You visited us 1 times! Enjoying our articles? Unlock Full Access!

Characteristics of Equilibrium Constant

AB2 dissociat...

Question

$A B_{2}$ dissociates as $A B_{2} (g) ⇌ A B (g) + B (g)$
When the initial pressure of AB is $500$ mm, the total pressure becomes $700$ mm when the equilibrium is attained. Calculte $K_{p}$ for the reaction assuming volume remains constant.

133.3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

125

No worries! We‘ve got your back. Try BYJU‘S free classes today!

750

No worries! We‘ve got your back. Try BYJU‘S free classes today!

375

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $133.3$
$A B_{2} (g) ⇌ A B (g) + B (g) 500 m m 00 500 - x x x 500 - 200200200$
Hence,
$500 - x + x + x = 700$
$x = 200$
$K_{p} = \frac{P_{A B . P_{b}}}{P_{A B_{2}}}$
$= \frac{200 \times 200}{300} = 133.3 m m$

Suggest Corrections

Similar questions

{A B}_{2}

{A B}_{2 (g)} ⇌ {A B}_{(g)} + B_{(g)}

{A B}_{2}

K_{p}

A B_{2}

A B_{2} (g) \Leftrightarrow A B (g) + B (g)

A B_{2}

K_{p}

A B

2 A B (g) ⇌ 2 A (g) + B_{2} (g)

A B

500 m m

625 m m

K_{P}

{A B}_{2} (g)

A B_{2} (g) ⇌ A B (g) + B (g)

{A B}_{2}

600 m m

H g

800 m m

H g

X Y_{2}

X Y_{2} (g) ⇌ X Y (g) + Y (g)

X Y_{2}

H g

H g

K

Join BYJU'S Learning Program