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!

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

it's all four options.