AB dissociates as 2 AB ( g ) leftharpoons 2 A ( g )+ B 2( g ) When the initial pressure of AB is 500 mm Hg, the pressure becomes 625 mm Hg when the equilibrium is attained. Calculate KP for the reaction assuming volume remains constant.

Question

AB dissociates as 2 AB ( g ) leftharpoons 2 A ( g )+ B 2( g ) When the initial pressure of AB is 500 mm Hg, the pressure becomes 625 mm Hg when the equilibrium is attained. Calculate KP for the reaction assuming volume remains constant. (A) 500 (B) 125 (C) 750 (D) 375

Tardigrade Admin · Accepted Answer

underset(500-p) mm 2 AB leftharpoons undersetP2 A + undersetP/2B 2 Equilibrium pressures (500-p)+p+p / 2=625 mm p / 2=125 mm p=250 mm pA B=(500-250)=250 mm pA=p=250 mm pB2=p / 2=(250/2) mm =125 mm Kp=(p B 2 × p A 2/p AB 2) =(125 ×(250)2/(250)2) =125 mm

Q. AB dissociates as 2AB(g)⇌2A(g)+B2​(g) When the initial pressure of AB is 500mmHg, the pressure becomes 625mmHg when the equilibrium is attained. Calculate KP​ for the reaction assuming volume remains constant.

500

125

750

375

(500−p)mm2AB​⇌P2A​+P/2B2​​ Equilibrium pressures (500−p)+p+p/2=625 mm p/2=125 mm p=250 mm pAB​=(500−250)=250 mm pA​=p=250 mm pB2​​=p/2=2250​ mm =125 mm Kp​=pAB2​pB2​​×pA2​​ =(250)2125×(250)2​ =125 mm

Q. $A B$ dissociates as
$2 A B (g) ⇌ 2 A (g) + B_{2} (g)$
When the initial pressure of $A B$ is $500 mm H g$ , the pressure becomes $625 mm H g$ when the equilibrium is attained. Calculate $K_{P}$ for the reaction assuming volume remains constant.