Consider the reaction:
[tex]\[ 2 HF(g) \longleftrightarrow H_2(g) + F_2(g) \][/tex]

At equilibrium at [tex]\(600 K\)[/tex], the concentrations are as follows:
[tex]\[
\begin{array}{l}
[HF] = 5.82 \times 10^{-2} M \\
[H_2] = 8.4 \times 10^{-3} M \\
[F_2] = 8.4 \times 10^{-3} M
\end{array}
\][/tex]

What is the value of [tex]\( K_{\text{eq}} \)[/tex] for the reaction expressed in scientific notation?

A. [tex]\( 2.1 \times 10^{-2} \)[/tex]

B. [tex]\( 2.1 \times 10^2 \)[/tex]

C. [tex]\( 1.2 \times 10^3 \)[/tex]

D. [tex]\( 1.2 \times 10^{-3} \)[/tex]



Answer :

To determine the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for the reaction:

[tex]\[ 2 HF (g) \longleftrightarrow H_2 (g) + F_2 (g), \][/tex]

we use the given equilibrium concentrations. The formula for the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for this reaction can be written as:

[tex]\[ K_{\text{eq}} = \frac{[H_2][F_2]}{[HF]^2}. \][/tex]

Given the equilibrium concentrations:
[tex]\[ [HF] = 5.82 \times 10^{-2} \text{ M}, \][/tex]
[tex]\[ [H_2] = 8.4 \times 10^{-3} \text{ M}, \][/tex]
[tex]\[ [F_2] = 8.4 \times 10^{-3} \text{ M}, \][/tex]

we substitute these values into the formula for [tex]\( K_{\text{eq}} \)[/tex]:

[tex]\[ K_{\text{eq}} = \frac{(8.4 \times 10^{-3})(8.4 \times 10^{-3})}{(5.82 \times 10^{-2})^2}. \][/tex]

Now, let's calculate it step-by-step:

1. Calculate the numerator:
[tex]\[ (8.4 \times 10^{-3}) \times (8.4 \times 10^{-3}) = 70.56 \times 10^{-6} = 7.056 \times 10^{-5}. \][/tex]

2. Calculate the denominator:
[tex]\[ (5.82 \times 10^{-2})^2 = 33.8724 \times 10^{-4} = 3.38724 \times 10^{-3}. \][/tex]

3. Divide the numerator by the denominator:
[tex]\[ K_{\text{eq}} = \frac{7.056 \times 10^{-5}}{3.38724 \times 10^{-3}} \approx 2.083 \times 10^{-2}. \][/tex]

Upon rounding to two significant figures, we get:

[tex]\[ K_{\text{eq}} \approx 2.1 \times 10^{-2}. \][/tex]

Thus, the value of [tex]\( K_{\text{eq}} \)[/tex] for the reaction is [tex]\( 2.1 \times 10^{-2} \)[/tex]. Therefore, the correct answer is:

[tex]\[ \boxed{2.1 \times 10^{-2}} \][/tex]