Question

The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to PNO in air at 25°C? Assume that PO2 5 0.21 atm and does not change.

Answer 1

Answer : The ratio of
`p_(NO_2)` to
`p_(NO)` is,
`6.87* 10^5`

Solution : Given,

`K_p=1.5* 10^6`

`p_(O_2)` = 0.21 atm

The given equilibrium reaction is,

`NO(g)+(1)/(2)O_2\rightleftharpoons NO_2(g)`

The expression of
`K_p` will be,

`K_p=\frac{(p_(NO_2))}{(p_(NO))* (p_(O_2))^{(1)/(2)}}`

Now put all the given values in this expression, we get:

`1.5* 10^6=\frac{(p_(NO_2))}{(p_(NO))* (0.21)^{(1)/(2)}}`

`((p_(NO_2)))/((p_(NO)))=(1.5* 10^6)* (0.21)^{(1)/(2)}`

`((p_(NO_2)))/((p_(NO)))=6.87* 10^5`

Therefore, the ratio of
`p_(NO_2)` to
`p_(NO)` is,
`6.87* 10^5`