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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?

Respuesta :

znk

Answer:

[tex]\boxed{3.6}[/tex]

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 $\boxed{\mathbf{3.6}}$}

Answer:

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Explanation:

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