Question

one mole of nitrogen reacts with 3 mole of hydrogen in a 4l vessel. If 25% of nitrogen converted into ammonia then find equilibrium constant for the given reaction............. Plz ans fast

Answer 1

from the given data, we can write ;

$N_2+3H_2\Leftrightarrow 2NH_3$

at t = 0, 1mol ; 3mol. ; 0

at eq , 1 - 0.25 ; 3 - 3 × 0.25 ; 2 × 0.25

at equilibrium,

$[N_2]=1-0.25=0.75$

$[H_2]=3-0.75=2.25$

and $[NH_3]=0.50$

hence, equilibrium constant, $K_C=\frac{[NH_3]^2}{[N_2][H_2]^3}$

= (0.5)²/{(0.75) × (2.25)³}

= 0.02926 $(mol L^{-1})^{-2}$

hence, equilibrium constant is 0.02926 $(mol L^{-1})^{-2}$

Answer 2

Solution:

quantity given initially-

N₂ - 1 mol

H₂ - 3 mol

NH3 - nill (beacuse ammonia form after RXⁿ)

since reactants and products are in gaseous form so instead of finding partial pressure of each term we will proceed in molar concentration and at last will find Kp  by this formula.

                          Kp = Kc (RT) Δn

  Δn=mol of product gas−mol of reactant gas

                               

So basic equation of the given reaction is as follows:-

                                                N₂ + H₂  => NH₃  (forward rxⁿ)

now after balancing the above equation at equilibrium it would look like this

                                               N₂ + 3H₂   <=>   2NH₃

ratio between N₂ , H₂ and NH₃ will be 1: 3: 2  at equilibrium

that means from 1 mole of nitrogen gas 2 moles of ammonia gas will be formed and 3 moles of H₂ will be consumed.

Quantity at equilibrium-

N₂-   3/4 mole                          (since 25% I.e 1/4 mole has been consumed)

H₂ -  3- 3(1/4)  = 9/4 mole       ( 3 times of N₂ quantity will be consumed)

NH₃ -  1/2 mole                       (2 times of N₂ consumption will be created in           form of ammonia as ratio of NH₃ is 2 times that of N₂)

But we need to find the moles/lit at equilibrium

N₂ -  3/4*4  = 3/16   Mol/Lit

H₂ -  9/4*4  = 9/16  Mol/Lit

NH₃ - 1/2*4  = 1/8   Mol/Lit

Reaction constant = [NH₃]²/ [ N₂] [H₂]³

so Kc = (1/8)² / (3/16) * (9/16)³

          =  0.46822

now  Kp = Kc (RT) Δn

                 Kc= 0.46822

                  R= gas constant = 0.8314

                  T= room temp i.e 298 K

                 Δn =  (- 10/16)

             Kp = 0.46822 *(0.8314 * 298)⁻ ¹⁰/₁₆

                    = 0.46822* 0.03189

                 = 0.0149