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The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the reaction. N2O4(g)⇌2NO2(g) 0.0005 atm 0.095 atm Express your answer using two significant figures.

Respuesta :

Answer:

[tex]K_{p}[/tex] for the reaction is 18.05

Explanation:

Equilibrium constant in terms of partial pressure ([tex]K_{p}[/tex]) for this reaction can be written as-

                [tex]K_{p}=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}[/tex]

where [tex]P_{NO_{2}}[/tex] and [tex]P_{N_{2}O_{4}}[/tex] are equilibrium partial pressure of [tex]NO_{2}[/tex] and [tex]N_{2}O_{4}[/tex] respectively

Hence [tex]K_{p}=\frac{(0.095)^{2}}{(0.0005)}[/tex] = 18.05

So, [tex]K_{p}[/tex] for the reaction is 18.05

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