Question

Consider the reaction N2(g) + O2(g) â 2 NO(g) K = 0.0025 A rigid container initially contains 8.00 atm of nitrogen and 5.00 atm of oxygen. What will be the partial pressure of nitrogen at equilibrium?

Answer 1

7.8458 atm

Step-by-step explanation:

For the reaction

`N_(2)(g) + O_(2)(g) - -> 2NO(g)`

Kp is defined as:

`Kp=((P_(NO))^(2))/(P(N_(2))*P(O_(2)))`

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

`Kp=((2x)^(2))/((5-x)*(8-x)) - - ->Kp(x^(2)-13x+40)=4x^(2)`

working with this expression

`Kp(x^(2)-13x+40)=4x^(2) ---> Kpx^(2)-13Kpx+40Kp=4x^(2)`

`0=4x^(2)-Kpx^(2)+13Kpx-40Kp - - -> (4-Kp)x^(2)+13Kpx-40Kp`

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:

`x=\frac{-b(+-)\sqrt{b^(2)-4ac}}{2a}`

we will get to values for X

`x_(1)= \frac{-0.0325+\sqrt{0.0325^(2)-4*3.9975*(-0.1)}}{2*3.9975}=0.1541`

`x_(2)= \frac{-0.0325-\sqrt{0.0325^(2)-4*3.9975*(-0.1)}}{2*3.9975}=-0.1622`

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

`P(N_(2))=8 atm - x = 8 atm - 0.1541 =7.8458 atm`