^2}$

Now put all the given values in this expression, we get :

$K_{eq}=\frac{(8.4\times 10^{-3})\times (5.4\times 10^{-3})}{(5.82\times 10^{-2})^2}$

$K_{eq}=0.021=2.1\times 10^{-2}$

Therefore, the value of $K_{eq}$ for the following reaction will be, $2.1\times 10^{-2}$