Question

What is the value of KP for the reaction of nitrogen and oxygen to make dinitrogen monoxide if the equilibrium partial pressures of nitrogen is 1.3241 atm, the partial pressure of oxygen is 1.0844 atm and the partial pressure of dinitrogen monoxide is 0.0339 atm?

Answer 1

The value of equilibrium constant is :

`K_p=6.04* 10^(-4)`

Step-by-step explanation:

Partial pressure of nitrogen at equilibrium,
`p_1= 1.3241 atm`

Partial pressure of oxygen at equilibrium,
`p_2= 1.0844 atm`

Partial pressure of dinitrogen monoxide at equilibrium,
`p_3= 0.0339 atm`

`2N_2(g)+O_2(g)\rightleftharpoons 2N_2O(g)`

The expression of
`K_p` will be given as:

`K_p=(p_3^(2))/(p_1^(2)* p_2)`

`K_p=((0.0339 atm)^2)/( (1.3241 atm)^2* 1.0844 atm)=6.04* 10^(-4)`

The value of equilibrium constant is :

`K_p=6.04* 10^(-4)`