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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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:

Therefore,

Hence, the value of K_p for the given equilibrium is 4.0.

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Similar questions

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. 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)

Q. 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)

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

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

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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 Kp for the given equilibrium?

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: Therefore, Hence, the value of Kp for the given equilibrium is 4.0.

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?

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:

Therefore,

Hence, the value of K_p for the given equilibrium is 4.0.