To calculate the equilibrium constant \( K_{\text{eq}} \) for the reaction:
\[ 2 \text{NOCl}(g) \leftrightarrow 2 \text{NO}(g) + \text{Cl}_2(g) \]
we use the expression for \( K_{\text{eq}} \):
\[ K_{\text{eq}} = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2} \]
We have the following concentrations at equilibrium:
- \( [\text{NOCl}] = 1.4 \cdot 10^{-2} , \text{M} \)
- \( [\text{NO}] = 1.2 \cdot 10^{-3} , \text{M} \)
- \( [\text{Cl}_2] = 2.2 \cdot 10^{-3} , \text{M} \)
Now, we can substitute these values into the \( K_{\text{eq}} \) expression:
\[ K_{\text{eq}} = \frac{(1.2 \cdot 10^{-3})^2 (2.2 \cdot 10^{-3})}{(1.4 \cdot 10^{-2})^2} \]
Calculating the numerator:
\[ (1.2 \cdot 10^{-3})^2 = 1.44 \cdot 10^{-6} \] \[ (1.44 \cdot 10^{-6}) \cdot (2.2 \cdot 10^{-3}) = 3.168 \cdot 10^{-9} \]
Now, calculating the denominator:
\[ (1.4 \cdot 10^{-2})^2 = 1.96 \cdot 10^{-4} \]
Now, substituting the values back into the \( K_{\text{eq}} \) expression:
\[ K_{\text{eq}} = \frac{3.168 \cdot 10^{-9}}{1.96 \cdot 10^{-4}} \]
Calculating this:
\[ K_{\text{eq}} = 1.6133 \cdot 10^{-5} \]
Rounding appropriately gives us:
\[ K_{\text{eq}} \approx 1.6 \cdot 10^{-5} \]
Thus, the value of \( K_{\text{eq}} \) expressed in scientific notation is:
\[ \boxed{1.6 \cdot 10^{-5}} \]
Therefore, the correct answer is option A. \( 1.6 \cdot 10^{-5} \).