To find the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for the reaction [tex]\( 2 NOCl (g) \rightleftharpoons 2 NO (g) + Cl_2 (g) \)[/tex], we can make use of the provided equilibrium concentrations and the given formula for [tex]\( K_{\text{eq}} \)[/tex]:
[tex]\[ K_{\text{eq}} = \frac{[NO]^2 [Cl_2]}{[NOCl]^2} \][/tex]
Step-by-step solution:
1. Identify the provided equilibrium concentrations:
- [tex]\([NO] = 0.02 \, \text{M}\)[/tex]
- [tex]\([Cl_2] = 0.01 \, \text{M}\)[/tex]
- [tex]\([NOCl] = 0.5 \, \text{M}\)[/tex]
2. Substitute these values into the equilibrium constant expression:
[tex]\[
K_{\text{eq}} = \frac{(0.02)^2 \times 0.01}{(0.5)^2}
\][/tex]
3. Calculate the numerator:
- [tex]\((0.02)^2 = 0.0004\)[/tex]
- [tex]\(0.0004 \times 0.01 = 0.000004\)[/tex]
4. Calculate the denominator:
- [tex]\((0.5)^2 = 0.25\)[/tex]
5. Divide the numerator by the denominator:
[tex]\[
K_{\text{eq}} = \frac{0.000004}{0.25} = 0.000016
\][/tex]
6. Therefore, the value of the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] is:
[tex]\[
K_{\text{eq}} = 1.6 \times 10^{-5}
\][/tex]
Hence, the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] for the given reaction under the specified conditions is [tex]\( 1.6 \times 10^{-5} \)[/tex].
