Answer :
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
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