When N 2 O 5 is heated at certain temperature, it dissociates as N 2 O 5( g ) leftharpoons N 2 O 3( g )+ O 2( g ) ; Kc =2 ⋅ 5. At the same time N 2 O 3 also decomposes as: N 2 O 3( g ) leftharpoons N 2 O ( g )+ O 2( g ). If initially 4.0 moles of N 2 O 5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O 2 at equilibrium is 2.5 M. Equilibrium concentration of N 2 O 5 is:

Question

When N 2 O 5 is heated at certain temperature, it dissociates as N 2 O 5( g ) leftharpoons N 2 O 3( g )+ O 2( g ) ; Kc =2 ⋅ 5. At the same time N 2 O 3 also decomposes as: N 2 O 3( g ) leftharpoons N 2 O ( g )+ O 2( g ). If initially 4.0 moles of N 2 O 5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O 2 at equilibrium is 2.5 M. Equilibrium concentration of N 2 O 5 is: (A) 1.0 M (B) 1.5 M (C) 2.166 M (D) 1.846 M

Tardigrade Admin · Accepted Answer

underset4 - xN2O5 leftharpoons undersetx - yN2O3 + undersetx + yO2 undersetx - yN2O3 rightharpoons undersetyN2O + undersety + xO2 ∵ [O2]= x + y =2 ⋅ 5 for N 2 O 5, Kc =[ N 2 O 5][ O 2] /[ N 2 O 5] and 2.5=((x+y)(x-y)/4-x) ∴ x=2.166 [ N 2 O 5]=4- x =1 ⋅ 846

1.0 M

1.5 M

2.166 M

1.846 M

4−xN2​O5​​⇌x−yN2​O3​​+x+yO2​​ x−yN2​O3​​⇋yN2​O​+y+xO2​​ ∵[O2​]=x+y=2⋅5 for N2​O5​,Kc=[N2​O5​][O2​]/[N2​O5​] and 2.5=4−x(x+y)(x−y)​ ∴x=2.166 [N2​O5​]=4−x=1⋅846

$4 - x N_{2} O_{5} ⇌ x - y N_{2} O_{3} + x + y O_{2}$
$x - y N_{2} O_{3} ⇋ y N_{2} O + y + x O_{2}$
$∵ [O_{2}] = x + y = 2 \cdot 5$
for $N_{2} O_{5}, Kc = [N_{2} O_{5}] [O_{2}] / [N_{2} O_{5}]$
and $2.5 = \frac{( x + y ) ( x - y )}{4 - x}$
$∴ x = 2.166$
$[N_{2} O_{5}] = 4 - x = 1 \cdot 846$