Question

Even at high temperatures, the formation of NO is not favored: (Kc = 4.10 × 10−4 at 2000°C) N2(g) + O2(g) ⇌ 2 NO(g) What is [NO] when a mixture of 0.20 mol of N2(g) and 0.15 mol of O2(g) reach equilibrium in a 1.0−L container at 2,000°C?

Answer 1

Answer: The equilibrium concentration of NO is 0.0034 M

Explanation:

We are given:

Initial moles of nitrogen gas = 0.20 moles

Initial moles of oxygen gas = 0.15 moles

Volume of container = 1.0 L

We know that:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

So, $\text{Molarity of }N_2=\frac{0.20}{1}=0.20M$

$\text{Molarity of }O_2=\frac{0.15}{1}=0.15M$

The given chemical reaction follows:

                     $N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initial:             0.20     0.15

At eqllm:      0.20-x    0.15-x      2x

The expression of $K_c$ follows:

$K_c=\frac{[NO]^2}{[N_2]\times [O_2]}$

We are given:

$K_c=4.10\times 10^{-4}$

Putting values in above equation, we get:

$4.10\times 10^{-4}=\frac{(2x)^2}{(0.20-x)\times (0.15-x)}\\\\x=-0.0018,0.0017$

Neglecting the negative value of 'x' because concentration cannot be negative.

So, equilibrium concentration of NO = 2 x = (2 × 0.0017) = 0.0034 M

Hence, the equilibrium concentration of NO is 0.0034 M