Answer:
7.8458 atm
Explanation:
For the reaction
[tex] N_{2}(g) + O_{2}(g) - -> 2NO(g)[/tex]
Kp is defined as:
[tex]Kp=\frac{(P_{NO})^{2}}{P(N_{2})*P(O_{2})}[/tex]
The conditions in the system are:
O2 N2 NO
initial 5 atm 8 atm 0
equilibrium 5-x 8-x 2x
From the stoichiometric relationship in the reaction we know that to produce 2x of NO we need x of N2 and O2.
Replacing the values in the expression for Kp we get
[tex]Kp=\frac{(2x)^{2}}{(5-x)*(8-x)} - - ->Kp(x^{2}-13x+40)=4x^{2}[/tex]
working with this expression
[tex]Kp(x^{2}-13x+40)=4x^{2} ---> Kpx^{2}-13Kpx+40Kp=4x^{2}[/tex]
[tex]0=4x^{2}-Kpx^{2}+13Kpx-40Kp - - -> (4-Kp)x^{2}+13Kpx-40Kp[/tex]
This a quadratic equation with
a=(4-Kp)
b=13*Kp
C=-40*Kp
Replacing the given value for Kp we get:
a=3.9975
b=0.0325
C=-0.1
the solution for this equation is given by the next equation:
[tex]x=\frac{-b(+-)\sqrt{b^{2}-4ac}}{2a}[/tex]
we will get to values for X
[tex]x_{1}= \frac{-0.0325+\sqrt{0.0325^{2}-4*3.9975*(-0.1)}}{2*3.9975}=0.1541[/tex]
[tex]x_{2}= \frac{-0.0325-\sqrt{0.0325^{2}-4*3.9975*(-0.1)}}{2*3.9975}=-0.1622[/tex]
Since negative partial pressures don't have physical meaning the solution for the system is x=0.1541.
So the partial pressure of nitrogen at the equilibrium is
[tex] P(N_{2})=8 atm - x = 8 atm - 0.1541 =7.8458 atm [/tex]
