Question

a reaction mixture initially contains 10.0 atm N2 and 10.0 atm H2. If the equilibrium pressure of NH3 is measured to be 6.0 atm, find the equilibrium constant (Kp) for the reaction. g

Answer 1

`Kp=5.14`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`N_2+3H_2\rightarrow 2NH_3`

Thus, the equilibrium expression is written as:

`Kp=(p_(NH_3)^2)/(p_(N_2)p_(H_2)^3)`

And in terms of the reaction extent:

`Kp=((2x)^2)/((10-x)(10-3x)^3)`

Thus, from the equilibrium pressure of ammonia we can compute the reaction extent:

`p_(NH_3)=2x=6.0 atm\\\\x=3.0atm`

Therefore, the equilibrium constant turns out:

`Kp=((2*3.0)^2)/((10.0-3.0)(10.0-3*3.0)^3)\\\\Kp=5.14`

Regards.