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Derivation of Kp and Kc

In reductionC...

Question

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 $C H_{3} C O C H_{3} (g)$ is $150 m m$ and at equilibrium the mole fraction of $C O (g)$ is $\frac{1}{3}$ then the value $K_{p}$ is

A

50 m m

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B

100 m m

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C

33.3 m m

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D

75 m m

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Solution

The correct option is D $75 m m$
$C H_{3} C O C H_{3} (g) ⇌ C H_{3} C H_{3} (g) + C O (g)$
$a t t = 015000$
$a t t = t_{e q} 150 - x x x$
So total pressure at equilibrium = $150 + x$
According to the question,
$\frac{x}{150 + x} = \frac{1}{3} \Rightarrow x = 75 m m$
Hence total pressure $(P^{o}) = 150 + 75 m m = 225 m m$
Now, $K_{p} = \frac{(P_{C H_{3} C H_{3}}) (P_{C O})}{P_{C H_{3} C O O H}} = \frac{\frac{1}{3} P^{o} \times \frac{1}{3} P^{o}}{\frac{1}{3} P^{o}}$
$\Rightarrow K_{p} = \frac{1}{3} \times 225 m m = 75 m m$

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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}

K_{p}

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

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}

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:

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:

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