Two moles of ammonia was introduced in an evacuated vessel of 1 L capacity. At high temperature, the gas undergoes partial dissociation according to the equation: 2 NH 3( g ) leftharpoons N 2( g )+3 H 2( g ) At equilibrium the concentration of ammonia was found to be 1 mol. What is the value of K ?

Question

Two moles of ammonia was introduced in an evacuated vessel of 1 L capacity. At high temperature, the gas undergoes partial dissociation according to the equation: 2 NH 3( g ) leftharpoons N 2( g )+3 H 2( g ) At equilibrium the concentration of ammonia was found to be 1 mol. What is the value of K ? (A) 3 / 4=0.75 mol 2 L -2 (B) 3 / 2=1.5 mol 2 L -2 (C) 27 / 16=1.7 mol 2 L -2 (D) 27 / 64=0.42 mol 2 L -2

Tardigrade Admin · Accepted Answer

Let α be the degree of dissociation: Hence we have: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/d1d0d544a5588838d9e6d2d45a530b4e-.png" /> At equilibrium: [ NH 3]=2-2 α=1 or α=1 / 2 [ N 2]=(1/2),[ H 2]=(3/2),[ NH 3]=1 ∴ K=([ N 2][ H 2]3/[ NH 3]2)=((1/2) × (3/2) × (3/2) × (3/2)/1 × 1)=(27/16)=1.7 mol 2 L -2

$3/4 = 0.75 m o l^{2} L^{- 2}$

$3/2 = 1.5 m o l^{2} L^{- 2}$

$27/16 = 1.7 m o l^{2} L^{- 2}$

$27/64 = 0.42 m o l^{2} L^{- 2}$

Q. Two moles of ammonia was introduced in an evacuated vessel of 1L capacity. At high temperature, the gas undergoes partial dissociation according to the equation: 2NH3​(g)⇌N2​(g)+3H2​(g) At equilibrium the concentration of ammonia was found to be 1mol. What is the value of K ?

3/4=0.75mol2L−2

3/2=1.5mol2L−2

27/16=1.7mol2L−2

27/64=0.42mol2L−2

Let α be the degree of dissociation: Hence we have : At equilibrium : [NH3​]=2−2α=1 or α=1/2 [N2​]=21​,[H2​]=23​,[NH3​]=1 ∴K=[NH3​]2[N2​][H2​]3​=1×121​×23​×23​×23​​=1627​=1.7mol2L−2

$3/4 = 0.75 m o l^{2} L^{- 2}$

$3/2 = 1.5 m o l^{2} L^{- 2}$

$27/16 = 1.7 m o l^{2} L^{- 2}$

$27/64 = 0.42 m o l^{2} L^{- 2}$