Question

name(s):

1. nitric oxide is a highly reactive, colourless and poisonous gas. it reacts with bromine as shown in the equation below. at equilibrium, concentration values of the compounds are also given below. what is the value of the equilibrium constant $k_{eq}$, for the reaction below? show all your work.

$2 no(g) + br_2(g) \leftrightarrow 2 nobr(g)$
$no = 0.0352$ mol/l
$br_2 = 0.0118$ mol/l
$nobr = 0.0518$ mol/l

Explanation:

Step1: Recall Equilibrium Constant Formula

For reaction \( aA + bB
ightleftharpoons cC + dD \), \( K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b} \). For \( 2NO(g)+Br_2(g)
ightleftharpoons 2NOBr(g) \), formula is \( K_{eq}=\frac{[NOBr]^2}{[NO]^2[Br_2]} \).

Step2: Substitute Concentrations

Given \([NO] = 0.0352\space mol/L\), \([Br_2]=0.0118\space mol/L\), \([NOBr]=0.0518\space mol/L\). Substitute into formula:
\( K_{eq}=\frac{(0.0518)^2}{(0.0352)^2\times0.0118} \)

Step3: Calculate Numerator and Denominator

Numerator: \((0.0518)^2 = 0.0518\times0.0518 = 0.00268324\)
Denominator: \((0.0352)^2\times0.0118 = 0.00123904\times0.0118 \approx 0.0000146207\)

Step4: Divide Numerator by Denominator

\( K_{eq}=\frac{0.00268324}{0.0000146207} \approx 183.5 \)

Answer:

The value of \( K_{eq} \) is approximately \( 184 \) (or \( 1.84\times10^2 \)) (depending on rounding during calculation; precise value from steps is ~183.5).