Question

For the gas phase homogenous equilibrium $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$, $K_c = 0.1$ at $1500\text{ K}$. If the initial concentrations of $N_{2(g)}$ and $O_{2(g)}$ are each $0.04\text{ mol L}^{-1}$ what is the equilibrium concentration of $NO_{(g)}$?

• A. $0.100\text{ mol L}^{-1}$
• B. $0.0100\text{ mol L}^{-1}$
• C. $0.022\text{ mol L}^{-1}$
• D. $0.02\text{ mol L}^{-1}$
• E. $0.011\text{ mol L}^{-1}$

Hint:

Since both reactants have the same initial concentration and stoichiometric coefficients, the denominator is a perfect square. Taking the square root of the whole equation simplifies it from a quadratic to a linear one.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We use the ICE (Initial, Change, Equilibrium) method to determine the concentrations of reactants and products at equilibrium.
Step 2: Key Formula or Approach:
\( K_C = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \).
Let \( 2x \) be the equilibrium concentration of NO.
Step 3: Detailed Explanation:
Reaction: N\textsubscript{2} + O\textsubscript{2} \(\rightleftharpoons\) 2NO
Initial: 0.04, 0.04, 0
Change: \(-x, -x, +2x \)
Equilibrium: \( 0.04-x, 0.04-x, 2x \)
\[ K_C = 0.1 = \frac{(2x)^2}{(0.04-x)^2} \] Take square root on both sides:
\[ \sqrt{0.1} = \frac{2x}{0.04-x} \implies 0.316 = \frac{2x}{0.04-x} \] \[ 0.316(0.04 - x) = 2x \implies 0.01264 - 0.316x = 2x \] \[ 0.01264 = 2.316x \implies x \approx 0.00545 \] Equilibrium concentration of NO = \( 2x = 2 \times 0.00545 \approx 0.011 \text{ mol L}^{-1} \).
Step 4: Final Answer:
The equilibrium concentration of NO is 0.011 mol L{-1}.