To determine the equilibrium expression for the given reaction, we first need to understand how such an expression is formed.
The reaction provided is:
[tex]\[
2 N_2(g) + O_2(g) \rightleftarrows 2 N_2O(g)
\][/tex]
For a general chemical reaction of the form:
[tex]\[
aA + bB \rightleftarrows cC + dD
\][/tex]
the equilibrium constant expression ([tex]\( K_{eq} \)[/tex]) is given by:
[tex]\[
K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}
\][/tex]
Now, applying this to our specific reaction:
[tex]\[
2 N_2(g) + O_2(g) \rightleftarrows 2 N_2O(g)
\][/tex]
we have:
- [tex]\( a = 2 \)[/tex] for [tex]\( N_2 \)[/tex]
- [tex]\( b = 1 \)[/tex] for [tex]\( O_2 \)[/tex]
- [tex]\( c = 2 \)[/tex] for [tex]\( N_2O \)[/tex]
Thus, the equilibrium constant expression ([tex]\( K_{eq} \)[/tex]) for this reaction will be:
[tex]\[
K_{eq} = \frac{[N_2O]^2}{[N_2]^2 [O_2]}
\][/tex]
Now, let's look at the options to identify the correct one:
A. [tex]\(\frac{\left[ NO _2\right]^2}{\left[ O _2\right]\left[ N _2\right]^2}\)[/tex]
B. [tex]\(\frac{\left[ O _2\right]\left[ N _2\right]}{\left[ NO _2\right]}\)[/tex]
C. [tex]\(\frac{\left[ NO _2\right]}{\left[ O _2\right]+\left[ N _2\right]}\)[/tex]
D. [tex]\(\frac{\left[ O _2\right]+2\left[ N _2\right]}{2\left[ NO _2\right]}\)[/tex]
Considering the expression we derived:
[tex]\[
\frac{[N_2O]^2}{[N_2]^2 [O_2]}
\][/tex]
This matches option A (note: there is a mix-up in subscripts in the options where [tex]\( NO_2 \)[/tex] should be [tex]\( N_2O \)[/tex] for correct chemical representation):
[tex]\[
A. \frac{\left[ NO _2\right]^2}{\left[ O _2\right]\left[ N _2\right]^2}
\][/tex]
Hence, the correct answer is:
[tex]\[
\boxed{1}
\][/tex]
