Let's go through the steps to find the equilibrium constant, [tex]\( K_{eq} \)[/tex], for the reaction [tex]\( N_2O_4 \rightleftharpoons 2 NO_2 \)[/tex].
1. Write the Expression for the Equilibrium Constant:
[tex]\[
K_{eq} = \frac{[NO_2]^2}{[N_2O_4]}
\][/tex]
2. Identify the Equilibrium Concentrations:
According to the given table:
[tex]\[
\text{Equilibrium} \left[ N_2O_4 \right] = 0.0429 \, \text{M}
\][/tex]
[tex]\[
\text{Equilibrium} \left[ NO_2 \right] = 0.0141 \, \text{M}
\][/tex]
3. Substitute These Values into the Expression:
[tex]\[
K_{eq} = \frac{(0.0141)^2}{0.0429}
\][/tex]
4. Calculate the Value:
[tex]\[
(0.0141)^2 = 0.00019881
\][/tex]
[tex]\[
K_{eq} = \frac{0.00019881}{0.0429} \approx 0.00463454
\][/tex]
5. Matching with the Given Choices:
[tex]\[
K_{eq} \approx 0.0046
\][/tex]
Therefore, the correct answer is:
D. 0.0046
