To determine the equilibrium concentration of [tex]\( N_2O_4 \)[/tex] given the equilibrium concentration of [tex]\( NO_2 \)[/tex] and the equilibrium constant [tex]\( K_{eq} \)[/tex], we need to use the equilibrium expression related to this reaction:
[tex]\[ N_2O_4(g) \leftrightarrow 2 NO_2(g) \][/tex]
The equilibrium expression for this reaction is:
[tex]\[ K_{eq} = \frac{[NO_2]^2}{[N_2O_4]} \][/tex]
Given:
- [tex]\( K_{eq} = 5.85 \times 10^{-3} \)[/tex]
- [tex]\( [NO_2] = 1.78 \times 10^{-2} \, M \)[/tex]
We can rearrange the equilibrium expression to solve for the equilibrium concentration of [tex]\( N_2O_4 \)[/tex]:
[tex]\[ [N_2O_4] = \frac{[NO_2]^2}{K_{eq}} \][/tex]
Substitute the known values into the equation:
[tex]\[ [N_2O_4] = \frac{(1.78 \times 10^{-2} \, M)^2}{5.85 \times 10^{-3}} \][/tex]
Calculate the numerator:
[tex]\[ (1.78 \times 10^{-2})^2 = 3.1684 \times 10^{-4} \, M^2 \][/tex]
Then divide by the equilibrium constant:
[tex]\[ [N_2O_4] = \frac{3.1684 \times 10^{-4} \, M^2}{5.85 \times 10^{-3}} \][/tex]
[tex]\[ [N_2O_4] \approx 0.0542 \, M \][/tex]
Therefore, the equilibrium concentration of [tex]\( N_2O_4 \)[/tex] is approximately [tex]\( 0.0542 \, M \)[/tex].
