Question

For a reaction \(\text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO}\) . Given that \([\text{N}_2] = 2 \times 10^{-3}\), \([\text{O}_2] = 3 \times 10^{-3}\), \([\text{NO}] = 6 \times 10^{-6}\) Find the value of \(\text{K}_{\text{c}}\) for the reaction

Hint:

Always double-check the stoichiometric coefficients in the balanced chemical equation. Forgetting to square the \([\text{NO}]\) concentration is the most common error in this specific problem.

Answer 1

Step 1: Understanding the Concept:
The equilibrium constant in terms of concentration (\(K_c\)) for a reversible reaction is defined by the law of mass action. It is the ratio of the product of the equilibrium concentrations of the products to the product of the equilibrium concentrations of the reactants, with each term raised to a power equal to its stoichiometric coefficient from the balanced equation.
Step 2: Key Formula or Approach:
For a general reversible reaction: \(a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\)
The formula for the equilibrium constant is:
\[ K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]
For the specific reaction \(\text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO}\), the expression is:
\[ K_c = \frac{[\text{NO}]^2}{[\text{N}_2]^1 [\text{O}_2]^1} \]
Step 3: Detailed Explanation:
The given equilibrium concentrations are:
\([\text{N}_2] = 2 \times 10^{-3} \text{ M}\)
\([\text{O}_2] = 3 \times 10^{-3} \text{ M}\)
\([\text{NO}] = 6 \times 10^{-6} \text{ M}\)
Substitute these values into the equilibrium constant expression:
\[ K_c = \frac{(6 \times 10^{-6})^2}{(2 \times 10^{-3}) \cdot (3 \times 10^{-3})} \]
Calculate the square of the numerator:
\[ (6 \times 10^{-6})^2 = 36 \times 10^{-12} \]
Calculate the product in the denominator:
\[ (2 \times 10^{-3}) \cdot (3 \times 10^{-3}) = 6 \times 10^{-6} \]
Now, divide the numerator by the denominator:
\[ K_c = \frac{36 \times 10^{-12}}{6 \times 10^{-6}} \]
\[ K_c = \left(\frac{36}{6}\right) \times 10^{-12 - (-6)} \]
\[ K_c = 6 \times 10^{-12 + 6} \]
\[ K_c = 6 \times 10^{-6} \]
Step 4: Final Answer:
The calculated value of \(K_c\) is \(6 \times 10^{-6}\).