Answered

Consider the reaction.

$2 NOCl(g) \longleftrightarrow 2 NO(g)+Cl_2(g)$



At equilibrium, the concentrations are as follows:

$\begin{array}{l}

{[NOCl]=1.4 \times 10^{-2} M} \\

{[NO]=1.2 \times 10^{-3} M} \\

{\left[Cl_2\right]=2.2 \times 10^{-3} M}

\end{array}$



What is the value of $K_{\text {eq }}$ for the reaction expressed in scientific notation?

A$6.2 \times 10^{-4}$
B$1.6 \times 10^{-5}$
C$6.2 \times 10^4$
D$1.6 \times 10^5$

Verified Answer :

  • Use the law of mass action to define the equilibrium constant expression: \(K_{\text{eq}} = \frac{[NO]^2 [Cl_2]}{[NOCl]^2}\).
  • Substitute the given molar concentrations into the formula.
  • Calculate the resulting value: \(K_{\text{eq}} \approx 1.6 \times 10^{-5}\).
  • Match the result with the given options, identifying the correct choice as \(\boxed{\text{b}}\).

Explanation

  1. Analysis of the Equilibrium Constant Expression

The problem requires calculating the equilibrium constant \(K_{\text{eq}}\) for the reaction:
\[2 NOCl(g) \rightleftharpoons 2 NO(g) + Cl_2(g)\]
Given the equilibrium concentrations:
\[[NOCl] = 1.4 \times 10^{-2} \, \text{M}\]
\[[NO] = 1.2 \times 10^{-3} \, \text{M}\]
\[[Cl_2] = 2.2 \times 10^{-3} \, \text{M}\]
The equilibrium constant expression for this reaction is given by:
\[K_{\text{eq}} = \frac{[NO]^2 [Cl_2]}{[NOCl]^2}\]
We need to substitute the given values into this expression to find \(K_{\text{eq}}\).

  1. Calculation of the Equilibrium Constant

Substitute the provided concentration values into the expression for \(K_{\text{eq}}\):
\[K_{\text{eq}} = \frac{(1.2 \times 10^{-3})^2 \times (2.2 \times 10^{-3})}{(1.4 \times 10^{-2})^2}\]
First, calculate the squared terms:
\[(1.2 \times 10^{-3})^2 = 1.44 \times 10^{-6}\]
\[(1.4 \times 10^{-2})^2 = 1.96 \times 10^{-4}\]
Now, substitute these back into the expression:
\[K_{\text{eq}} = \frac{(1.44 \times 10^{-6}) \times (2.2 \times 10^{-3})}{1.96 \times 10^{-4}}\]
Perform the multiplication in the numerator:
\[(1.44 \times 10^{-6}) \times (2.2 \times 10^{-3}) = 3.168 \times 10^{-9}\]
Finally, divide by the denominator:
\[K_{\text{eq}} = \frac{3.168 \times 10^{-9}}{1.96 \times 10^{-4}} \approx 1.616 \times 10^{-5}\]

  1. Conclusion

The calculated value of \(K_{\text{eq}} \approx 1.6 \times 10^{-5}\) matches option (b). Therefore, the value of the equilibrium constant for the given reaction is \(1.6 \times 10^{-5}\), expressed in scientific notation. Thus, the correct option is \(\boxed{\text{b}}\).

Examples

The concept of an equilibrium constant is fundamental in chemistry for predicting the extent of a reaction. For example, in industrial chemical production, engineers use \(K_{\text{eq}}\) to determine the ideal temperature and pressure conditions to maximize the yield of a product, such as the synthesis of ammonia from nitrogen and hydrogen gases. By understanding how concentrations at equilibrium relate to one another, scientists can optimize processes to be more cost-effective and efficient, ensuring that the desired product is produced in the highest possible concentration.

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