A sample of $H I (g)$ is placed in a flask at a pressure of $0.2 a t m$ . At equilibrium the partial pressure of $H I (g)$ is $0.04 a t m$ . What is $K_{p}$ for the given equilibrium?

$2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$

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Solution

Change in the pressure of HI is $0.2 - 0.04 = 0.16$ atm.

Equilibrium prtial pressure of hydrogen is $\frac{0.16}{2} = 0.08 a t m$ .

Equilibrium partial pressure of iodine is 0.08 atm.

$K_{p} = \frac{P_{H_{2}} P_{I_{2}}}{P_{H I}^{2}} = \frac{0.08 \times 0.08}{(0.04)^{2}} = 4.0$

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2 H I (g) ⇌ H_{2} (g) + I_{2} (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?
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is initiated within a flask with 0.2 atm pressure of HI. If the partial pressure of

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

Change in the pressure of HI is 0.2−0.04=0.16 atm.Equilibrium prtial pressure of hydrogen is 0.162=0.08 atm.Equilibrium partial pressure of iodine is 0.08 atm.Kp=PH2PI2P2HI=0.08×0.08(0.04)2=4.0

A sample of $H I (g)$ is placed in a flask at a pressure of $0.2 a t m$ . At equilibrium the partial pressure of $H I (g)$ is $0.04 a t m$ . What is $K_{p}$ for the given equilibrium?

$2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$

Change in the pressure of HI is $0.2 - 0.04 = 0.16$ atm.

Equilibrium prtial pressure of hydrogen is $\frac{0.16}{2} = 0.08 a t m$ .

Equilibrium partial pressure of iodine is 0.08 atm.

$K_{p} = \frac{P_{H_{2}} P_{I_{2}}}{P_{H I}^{2}} = \frac{0.08 \times 0.08}{(0.04)^{2}} = 4.0$