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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

The reaction ...

Question

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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Solution

The correct option is A 4
at equilibrium ,
pressure of HI = 0.2 - 2x = 0.04
2x = 0.2 - 0.04 = 0.16
x = 0.08 atm

so, pressure of H2 and I2 at equilibrium is 0.08 atm .
e.g., at equilibrium ,

Kp = 0.08atm × 0.08atm/(0.04atm)²= 4

hence, Kp = 4

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Similar questions

Q. A sample of

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The reaction 2HI(g)⇌H2(g)+I2(g) is initiated within a flask with 0.2 atm pressure of HI. If the partial pressure of HI(g) at equilibrium is 0.04 atm the equilibrium constant of the reaction is

The correct option is A 4at equilibrium , pressure of HI = 0.2 - 2x = 0.04 2x = 0.2 - 0.04 = 0.16 x = 0.08 atm so, pressure of H2 and I2 at equilibrium is 0.08 atm .e.g., at equilibrium , Kp = 0.08atm × 0.08atm/(0.04atm)²= 4 hence, Kp = 4

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

The correct option is A 4
at equilibrium ,
pressure of HI = 0.2 - 2x = 0.04
2x = 0.2 - 0.04 = 0.16
x = 0.08 atm

so, pressure of H2 and I2 at equilibrium is 0.08 atm .
e.g., at equilibrium ,

Kp = 0.08atm × 0.08atm/(0.04atm)²= 4

hence, Kp = 4