For the equilibrium N 2 O 4( g ) leftharpoons 2 NO 2( g ), the equilibrium constant KP=0.112 at 298 K. If the partial pressure of NO 2 in the equilibrium mixture is 0.40 atm the partial pressure of N 2 O 4 would be

Question

For the equilibrium N 2 O 4( g ) leftharpoons 2 NO 2( g ), the equilibrium constant KP=0.112 at 298 K. If the partial pressure of NO 2 in the equilibrium mixture is 0.40 atm the partial pressure of N 2 O 4 would be (A) 0.28 atm (B) 0.56 atm (C) 2.8 atm (D) 1.43 atm

Tardigrade Admin · Accepted Answer

Kp= ( p NO 22/ p N 2 O 4) p N 2 O 4=(p NO 22/Kp)=((0.4)2/0.112)=1.43 atm

Q. For the equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g),$ the equilibrium constant $K_{P} = 0.112$ at $298 K$ . If the partial pressure of $N O_{2}$ in the equilibrium mixture is $0.40 a t m$ the partial pressure of $N_{2} O_{4}$ would be

0.28 atm

0.56 atm

2.8 atm

1.43 atm

$K_{p} = \frac{p _{N O_{2}}^{2}}{p _{N_{2} O_{4}}}$
$p N_{2} O_{4} = \frac{p _{NO 2}^{2}}{K _{p}} = \frac{( 0.4 ) ^{2}}{0.112} = 1.43 a t m$

Q. For the equilibrium N2​O4​(g)⇌2NO2​(g), the equilibrium constant KP​=0.112 at 298K. If the partial pressure of NO2​ in the equilibrium mixture is 0.40atm the partial pressure of N2​O4​ would be