The equilibrium constant at 25° C for the reaction, 2NO(.g.)+Br2(.g.)leftharpoons2NOBr(.g.) is equal to 160 atm- 1 . If the partial pressures of NO, Br2 and NOBr in a flask at 25° C are 0.01, 0.1 and 0.04 atm respectively. It can be said that

Question

The equilibrium constant at 25° C for the reaction, 2NO(.g.)+Br2(.g.)leftharpoons2NOBr(.g.) is equal to 160 atm- 1 . If the partial pressures of NO, Br2 and NOBr in a flask at 25° C are 0.01, 0.1 and 0.04 atm respectively. It can be said that (A) there is equilibrium in the flask (B) there reaction will proceed in the forward direction (C) the reaction will proceed in the backward direction (D) the partial pressure of NOBr finally will be 0.05 atm

Tardigrade Admin · Accepted Answer

Q p =( p NOBrig 2/ p NOig 2 ⋅ p Br 2( g )) =((0.04 atm )2/(0 . operatornamelatm)2 ×(0.1 atm )) =(16 × 10-4 atm 2/1 × 10-4 atm × 1 × 10-1 atm ) =160 atm -1 Since Qp=Kp, therefore there is equilibrium in the flask.

Q. The equilibrium constant at 25∘C for the reaction, 2NO(g)+Br2​(g)⇌2NOBr(g) is equal to 160 atm−1 . If the partial pressures of NO, Br2​ and NOBr in a flask at 25∘ C are 0.01, 0.1 and 0.04 atm respectively. It can be said that

there is equilibrium in the flask

there reaction will proceed in the forward direction

the reaction will proceed in the backward direction

the partial pressure of NOBr finally will be 0.05 atm

Qp​=pNOig2​⋅pBr2​(g)​pNOBrig2​​ =(0.latm)2×(0.1atm)(0.04atm)2​ =1×10−4atm×1×10−1atm16×10−4atm2​ =160atm−1 Since Qp​=Kp​, therefore there is equilibrium in the flask.

Q. The equilibrium constant at $2 5^{\circ} C$ for the reaction, $2 NO (g) + B r_{2} (g) ⇌ 2 NOB r (g)$ is equal to 160 $a t m^{- 1}$ . If the partial pressures of NO, $B r_{2}$ and NOBr in a flask at $2 5^{\circ}$ C are 0.01, 0.1 and 0.04 atm respectively. It can be said that