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:
For a homogenous gas equilibrium, the equilibrium constant $K_c$ relates the concentrations of products to reactants. We can use an ICE (Initial, Change, Equilibrium) table to find the unknown concentrations.
Key Formula or Approach:
For the reaction $N_2 + O_2 \rightleftharpoons 2NO$, $K_c = \frac{[NO]^2}{[N_2][O_2]}$.

Step 2: Detailed Explanation:
Let the change in concentration be $x$.
Initial: $[N_2] = 0.04$, $[O_2] = 0.04$, $[NO] = 0$.
At Equilibrium: $[N_2] = 0.04 - x$, $[O_2] = 0.04 - x$, $[NO] = 2x$.
\[ K_c = \frac{(2x)^2}{(0.04 - x)(0.04 - x)} = 0.1 \]
Taking the square root of both sides:
\[ \frac{2x}{0.04 - x} = \sqrt{0.1} \approx 0.316 \]
\[ 2x = 0.316(0.04) - 0.316x \]
\[ 2x + 0.316x = 0.01264 \]
\[ 2.316x = 0.01264 \implies x \approx 0.00545 \]
The equilibrium concentration of $NO$ is $2x$:
\[ [NO]_{eq} = 2(0.00545) = 0.0109 \approx 0.011\text{ mol L}^{-1} \]

Step 3: Final Answer:
The equilibrium concentration of $NO$ is $0.011\text{ mol L}^{-1}$.