Question

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.

Answer 1

`K_(p)` for the reaction is 18.05

Step-by-step explanation:

Equilibrium constant in terms of partial pressure (
`K_(p)`) for this reaction can be written as-

`K_(p)=\frac{P_{NO_(2)}^(2)}{P_{N_(2)O_(4)}}`

where
`P_{NO_(2)}` and
`P_{N_(2)O_(4)}` are equilibrium partial pressure of
`NO_(2)` and
`N_(2)O_(4)` respectively

Hence
`K_(p)=((0.095)^(2))/((0.0005))` = 18.05

So,
`K_(p)` for the reaction is 18.05