Question

A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm NO2(g) at 25°C, and the following equilibrium is established. 2 NO2(g) ⇌ N2O4(g) At equilibrium, the partial pressure of NO2(g) is 0.512 atm. Calculate the equilibrium constant Kp for this reaction.

Answer 1

Answer: The equilibrium constant,
`K_p` for the given reaction is 6.653

Step-by-step explanation:

We are given:

Initial partial pressure of nitrogen dioxide = 1.00 atm

Initial partial pressure of dinitrogen tetraoxide = 1.500 atm

Equilibrium partial pressure of nitrogen dioxide = 0.512 atm

For the given chemical equation:

`2NO_2(g)\rightleftharpoons N_2O_4(g)`

Initial: 1.00 1.500

At eqllm: 1.00-2x 1.500+x

Evaluating the value of 'x'

`\Rightarrow (1.00-2x)=0.512\\\\x=0.244`

So, equilibrium partial pressure of dinitrogen tetraoxide = (1.500 + x) = [1.500 + 0.244] = 1.744 M

The expression of
`K_p` for above equation follows:

`K_p=(p_(N_2O_4))/((p_(NO_2))^2)`

Putting values in above equation, we get:

`K_p=(1.744)/((0.512)^2)\\\\K_p=6.653`

Hence, the equilibrium constant,
`K_p` for the given reaction is 6.653