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Byju's Answer

Standard XII

Chemistry

Characteristics of Chemical Equilibrium

In reaction C...

Question

In reaction $C H_{3} C O C H_{3_{(} g)} ⇌ C H_{3} C H_{3_{(} g)} + C O_{(g)}$ If the initial pressure of $C H_{3} C O C H_{3 (g)}$ is 150 mm and at Equilibrium the mole fraction of CO is $\frac{1}{3}$ then the value of Kp is

50mm

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100mm

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33.3mm

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75mm

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Solution

The correct option is A 50mm
mole fraction of each will be $= \frac{1}{3}$
Therefore partial pressure of each $= \frac{150}{3} = 50 m m$
Equilibrium constant $= \frac{50 \times 50}{50} = 50 m m$

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Similar questions

Q. In reduction

C H_{3} C O C H_{3} (g) ⇌ C H_{3} C H_{3} (g) + C O (g)

, if the initial pressure of

C H_{3} C O C H_{3} (g)

150 m m

and at equilibrium the mole fraction of

C O (g)

\frac{1}{3}

then the value

K_{p}

Q. In reduction

C H_{3} C O C H_{3} (g) ⇌ C H_{3} C H_{3} (g) + C O (g)

, if the initial pressure of

C H_{3} C O C H_{3} (g)

150 m m

and at equilibrium the mole fraction of

C O (g)

\frac{1}{3}

then the value

K_{p}

In the reaction,
$C H_{3} C O C H_{3_{(} g)} ⇌ C H_{3} C H_{3 (g)} + C O_{(g)}$
The initial pressure of $C H_{3} C O$ $C H_{3 (g)}$ is $150 m m H g$ and at equilibrium the mole fraction of $C O$ is $\frac{1}{3}$ then the value of $K_{p}$ is:

C H_{3} C O C H_{3} (g) ⇌ C_{2} H_{6} (g) + C O (g)

; The initial pressure of

C H_{3} C O C H_{3}

is 50 mm. At equilibrium the mole fraction of

C_{2} H_{6}

is 1/5 then

K_{p}

will be:

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In reaction CH3COCH3(g)⇌CH3CH3(g)+CO(g) If the initial pressure of CH3COCH3(g) is 150 mm and at Equilibrium the mole fraction of CO is 13 then the value of Kp is

The correct option is A 50mmmole fraction of each will be =13 Therefore partial pressure of each =1503=50mm Equilibrium constant =50×5050=50mm

In reaction $C H_{3} C O C H_{3_{(} g)} ⇌ C H_{3} C H_{3_{(} g)} + C O_{(g)}$ If the initial pressure of $C H_{3} C O C H_{3 (g)}$ is 150 mm and at Equilibrium the mole fraction of CO is $\frac{1}{3}$ then the value of Kp is

The correct option is A 50mm
mole fraction of each will be $= \frac{1}{3}$
Therefore partial pressure of each $= \frac{150}{3} = 50 m m$
Equilibrium constant $= \frac{50 \times 50}{50} = 50 m m$