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Standard XII

Chemistry

Derivation of Kp and Kc

XY2 dissociat...

Question

$X Y_{2}$ dissociates as, $X Y_{2} (g) ⇌ X Y (g) + Y (g)$
Initial pressure of $X Y_{2}$ is $600 m m H g$ . The total pressure at equilibrium is $800 m m H g$ . Assuming volume of system to remain constant, the value of $K_{p}$ is:

50

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100

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20

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400

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Solution

The correct option is B $100$
${X Y}_{2} (g) ⇌ X Y (g) + Y (g)$
$t = 0$ $P_{0}$ $0$ $0$
$t = t_{e q}$ $P_{0} (1 - α)$ $P_{0} α$ $P_{0} α$
$↓$ $↓$ $↓$
$400 m n$ $200$ $200 m n$
Given $P_{0} = 600 m n$
$P_{0} (1 + α) = 800 m n$
$1 + α = 4 / 3$
$α = 1 / 3$
$K_{P} = \frac{200 \times 200}{400} = 100$

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Similar questions

{X Y}_{2}

dissociates as

{X Y}_{2} (g) ⇌ X Y (g) + Y (g)

Initial pressure of

{X Y}_{2}

600 m m

H g

. The total pressure at equilibrium is

800 m m

H g

Assuming volume of system to remain constant, the value of

K_{p}

is:

$X Y_{2}$ dissociates as,
$X Y_{2} (g) ⇌ X Y (g) + Y (g)$
Initial pressure of $X Y_{2}$ is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of system to remain constant, the value of $K_{P}$ is

{X Y}_{2}

dissociation as-

{X Y}_{2} (g) ⇌ X Y (g) Y (g)

Initial pressure of

{X Y}_{2}

600 m m H g

.The total pressure st equilibrium is

800 m m H g

Assuming volume of system to remain constant. the value of

K_{p}

is:

X Y_{2}

dissociates as:

X Y_{2} (g) ⇌ X Y (g) + Y (g)

Initial pressure of

X Y_{2}

600 m m

H g

. The total pressure at equilibrium is

800 m m

H g

. Assuming volume of system to remain constant, the value of

K_{p}

is:

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 of Hg, the total equilibrium pressure is 800 mm of Hg. FInd the value of Equilibrium constant for the reaction assuming that the volume of the system remains unchanged.

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XY2 dissociates as, XY2(g)⇌XY(g)+Y(g)Initial pressure of XY2 is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of system to remain constant, the value of Kp is:

The correct option is B 100 XY2(g)⇌XY(g)+Y(g)t=0 P0 0 0t=teq P0(1−α) P0α P0α ↓ ↓ ↓ 400mn 200 200mnGiven P0=600mnP0(1+α)=800mn1+α=4/3α=1/3KP=200×200400=100

$X Y_{2}$ dissociates as, $X Y_{2} (g) ⇌ X Y (g) + Y (g)$
Initial pressure of $X Y_{2}$ is $600 m m H g$ . The total pressure at equilibrium is $800 m m H g$ . Assuming volume of system to remain constant, the value of $K_{p}$ is: