Question

question 14 (1 point) saved
hydrogen is the lightest of elements with atomic number one. when hydrogen gas reacts with bromine gas, they form hydrogen bromide gas, i.e., $\ce{h_{2}(g) + br_{2}(g) \leftrightarrow 2hbr(g)}$
at equilibrium, concentration values of the compounds are:

compound	$\ce{h_{2}(g)}$	$\ce{br_{2}(g)}$	$\ce{hbr(g)}$

the value of equilibrium constant $k_{\text{eq}}$, for the above reaction, is
$\circ$ 2.6
$\circ$ 3.1
$\bullet$ 4.3
$\circ$ 5.2
$\circ$ 8.7

Explanation:

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

For reaction \( aA + bB
ightleftharpoons cC + dD \), \( K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b} \). Here, \( \text{H}_2(\text{g}) + \text{Br}_2(\text{g})
ightleftharpoons 2\text{HBr}(\text{g}) \), so \( K_{eq}=\frac{[\text{HBr}]^2}{[\text{H}_2][\text{Br}_2]} \).

Step2: Substitute values

Given \( [\text{H}_2] = 0.024 \, \text{mol/L} \), \( [\text{Br}_2] = 0.024 \, \text{mol/L} \), \( [\text{HBr}] = 0.05 \, \text{mol/L} \).
Substitute into formula: \( K_{eq}=\frac{(0.05)^2}{(0.024)(0.024)} \).
Calculate numerator: \( (0.05)^2 = 0.0025 \).
Calculate denominator: \( 0.024 \times 0.024 = 0.000576 \).
Then \( K_{eq}=\frac{0.0025}{0.000576} \approx 4.3 \).

Answer:

4.3 (corresponding to the option with 4.3, e.g., the option labeled 4.3 in the multiple - choice list)