A sample of displaystyle HIleft- g right- is placed in a flask a pressure of 0-2 atm- At equilibrium the partial pressure of displaystyle HIleft- g right- is 0-04 atm- What is displaystyle - K - p - the given equilibrium-displaystyle 2HIleft- g right- rightleftharpoons - H - 2 -left- g right- - I - 2 -left- g right-

Question

A sample of displaystyle HIleft- g right-  is placed in a flask a pressure of 0-2 atm- At equilibrium the partial pressure of displaystyle HIleft- g right-  is 0-04 atm- What is displaystyle - K - p - the given equilibrium-displaystyle 2HIleft- g right- rightleftharpoons - H - 2 -left- g right- - I - 2 -left- g right-

Toppr Admin · Accepted Answer

Change in the pressure of HI is 0-2-x2212-0-04-0-16 atm-Equilibrium prtial pressure of hydrogen is 0-162-0-08-xA0-atm-Equilibrium partial pressure of iodine is 0-08 atm-Kp-PH2PI2P2HI-0-08-xD7-0-08-0-04-2-4-0

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.

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

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.

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