XY 2 dissociates as, XY 2( g ) leftharpoons XY ( g )+ Y ( g ) when the initial pressure of X Y2 is 600 mm Hg , the total equilibrium pressure is 800 mm Hg. Calculate K for the reaction assuming that, the volume of the system remains unchanged

Question

XY 2 dissociates as, XY 2( g ) leftharpoons XY ( g )+ Y ( g ) when the initial pressure of X Y2 is 600 mm Hg , the total equilibrium pressure is 800 mm Hg. Calculate K for the reaction assuming that, the volume of the system remains unchanged (A) 50 (B) 100 (C) 166.6 (D) 400

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c601b675f62945de22403a2c87e5b810-.png" /> Total pressure =600-P+P+P=800 ∴ P=200 mm ∴ Kp=(P × P/600-P)=(200 × 200/400)=100

Q. $X Y_{2}$ dissociates as, $X Y_{2 (g)} ⇌ X Y_{(g)} + Y_{(g)}$ when the initial pressure of $X Y_{2}$ is $600 mm H g,$ the total equilibrium pressure is $800 mm H g$ . Calculate $K$ for the reaction assuming that, the volume of the system remains unchanged

50

100

166.6

400

Total pressure $= 600 - P + P + P = 800$

$∴ P = 200 mm$

$∴ K_{p} = \frac{P \times P}{600 - P} = \frac{200 \times 200}{400} = 100$

Q. XY2​ dissociates as, XY2(g)​⇌XY(g)​+Y(g)​ when the initial pressure of XY2​ is 600mmHg, the total equilibrium pressure is 800mmHg. Calculate K for the reaction assuming that, the volume of the system remains unchanged

50

100

166.6

400

Total pressure =600−P+P+P=800 ∴P=200mm ∴Kp​=600−PP×P​=400200×200​=100

Q. $X Y_{2}$ dissociates as, $X Y_{2 (g)} ⇌ X Y_{(g)} + Y_{(g)}$ when the initial pressure of $X Y_{2}$ is $600 mm H g,$ the total equilibrium pressure is $800 mm H g$ . Calculate $K$ for the reaction assuming that, the volume of the system remains unchanged

Total pressure $= 600 - P + P + P = 800$

$∴ P = 200 mm$

$∴ K_{p} = \frac{P \times P}{600 - P} = \frac{200 \times 200}{400} = 100$