For the equilibrium reaction 2 NOBr ( g ) leftharpoons 2 NO ( g )+ Br 2( g ) If PB r2=(P/9), where P is total pressure at equilibrium. The ratio (Kp/P) is equal to

Question

For the equilibrium reaction 2 NOBr ( g ) leftharpoons 2 NO ( g )+ Br 2( g ) If PB r2=(P/9), where P is total pressure at equilibrium. The ratio (Kp/P) is equal to (A) (1/3) (B) (1/81) (C) (1/9) (D) (1/27)

Tardigrade Admin · Accepted Answer

undersetx2 NOBr ( g ) leftharpoons underset(2P/9)2 NO ( g ) + underset(P/9)Br 2( g ) Since total pressure is P. So, x+(2 P/9)+(P/9)=P ⇒ x=(6 P/9) ∴ Kp=(p NO 2 × p Br 2/p NOBr 2)=(((2 P/9))2 × (P/9)/((6 P)9)2) ∴ (Kp/P)=(1/81)

Q. For the equilibrium reaction
$2 NOB r (g) ⇌ 2 NO (g) + B r_{2} (g)$
If $P_{B r_{2}} = \frac{P}{9}$ , where $P$ is total pressure at equilibrium. The ratio $\frac{K _{p}}{P}$ is equal to

$\frac{1}{3}$

$\frac{1}{81}$

$\frac{1}{9}$

$\frac{1}{27}$

Q. For the equilibrium reaction 2NOBr(g)⇌2NO(g)+Br2​(g) If PBr2​​=9P​, where P is total pressure at equilibrium. The ratio PKp​​ is equal to

31​

811​

91​

271​

x2NOBr(g)​⇌92P​2NO(g)​+9P​Br2​(g)​ Since total pressure is P. So, x+92P​+9P​=P⇒x=96P​ ∴Kp​=pNOBr2​pNO2​×pBr2​​​=(96P​)2(92P​)2×9P​​ ∴PKp​​=811​

Q. For the equilibrium reaction
$2 NOB r (g) ⇌ 2 NO (g) + B r_{2} (g)$
If $P_{B r_{2}} = \frac{P}{9}$ , where $P$ is total pressure at equilibrium. The ratio $\frac{K _{p}}{P}$ is equal to

$\frac{1}{3}$

$\frac{1}{81}$

$\frac{1}{9}$

$\frac{1}{27}$