A sample of $H I (g)$ is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of $H I (g)$ is 0.04 atm. What is $K_{p}$ for the given equilibrium?
Given reaction is : $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$

Open in App

Solution

Given:- Initial pressure $= 0.2$ atm,
At equilibrium $= 0.04$ atm

$∴$ Decrease in pressure of HI $= 0.2 - 0.04 = 0.16 a t m$

$∴$ At equilibrium pressure of $H_{2}$ & $I_{2} = \frac{0.16}{2} = 0.08$ atm
$2 H I ⇌ H_{2} + I_{2}$

Initial $0.2$ $0$ $0$

At equi $0.04$ $0.08$ $0.08$

$∴ K_{p} = \frac{[H_{2}] [I_{2}]}{[H I]^{2}} = \frac{0.08 \times 0.08}{(0.04)^{2}} = 4$ .

Suggest Corrections

Similar questions

Q. 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 of

K_{P}

for the given equilibrium?

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?

Q. 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)

0.04 a t m

. What is

K_{p}

for the given equilibrium?

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

Q. The reaction

{2 H I}_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}

is initiated within a flask with 0.2 atm pressure of HI. If the partial pressure of

{H I}_{(g)}

at equilibrium is 0.04 atm the equilibrium constant of the reaction is

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?
$2 H I (g) \leftrightarrow H_{2} (g) + I_{2} (g)$

Join BYJU'S Learning Program

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

Given:- Initial pressure=0.2 atm,At equilibrium=0.04 atm∴ Decrease in pressure of HI=0.2−0.04=0.16 atm∴ At equilibrium pressure of H2 & I2=0.162=0.08 atm 2HI⇌H2+I2Initial 0.2 0 0At equi 0.04 0.08 0.08∴Kp=[H2][I2][HI]2=0.08×0.08(0.04)2=4.

A sample of $H I (g)$ is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of $H I (g)$ is 0.04 atm. What is $K_{p}$ for the given equilibrium?
Given reaction is : $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$