Question

The \( K_c \) of the reaction \[ \text{NH}_3(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g) \] at 1500 K is 0.1. What is the concentration of NO, when the initial concentration of \( \text{N}_2 \) and \( \text{O}_2 \) is 0.04 mol?

• A. \( 1.09 \times 10^{-2} \, \text{M} \)
• B. \( 10.9 \times 10^{-2} \, \text{M} \)
• C. \( 2.18 \times 10^{-2} \, \text{M} \)
• D. \( 1.09 \times 10^{-4} \, \text{M} \)
• E. \( 2.18 \times 10^{-4} \, \text{M} \)

Hint:

To solve for equilibrium concentrations, use the stoichiometric relationships and the equilibrium constant expression to form and solve a quadratic equation.

Answer 1

The Correct Option is A

The balanced reaction is: \[ \text{NH}_3(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g) \] Given: - \( K_c = 0.1 \) at 1500 K - Initial concentrations of \( \text{NH}_3 \) and \( \text{O}_2 \) are \( 0.04 \, \text{mol} \) - The concentration of NO is to be found at equilibrium. We assume that the change in the concentration of \( \text{NH}_3 \) and \( \text{O}_2 \) is \( x \). From the stoichiometry of the reaction: \[ [\text{NO}] = 2x \] The equilibrium concentrations are: \[ [\text{NH}_3] = 0.04 - x \] \[ [\text{O}_2] = 0.04 - x \] \[ [\text{NO}] = 2x \] Substitute these into the equilibrium expression for \( K_c \): \[ K_c = \frac{[\text{NO}]^2}{[\text{NH}_3][\text{O}_2]} \] \[ 0.1 = \frac{(2x)^2}{(0.04 - x)(0.04 - x)} \] \[ 0.1 = \frac{4x^2}{(0.04 - x)^2} \] Now, solve this equation for \( x \): \[ 0.1 = \frac{4x^2}{(0.04 - x)^2} \] \[ (0.04 - x)^2 = \frac{4x^2}{0.1} \] \[ (0.04 - x)^2 = 40x^2 \] Taking the square root of both sides: \[ 0.04 - x = \sqrt{40}x \] Now solve for \( x \) using algebra, and calculate the concentration of \( \text{NO} \): \[ x = 1.09 \times 10^{-2} \, \text{M} \] Therefore, the concentration of \( \text{NO} \) is: \[ [\text{NO}] = 2x = 2 \times 1.09 \times 10^{-2} = 1.09 \times 10^{-2} \, \text{M} \]
Final Answer: (A) \( 1.09 \times 10^{-2} \, \text{M} \)