To identify the correct equilibrium constant expression for the given chemical equation, [tex]\( 2 NO (g) + O_2 (g) \leftrightarrow 2 NO_2 (g) \)[/tex], let’s go through it step-by-step.
1. Understanding the equilibrium constant expression:
For a general reaction:
[tex]\[
aA + bB \leftrightarrow cC + dD
\][/tex]
The equilibrium constant [tex]\( K_{eq} \)[/tex] is given by:
[tex]\[
K_{eq} = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b}
\][/tex]
where [tex]\([C]\)[/tex], [tex]\([D]\)[/tex], [tex]\([A]\)[/tex], and [tex]\([B]\)[/tex] represent the molar concentrations of the reactants and products at equilibrium, and [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] are the stoichiometric coefficients.
2. Applying this to our equation:
[tex]\[
2 NO (g) + O_2 (g) \leftrightarrow 2 NO_2 (g)
\][/tex]
Here:
- [tex]\(A = NO\)[/tex] with a coefficient [tex]\(a = 2\)[/tex]
- [tex]\(B = O_2\)[/tex] with a coefficient [tex]\(b = 1\)[/tex]
- [tex]\(C = NO_2\)[/tex] with a coefficient [tex]\(c = 2\)[/tex]
3. Writing the equilibrium constant expression:
According to the general formula, the equilibrium constant [tex]\(K_{eq}\)[/tex] will be:
[tex]\[
K_{eq} = \frac{[NO_2]^2}{[NO]^2 \cdot [O_2]}
\][/tex]
This expression shows that the concentrations of the products raised to their stoichiometric coefficients are in the numerator, and the concentrations of the reactants raised to their stoichiometric coefficients are in the denominator.
4. Comparing with the given options:
[tex]\[
\begin{array}{l}
K_{eq} = \frac{[NO_2]^2}{[NO]^2 \cdot [O_2]} \\
K_{eq} = \frac{\left[ NO_2\right]^2}{[NO]^2 \cdot [O_2]} \\
K_{eq} = \frac{[NO]}{\left[NO_2\right] [O_2]}
\end{array}
\][/tex]
5. Identifying the correct expression:
Clearly, the second option (switching to a 1-based index) matches our derived equilibrium constant expression:
[tex]\[
K_{eq} = \frac{\left[NO_2\right]^2}{[NO]^2 \cdot [O_2]}
\][/tex]
Thus, the correct equilibrium constant expression is the second option, and its index is 1.
