Question

The equilibrium constant for the reaction 2NO2(g) N2O4(g) is Keq = . If a sample at equilibrium was found to contain 0.058 M NO2 and 0.012 M N2O4, what would the Keq value for this reaction at that temperature be?

Answer 1

hey can someone pls help me with this

Step-by-step explanation:

Answer 2

`\boxed{3.6}`

Step-by-step explanation:

2NO₂ ⇌ N₂O₄

E/mol·L⁻¹: 0.058 0.012

K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\

\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}