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Standard XII

Chemistry

Derivation of Kp and Kc

For X Y 2g ...

Question

For $X Y_{2 (g)} ⇌ X Y_{(g)} + Y_{(g)}$ , intially 1 mole each of $X Y_{2}$ and y are present in 10L flask at 500mm.If the equilibrium pressure of XY is 150mm, $K_{p}$ is?

500mm Hg

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\sqrt{500}

mm Hg

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300mm Hg

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600mm Hg

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Solution

The correct option is B 600mm Hg
${X Y_{2}}_{(g)} ⟶ X Y_{(g)} + Y_{(g)}$
$1 m o l e$ $1 m o l e$ at $500 m m$
i.e. $X Y_{2}$ & $Y$ have same partial pressure i.e. $250 m m$ .
$∴$ At equilibrium, $P_{X Y_{2}} = (250 - p) m m$ $P_{X Y} = (P) m m$ & $P_{Y} = (250 + p) m m$
Given $P = 150 m m$ is equilibriummm pressure of $X Y$
$\Rightarrow P_{X Y_{2}} = 250 - 150 = 100 m m$
$P_{Y} = 250 + 150 = 400 m m$
$∴ K_{P} = \frac{150 \times 400}{100} = 600 m m H g$

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Similar questions

X Y_{2}

dissociates as

X Y_{2 (g)} ⇌ X Y_{(g)} + Y_{(g)}

. When the initial pressure is

600

mm of Hg, the total pressure at equilibrium developed is

800

mm of Hg.

Therefore, the pressure of

Y

at equilibrium 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

X Y_{2} (g) ⇌ X Y (g) + Y (g)

. When the initial pressure of

X Y_{2}

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 :

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.

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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For XY2(g)⇌XY(g)+Y(g), intially 1 mole each of XY2 and y are present in 10L flask at 500mm.If the equilibrium pressure of XY is 150mm, Kp is?

The correct option is B 600mm HgXY2(g)⟶XY(g)+Y(g) 1mole 1mole at 500mmi.e. XY2 & Y have same partial pressure i.e. 250mm.∴ At equilibrium, PXY2=(250−p)mm PXY=(P)mm & PY=(250+p)mmGiven P=150mm is equilibriummm pressure of XY⇒PXY2=250−150=100mm PY=250+150=400mm∴KP=150×400100=600mmHg

For $X Y_{2 (g)} ⇌ X Y_{(g)} + Y_{(g)}$ , intially 1 mole each of $X Y_{2}$ and y are present in 10L flask at 500mm.If the equilibrium pressure of XY is 150mm, $K_{p}$ is?