CBSE Class 11-science Answered

                                                                   N2   +  3H2................> 2 NH3

Initially the no. Of moles are:                    1          3                    0

At equilibrium the no. Of moles are:          1-a   3(1-a)                 2a

We know that,

Partial pressure(p)  = mole fraction * total pressure(P)

Hence,

pN₂=(1-a)*P/2(2-a),

pH₂=3(1-a)*P/2(2-a),

pNH₃=2a*P/2(2-a)

Here,

pN2= .80==(1-a)*2.80/2(2-a)

a=(4-2a)*.80=2.80-2.80a

3.20-1.60a=2.80-2.80a

1.20a =-.40

 a= -.33

Kp= [pNH₃ ]² / [pN₂ ][ pH₂ ]³

Kp=[2a*P/2(2-a)]² /[ (1-a)*P/2(2-a)] [3(1-a)*P/2(2-a)]³

Kp=64a²/9P² (after negelecting the a<<<<1)

Kp=64(.33)² /9(2.80)²

=.098

Kp=KcRT^{(Δ n)}

.098=Kc*[8.314*300]^(-2)

Kc=.098/[.0821*400]^(-2)

⁼105.68