Question

Given the reaction: N2 + O2 = 2NO for which the Keq at 2273 K is 1.2 x 10-4

Answer 1

Answer:

(a): The expression of equilibrium constant is $K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$

(b): The equation to solve the concentration of NO is $[NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}$

(c): The concentration of NO is 0.0017 M.

Explanation:

The equilibrium constant is defined as the ratio of the concentration of products to the concentration of reactants raised to the power of the stoichiometric coefficient of each. It is represented by the term $K_{eq}$

(a):

The given chemical equation follows:

$N_2+O_2\rightarrow 2NO$

The expression for equilbrium constant will be:

$K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$

(b):

The equation to solve the concentration of NO follows:

$[NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}$            ......(1)

(c):

Given values:

$K_{eq}=1.2\times 10^{-4}$

$[N_2]_{eq}=0.166M$

$[O_2]_{eq}=0.145M$

Plugging values in equation 1, we get:

$[NO]=\sqrt{(1.2\times 10^{-4})\times 0.166\times 0.145}$

$[NO]=\sqrt{2.88\times 10^{-6}}$

$[NO]=0.0017 M$

Hence, the concentration of NO is 0.0017 M.