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Select the correct answer.

\begin{tabular}{|c|c|c|c|}
\hline
\multicolumn{2}{|c|}{Initial} & \multicolumn{2}{c|}{Equilibrium} \\
\hline
[tex]$\left[ N _2 O _4 \right]$[/tex] & [tex]$\left[ NO _2 \right]$[/tex] & [tex]$\left[ N _2 O _4 \right]$[/tex] & [tex]$\left[ NO _2 \right]$[/tex] \\
\hline
0.0200 & 0.0600 & 0.0429 & 0.0141 \\
\hline
\end{tabular}

Consider the following equilibrium and the table of experimental data:

[tex]\[
N _2 O _{4(g)} \rightleftharpoons 2 NO _{2(g)}
\][/tex]

Which of the following represents the [tex]$K _{\text{eq}}$[/tex] value?

A. 0.18

B. 0.0067

C. 0.131

D. 0.0046

Sagot :

GinnyAnsweridn
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
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