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The table below gives the equilibrium concentrations for this reaction at acertain temperature:N₂(g) + O₂(g) →→→→2NO(g)0.69 M[N₂]0.98 MOA. 20OB. 1.7 x 10-3OC. 5.0 × 10-2OD. 9.9[0₂]What is the equilibrium constant for the reaction?0.034 M[NO]

The table below gives the equilibrium concentrations for this reaction at acertain temperatureNg Og 2NOg069 MN098 MOA 20OB 17 x 103OC 50 102OD 990What is the eq class=

Respuesta :

Answer:

Option B is correct

[tex]1.7\times10^{-3}[/tex]

Explanations:

Given the chemical reaction below;

[tex]N_2(g)+O_2(g)\rightarrow2NO(g)[/tex]

The equilibrium constant for the reaction is given as;

[tex]k=\frac{[NO]^2}{[O_2][N_2]}[/tex]

Given the following parameters

[tex]\begin{gathered} [NO]=0.034M \\ [N_2]=0.69M \\ [O_2]=0.98M \end{gathered}[/tex]

Substitute

[tex]\begin{gathered} k=\frac{(0.034)^2}{(0.69)(0.98)} \\ k=\frac{0.001156}{0.6762} \\ k=0.001709=1.7\times10^{-3} \end{gathered}[/tex]

Therefore the equilibrium constant for the reaction is 1.7 * 10^-3

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