Question

At equilibrium, the concentration of \( N_2 = 5 \times 10^{-3} \, \text{M} \), \( O_2 = 2.8 \times 10^{-3} \, \text{M} \), and \( NO = 1.4 \times 10^{-3} \, \text{M} \) in a sealed vessel at 800 K. What is the value of \( K_c \) for the reaction at the same temperature? \[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]

• A. 0.41
• B. 0.14
• C. 0.18
• D. 0.5
• E. 0.28

Hint:

To calculate \( K_c \), remember to use the concentrations of products and reactants raised to the power of their stoichiometric coefficients.

Answer 1

The Correct Option is B

The equilibrium constant \( K_c \) for the reaction is calculated using the formula: \[ K_c = \frac{[NO]^2}{[N_2][O_2]} \] Substitute the given concentrations into the equation: \[ K_c = \frac{(1.4 \times 10^{-3})^2}{(5 \times 10^{-3})(2.8 \times 10^{-3})} \] First, calculate the numerator and denominator: - Numerator: \( (1.4 \times 10^{-3})^2 = 1.96 \times 10^{-6} \) - Denominator: \( (5 \times 10^{-3})(2.8 \times 10^{-3}) = 1.4 \times 10^{-5} \) Now, divide the numerator by the denominator: \[ K_c = \frac{1.96 \times 10^{-6}}{1.4 \times 10^{-5}} = 0.14 \] Thus, the correct answer is (B) 0.14.