To find the equilibrium constant ([tex]\( K_{\text{eq}} \)[/tex]) for the reaction:
[tex]\[ 2 \text{NOCl} (g) \leftrightarrow 2 \text{NO} (g) + \text{Cl}_2 (g) \][/tex]
we use the given equilibrium concentrations:
[tex]\[
\begin{array}{l}
[ \text{NOCl} ] = 1.4 \times 10^{-2} \, \text{M} \\
[ \text{NO} ] = 1.2 \times 10^{-3} \, \text{M} \\
[ \text{Cl}_2 ] = 2.2 \times 10^{-3} \, \text{M}
\end{array}
\][/tex]
The expression for the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for this reaction is:
[tex]\[ K_{\text{eq}} = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2} \][/tex]
Let's plug in the given equilibrium concentrations into this expression:
[tex]\[ K_{\text{eq}} = \frac{(1.2 \times 10^{-3})^2 (2.2 \times 10^{-3})}{(1.4 \times 10^{-2})^2} \][/tex]
Now, let's calculate step-by-step:
1. Calculate [tex]\( (1.2 \times 10^{-3})^2 \)[/tex]:
[tex]\[ (1.2 \times 10^{-3})^2 = 1.44 \times 10^{-6} \][/tex]
2. Multiply this result by [tex]\( 2.2 \times 10^{-3} \)[/tex]:
[tex]\[ 1.44 \times 10^{-6} \times 2.2 \times 10^{-3} = 3.168 \times 10^{-9} \][/tex]
3. Calculate [tex]\( (1.4 \times 10^{-2})^2 \)[/tex]:
[tex]\[ (1.4 \times 10^{-2})^2 = 1.96 \times 10^{-4} \][/tex]
4. Divide the numerator by the denominator:
[tex]\[ K_{\text{eq}} = \frac{3.168 \times 10^{-9}}{1.96 \times 10^{-4}} \][/tex]
Now, perform the division:
[tex]\[ K_{\text{eq}} = 1.616326530612245 \times 10^{-5} \][/tex]
Therefore, the value of [tex]\( K_{\text{eq}} \)[/tex] for the reaction expressed in scientific notation is approximately:
[tex]\[ 1.6 \times 10^{-5} \][/tex]
Hence, the correct answer is:
[tex]\[ 1.6 \times 10^{-5} \][/tex]
