Question

For the equilibrium, N2O4(g)<----> 2NO2(g), the equilibrium constant Kp = 0.316. Calculate the equilibrium partial pressure of NO2 if the equilibrium partial pressure of N2O4(g) is 3.48 atm.
1.05 atm
1.10 atm
3.32 atm
0.301 atm

Answer 1

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 2

Answer:- Equilibrium partial pressure of $NO_2$ is 1.05 atm.

Solution:- The given balanced equation is:

$N_2O_4(g)\leftrightarrow 2NO_2(g)$

Equilibrium expression for the equation is written as:

$Kp=\frac{p(NO_2)^2}{p(N_2O_4)}$

(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:

$0.316=\frac{(2n)^2}{3.48}$

On cross multiply:

$(2n)^2=0.316(3.48)$

Taking square root to both sides:

$\sqrt{(2n)^2}=\sqrt{0.316(3.48)}$

$2n=1.05$

So, the equilibrium partial pressure of $NO_2$ is 1.05 atm.