The equilibrium reaction is
N2O4(g)<----> 2NO2(g)
For reaction
Kp = (pNO2)^2 / pN2O4
Given:
Kp = 0.316
pN2O4 = 3.48 atm
To calculate
pNO2 = ?
0.316 = (pNO2)^2 / 3.48
(pNO2)^2 = 1.0997
pNO2 = 1.049 atm
Answer:- Equilibrium partial pressure of [tex]NO_2[/tex] is 1.05 atm.
Solution:- The given balanced equation is:
[tex]N_2O_4(g)\leftrightarrow 2NO_2(g)[/tex]
Equilibrium expression for the equation is written as:
[tex]Kp=\frac{p(NO_2)^2}{p(N_2O_4)}[/tex]
(In this expression p stands for partial pressure.)
If we consider the initial partial pressure for the reactant gas as m and the product gas as 0 since initially there is no product. Let's say the change in pressure is n. Then equilibrium partial pressure of reactant gas will be (m-n) and reactant gas equilibrium partial pressure be 2n since it's coefficient is two.
Equilibrium partial pressure of reactant gas is given as 3.48 atm. It means (m-n) = 3.48
Let's plug in the values in the equilibrium expression:
[tex]0.316=\frac{(2n)^2}{3.48}[/tex]
On cross multiply:
[tex](2n)^2=0.316(3.48)[/tex]
Taking square root to both sides:
[tex]\sqrt{(2n)^2}=\sqrt{0.316(3.48)}[/tex]
[tex]2n=1.05[/tex]
So, the equilibrium partial pressure of [tex]NO_2[/tex] is 1.05 atm.
