Question

consider the reaction.
2hf(g)⇌h₂(g) + f₂(g)
at equilibrium at 600 k, the concentrations are as follows.
hf = 5.82 × 10⁻² m
h₂ = 8.4 × 10⁻³ m
f₂ = 8.4 × 10⁻³ m
what is the value of ( k_{\text{eq}} ) for the reaction expressed in scientific notation?
○ ( 2.1 \times 10^{-2} )
○ ( 2.1 \times 10^{2} )
○ ( 1.2 \times 10^{3} )
○ ( 1.2 \times 10^{-3} )

Explanation:

Step1: Recall the formula for \( K_{eq} \)

For the reaction \( aA + bB
ightleftharpoons cC + dD \), the equilibrium constant \( K_{eq} \) is given by \( K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b} \). For the reaction \( 2HF(g)
ightleftharpoons H_2(g)+F_2(g) \), the formula becomes \( K_{eq}=\frac{[H_2][F_2]}{[HF]^2} \).

Step2: Substitute the given concentrations

We know \( [HF] = 5.82\times 10^{-2}\ M \), \( [H_2]=8.4\times 10^{-3}\ M \), and \( [F_2]=8.4\times 10^{-3}\ M \). Substituting these values into the formula:
\[

$$\begin{align*} K_{eq}&=\frac{(8.4\times 10^{-3})(8.4\times 10^{-3})}{(5.82\times 10^{-2})^2}\\ &=\frac{(8.4\times 8.4)\times 10^{-6}}{(5.82\times 5.82)\times 10^{-4}}\\ &=\frac{70.56\times 10^{-6}}{33.8724\times 10^{-4}}\\ &=\frac{70.56}{33.8724}\times\frac{10^{-6}}{10^{-4}}\\ &\approx 2.083\times 10^{-2}\\ &\approx 2.1\times 10^{-2} \end{align*}$$

\]

Answer:

A. \( 2.1\times 10^{-2} \)