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Standard XII

Chemistry

Relation between Kp and Kc

In the reacti...

Question

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:

$50 m m H g$

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$100 m m H g$

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$33.33 m m H g$

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$75 m m H g$

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Solution

The correct option is A
$50 m m H g$

Mole fraction of each will be = $\frac{1}{3}$

Therefore partial pressure of each = $\frac{150}{3}$ = 50 mmHg

Equilibrium constant= $50 \times \frac{50}{50} = 50 m m H g$

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Similar questions

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:

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 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

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}

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 the reaction, CH3COCH3(g)⇌CH3CH3 (g)+CO(g) The initial pressure of CH3CO CH3 (g) is 150 mm Hg and at equilibrium the mole fraction of CO is 13 then the value of Kp is:

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

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:

The correct option is A
$50 m m H g$

Mole fraction of each will be = $\frac{1}{3}$

Therefore partial pressure of each = $\frac{150}{3}$ = 50 mmHg

Equilibrium constant= $50 \times \frac{50}{50} = 50 m m H g$