For the reaction of the habers process at 127 c the partial pressures of nitrogen and hydrogen are 0.80 and 0.40 at equilibrium respectivly.The total pressure of the system is 2.80 atm . Calculate kp and kc for the reaction.

Asked by saksham pathak | 17th Oct, 2010, 12:26: AM

Expert Answer:

                                                                   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

 

 

 

 

Answered by  | 20th Oct, 2010, 10:36: PM

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