To solve the equilibrium problem, let’s break it down step by step.

Problem Statement:

The reaction is:
$N_2(g) + O_2(g) \leftrightharpoons 2NO(g)$
The equilibrium constant is $K_c = 0.10$ at $2000^\circ C$.
Initial concentrations:
[ [N_2]_0 = 0.040 , M, \quad [O_2]_0 = 0.040 , M, \quad [NO]_0 = 0 ]

We need to find the equilibrium concentration of NO.

Step 1: Define the Change in Concentrations

Let the change in concentration of $NO$ be $+2x$, since 2 moles of $NO$ are produced for every mole of $N_2$ and $O_2$ consumed. The changes for $N_2$ and $O_2$ will be $-x$ each.

At equilibrium, the concentrations are:
[ [N_2] = 0.040 - x, \quad [O_2] = 0.040 - x, \quad [NO] = 2x ]

Step 2: Write the Expression for $K_c$

The equilibrium constant $K_c$ is given by:
[ K_c = \frac{[NO]^2}{[N_2][O_2]} ]

Substitute the equilibrium concentrations:
$0.10 = \frac{(2x)^2}{(0.040 - x)(0.040 - x)}$

Step 3: Simplify the Equation

$0.10 = \frac{4x^2}{(0.040 - x)^2}$

Take the square root of both sides:
$\sqrt{0.10} = \frac{2x}{0.040 - x}$

$0.316 = \frac{2x}{0.040 - x}$

Multiply through by $(0.040 - x)$:
$0.316(0.040 - x) = 2x$

$0.01264 - 0.316x = 2x$

Combine like terms:
$0.01264 = 2.316x$

Solve for $x$:
$x = \frac{0.01264}{2.316} = 0.00546 \, M$

Step 4: Find the Equilibrium Concentration of $NO$

The equilibrium concentration of $NO$ is:
$[NO] = 2x = 2(0.00546) = 0.01092 \, M$

Final Answer:

The equilibrium concentration of $NO$ is approximately 0.010 M, matching the third option.

Let me know if you have further questions or would like a deeper explanation!

Related Questions:

1. How do you calculate equilibrium concentrations when $K_c$ is very large or very small?
2. What is the effect of temperature on $K_c$ for an exothermic reaction?
3. How can you simplify equilibrium problems with limiting reactants?
4. How does Le Chatelier’s Principle apply to this system