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Standard XII

Chemistry

Derivation of Kp and Kc

A sample of H...

Question

A sample of HI_{(g)} is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI_{(g)} is 0.04 atm. What is K_p for the given equilibrium?
$2 H I (g) \leftrightarrow H_{2} (g) + I_{2} (g)$

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Solution

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 - 0.04 = 0.16. The given reaction is:
$\begin{matrix} 2 H I_{(g)} & \leftrightarrow & H_{2 (g)} & + & I_{2 (g)} I n i t i a l c o n c . & 0.2 a t m & 0 & 0 A t e q u i l i b r i u m & 0.04 a t m & \frac{0.16}{2} & \frac{2.15}{2} = 0.08 a t m & = 0.08 a t m \end{matrix}$
Therefore,
$K_{p} = \frac{P_{H_{2}} \times P_{I_{2}}}{P_{H I}} = \frac{0.08 \times 0.08}{(0.04)^{2}} = \frac{0.0064}{0.0016} = 4.0$
Hence, the value of $K_{p}$ for the given equilibrium is 4.0.

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Derivation of Kp and Kc

Standard XII Chemistry

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A sample of HI_{(g)} is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI_{(g)} is 0.04 atm. What is K_p for the given equilibrium? 2HI(g)↔H2(g)+I2(g)

A sample of HI_{(g)} is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI_{(g)} is 0.04 atm. What is K_p for the given equilibrium?
$2 H I (g) \leftrightarrow H_{2} (g) + I_{2} (g)$