Question

n2 and h2 are taken in molar ratio 1:3 molar ratio in a closed vessel to attain the following equilibrium n2 + 3h2 -->2nh3 . find kp for the reaction at total pressure of 2p if p of n2 at epb is p/3. ans=4/3p^2 explain plz

Answer 1

The reaction of nitrogen with hydrogen can be shown as:

$N_2   + 3 H_2   \rightarrow   2  NH_3$

$P        3 P$

P –x = P /3  

$X = 2P /3$

                           

                            $3 P – 3x= 3 P – 6 P /3$

$P /3   P$

So, pressure of dinitrogen at equilibrium is P /3 and for hydrogen is P

Now, total pressure at equilibrium is 2 P

$P_N_2 + P_H_2 + P_N_H_3 = 2 P$

$P/3 + P + P_N_H_3 = 2 P$

$P_N_H_3 = 2 P/3$

The equilibrium constant for pressure can be shown as:

$Kp = P_N_H_3^2 / P_N  P_H^3$

$K =(4 P^2 / 9 ) / (P /3) (P^3)$

$= (4) / (3) (P^2)$

Thus, value of Kp is  (4) / (3) (P^2)

Answer 2

Answer:

Let the initial moles of N2 and H2 be 1 and 3.

                         N2+3H2⟶2NH3

Initial                  1        3                  0

at Eqb.            1−x 3−3x          2x

% of volume is same as % by mole NH3

4−2x2x=0.178

x=2+(2×0.178)4×0.178

x=0.302

mole fraction of H2 at eqb.=4−2x3−3x=0.6165

Mole fraction of N2 at eqb.=1−0.6165=0.178=0.2055

Kp=(XNH3.PT)2/(XN2)(XH2.PT)3

     =(0.2055×30)(0.6165×30)3(0.178×30)2

=7.31×10−4atm−2