Even at high T, the formation of NO is not favored:
N2(g) + O2(g) reverse reaction arrow 2 NO(g) Kc = 4.10 x 10−4 at 2000°C.
What is [NO] when a mixture of 0.25 mol of N2(g) and 0.10 mol of O2(g) reach equilibrium in a 1.0-L container at 2000°C?
M

3,16x10⁻³M

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g). The kc is defined as:

kc = [NO]² / [N₂] [O₂] = 4,10x10⁻⁴ (1)

If you add in a 1,0L container 0,25 mol of N₂ and 0,10 mol of O₂, concentrations in equilibrium will be:

[N₂] = 0,25M - x

[O₂] = 0,10M - x

[NO] = 2x

Replacing in (1):

[2X]² / [0,25-x] [0,10-x] = 4,10x10⁻⁴

[2X]² / 0,025 - 0,35x + x²= 4,10x10⁻⁴

4X² = 4,10x10⁻⁴x² - 1,435x10⁻⁴x + 1,025x10⁻⁵

3,99959x² + 1,435x10⁻⁴x - 1,025x10⁻⁵ = 0

Solving for x:

x = -0,0016 (Wrong answer, there is no negative concentrations)

x = 0,00158 (Right answer)

As molar concentration of NO in equilibrium is 2x:

[NO] = 2x = 2×0,00158 = 3,16x10⁻³M

I hope it helps!

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.