Question

the concentration of NO2 is twice the concentration of O2 gas? For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when the concentration of NO2 is twice the concentration of O2 gas? 4.2 × 10-10 M 2.1 × 10-10 M 2.4 × 109 M 4.8 × 109 M 1.7 x 10 -9 M

Answer 1

Answer: Concentration of N₂ is 4.8.$10^{9}$ M.

Explanation: $K_{c}$ is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For

N2(g) + 2 O2(g) ⇄ 2 NO2(g)

$K_{c}$ = $\frac{[NO2]^{2} }{[N2][O2]^{2} }$

From the question concentration of NO2 is twice of O2:

[NO2] = 2[O2]

Substituting this into $K_{c}$:

$K_{c}$ = $\frac{[2O2]^{2} }{[N2][O2]^{2} }$

8.3.$10^{-10}$ = $\frac{4O2^{2} }{[N2].O2^{2} }$

[N2] = $\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }$

[N2] = $\frac{4}{8.3.10^{-10} }$

[N2] = 4.8.$10^{9}$

The concentration of N2 in the equilibrium is [N2] = 4.8.$10^{9}$M.